INTEGRAL CALCULUS

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1 Part I, Section I,Chapter I. Translated and annotated by Ian Bruce.

page 22

INTEGRAL CALCULUS BOOK ONE PART ONE OR

A METHOD FOR FINDING FUNCTIONS OF ONE VARIABLE FROM SOME GIVEN RELATION OF THE DIFFERENTIALS OF THE FIRST ORDER FIRST SECTION CONCERNING THE

INTEGRATION OF DIFFERENTIAL FORMULAS.

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1 Part I, Section I,Chapter I. page 23

Translated and annotated by Ian Bruce.

CHAPTER I

CONCERNING THE INTEGRATION OF RATIONAL DIFFERENTIAL FORMULAS DEFINITION 40. The formula of a differential is rational, when the differential dx of the variable x of which the function is sought is multiplied by a rational function of x; or if X designates a rational function of x, this formula Xdx of the differential is said to be rational. COROLLARY 1 41. Hence in this chapter a function of x is sought of this kind, which if it should be put as y, in order that to X, so that

dy dx

dy dx

is equal to a rational function of x, or with such a function put equal

= X.

COROLLARY 2 42. Hence a function of x of this kind is sought, of which the differential is equal to Xdx; hence the integral of this, which now is accustomed to be indicated by Xdx , provides

∫

the function sought. COROLLARY 3 43. But if P should be a function of x of this kind, so that the differential of this is dP = Xdx, since the quantity P + C likewise is differential, the complete integral of the proposed formula Xdx is P + C [, for some arbitrary constant C.] SCHOLION 1 44. Questions of this kind are referred to in the first part of the first book, in which functions of the single variable x are sought from a given relation of the differentials of the first degree. It is evident that if the function sought is equal to y and

dy dx

= p, it is

required to prevail, in order that for some proposed equation between the three quantities x, y and p, from this the natural form of the function y is found, or an equation between x and y, with the letter p elicited. But the question thus proposed in general is considered to outdo the analytical powers so far, so that at no time can a solution be expected.

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1 Part I, Section I,Chapter I. page 24 Therefore our strengths are to be exercised in the simpler cases, among which the case Translated and annotated by Ian Bruce.

occurs, in which p is equal to some function of x, for example X, so that

dy dx

= X or

∫

dy = Xdx and thus the integral y = Xdx is required, as we arrange in the first section. Now this case extends widely to various natural forms of the function X and also it is involved in many more difficult cases, from which we put in place only questions to be resolved in this chapter, in which that function X is rational, and then to be progressing to these irrational functions. Hence this part is subdivided conveniently into two parts, in the first of which the integration of the simple formulas, in which p =

dy dx

is equal to a

function of x only, is to be treated, but in the other it is appropriate to give an account of the integration, when the equation proposed is some function of x, y and p. And since in these two sections and chiefly in the first, most is an elaboration of the geometers, that occupy the greater part of the whole work. SCHOLION 2 45. But first the principles of integration are to be desired from the differential calculus, and in the same way as the principles of division from multiplication, and the principles of the extraction of roots from an account of raising to powers, are accustomed to be taken. Since therefore, if a magnitude to be differentiated depends on several parts, such as P + Q – R, the differential of this is dP + dQ – dR, thus in turn, if the formula of the differential depends on several parts as Pdx +Qdx – Rdx, the integral is

∫ Pdx + ∫ Qdx – ∫ Rdx obviously from the integrals of the individual parts separately. Then, since the differential of the magnitude aP is adP, the integral of the differential formula aPdx is a Pdx ,

∫

clearly as the differential formula is multiplied by that constant magnitude, the integral must be multiplied by the same. Thus if the formula of the differentials is aPdx + bQdx + cRdx, whatever functions of x are designated by P, Q, R, the integral is

∫

∫

∫

a Pdx + b Qdx + c Rdx , thus in order that the integration is to be put in place of the forms Pdx, Qdx and Rdx, and thus the above must be increased by the addition of an arbitrary constant C, in order that the complete integral is obtained. PROBLEM 1 46. To find the function of x, in order that the differential of this is equal to ax n dx , or to integrate the differential formula ax n dx .



SOLUTION Since the differential of the power x is mx m −1dx , there is in turn m

∫ mx

m −1

dx = m x m −1dx = x m

∫

and thus

∫x

m −1

dx = 1 x m ; m

on making m – 1 = n or m = n +1; there becomes

∫ x dx = n1+1 x n

n +1

and a x n dx = na+1 x n +1 .

∫

From which the completed integral of the proposed differential formula is a x n +1 + C , n +1

the reason or from which thus is apparent, because the differential actually is equal to

ax n dx . And this integration always has a place, whatever number is attributed to the exponent n, either positive or negative, either an integer or a fraction, or even irrational. Hence a single case is excepted, in which the exponent n = –1 or it is proposed to integrate this formula adx . Now in the differential calculus I have shown, if lx denotes x

the hyperbolic logarithm of x, the differential of this is equal to dx , from which in turn x

we can conclude that

∫ dxx = lx

and

∫ adxx = alx .

Whereby with an arbitrary constant added the complete integral of the formula adx is x

equal to

alx + C= lxa+ C, that also on putting lc for C can thus be expressed: lcxa. COROLLARY 1 47. Hence the integral of the differential formulas ax n dx is always algebraic with the only excepted case, in which n = –1 and the integral is expressed by logarithms, which are to be referred to as transcendent functions. Obviously, it is given by

∫ adxx = alx + C = lcx

n

.



COROLLARY 2 48. If the exponent n denotes a positive number, then the following integrations are to be expressed properly, in as much as they are especially easy:

∫ adx = ax + C, ∫ axdx = a2 xx + C , ∫ ax dx = a3 x + C, 3 4 4 5 5 6 ∫ ax dx = a4 x + C, ∫ ax dx = a5 x + C, ∫ ax dx = a6 x + C. 2

3

COROLLARY 3 49. If n is a negative number, on putting n = – m there becomes

∫ adx x m

= a x1− m + C = 1− m

−a ( m −l ) x m −1

+C ,

from which these simpler cases can be noted :

= − a + C, ∫ adx = − a + C, ∫ adx = − a + C, ∫ adx x 2 xx x x x 3x = − a + C, ∫ adx = − a + C, etc. ∫ adx x x 4x 5x 2

3

5

4

4

6

3

5

COROLLARY 4 50. But also if n denotes a fractional number, hence the integrals can be obtained. First let n = m ; then 3

∫ adx

x m = 2a x x m + C, m+2

from which the cases are to be noted :

∫ adx x = 23a x x + C, ∫ axdx x = 25a x x + C, 3 3 4 ∫ axxdx x = 27a x x + C, ∫ ax dx x = 29a x x + C. 2

COROLLARY 5

51. Also there can be put n = − m and there is had 2 adx = 2a −2a +C = +C, m m x 2−m x ( m − 2) xm−2

∫

from which these cases are to be noted :

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1 Part I, Section I,Chapter I.

∫

Translated and annotated by Ian Bruce. adx = 2a x + C, adx = −2 a + C, adx = −2 a + C, x x x x xx x 3 x x

∫

page 27

∫

∫ xadxx = 5 x−2ax + C. 3

2

COROLLARIUM 6

52. If in general we put n =

μ , then there arises ν

∫

μ

ax ν dx = ν a x μ +ν

μ +ν ν

+C

or through roots, ν

ν x μ = ν a x μ +ν + C ;

∫ adx but if there is put n =

−μ

ν

μ +ν

then there is had

∫x

= νν−aμ x

adx μ ν

ν −μ ν

+C

or through roots,

∫ adxx = νν−aμ ν

μ

ν ν −μ

x

+C .

SCHOLIUM1 53. Nevertheless I have decided in this chapter to treat only rational functions, yet these irrationals can still be presented at once, as likewise they can be treated as rationals. Hence the remaining more complicated formulas are able to be integrated also, if functions of a certain variable z are taken for x. Just as if we put x = f +gz, then dx = gdz; whereby if for a we write ag , there is had

∫ adz ( f + gz )

n

=

a ( n +1) g (

f + gz )

n +1

+C ,

but in the particular case, in which n = –1,

= a l ( f + gz ) + C . ∫ ( fadz + gz ) g Then if there shall be n = – m, the integral becomes

∫ ( f adz + gz ) μ

m

=

But on putting n = ν there is produced

−a

( m −1) g ( f + gz )

m −1

+C.


∫

Translated and annotated by Ian Bruce. μ μ +1 adz ( f + gz ) ν = ν a ( f + gz ) ν + C ; (ν + μ ) g

page 28

μ

but on putting n = − ν there is obtained

∫ ( f +adzgz )

μ ν

=

ν a( f + gz ) μ

(ν − μ ) g ( f + gz )ν

+C .

SCHOLIUM 2 54. Here another significant property deserves to be noted. Since here a function y is dy

sought, in order that dy = ax n dx , if we put dx = p , this relation is obtained p = ax n , from which the function y must be investigated. Therefore since

y = na+1 x n +1 + C , on account of ax n = p there is also px

y = n +1 + C and thus we have the case, where the relation of the differentials through a certain equation is proposed between x, y and p and now we know for each to be satisfied by the

equation y = na+1 x n +1 + C . Now this no further is the complete integral for the relation px

held in the equation y = n +1 + C , but only a particular one, because the new integral does not involve a constant, which is not in the relation of the differentials. Moreover the complete integral is y = naD x n +1 + C +1

involving a new constant D; hence there becomes dy = aDx n = p dx

and thus

px

y = n +1 + C.

If this is not relevant to the present situation, yet it is helpful to be noted.

PROBLEM 2 55. To find a function of x, of which the differential is equal to Xdx, with X denoting some rational integer function of x, or to definite the integral Xdx .

∫

SOLUTION Since X is a function of x of the rational integers [e. g. a polynomial], by necessity it contained in this form :

X = α + β x + γ x 2 + δ x3 + ε x 4 + etc.,


page 29

from which by the previous problem the integral sought is

∫ Xdx = C + α x + 12 β x

+ 13 γ x3 + 14 δ x 4 + 15 ε x5 + 15 ζ x6 + etc.,

2

And in general if there should be

X = α xλ + β x μ + γ xν + etc., then there becomes

∫ Xdx = C + λα+1 x

λ +1

β

γ

+ β +1 x μ +1 + γ +1 xν +1 + etc.,

where the exponents α , β ,ν etc. are able to signify both positive and negative numbers as well as fractions, provided it is noted, if there should be λ = −1 , then there becomes

∫ αxdx = α lx , which is the single case to be referred to the order of transcendent functions. PROBLEM 3 56. If X denotes some rational fractional function of x, to describe the method, with the help of which it is convenient to investigate the integral of the formula Xdx.

Therefore let

X=M N

SOLUTION , thus in order that M and N are to become integral functions of x,

and in the first place it is to be considered, whether the sum of the powers of x in the numerator M is either greater or less than in the denominator N, in which case from the , integral [i. e. whole] parts can be elicited by division; the integration of fraction M N which can be obtained without difficulty, and the whole calculation is reduced to a , in which the sum of the powers of x in the numerator M is less fraction of this kind M N than in the denominator N. Then all the factors of the denominator N itself are sought, both simple if they are real, as well as real squares, clearly in turn of the simple squares of [complex conjugate] imaginary factors arising ; likewise it is to be considered whether these factors are all unequal or not ; for indeed with the equality of the factors another method for the into simple fractions has to be put in place, since they arise resolution of the fraction M N from the factors of individual fractions, of which the sum of the fractions proposed is . Certainly from the simple factor a + bx the fraction arises : equal to M N A a + bx

;

if two are equal or the denominator N should have the factor (a + bx)2 , hence there arises the fractions : A B ; 2 + a +bx (a + bx )


Translated and annotated by Ian Bruce. 3

moreover from the factor (a + bx) of this kind these three fractions arise: A (a + bx )3

+

B (a +bx ) 2

+ a +Cbx :

and thus henceforth. But a double factor, the form of which is aa − 2abxcos.ζ + bbxx , unless the two should be equal, gives a fraction of the form A+ Bx ; aa − 2 abxcos .ζ + bbxx but if the N involves two equal factors of this kind, from this the two partial fractions of this kind arise : A+ Bx C + Dx 2 + aa − 2abxcos .ζ +bbxx

( aa −2abxcos.ζ +bbxx )

but if the cube, thus ( aa − 2abxcos.ζ + bbxx ) should be a factor of the denominator N, 3

from that there arises the three partial fractions : A+ Bx 3 2 cos.ζ +bbxx ) aa abx − (

+

C + Dx 2 2 cos .ζ +bbxx ) aa abx − (

E + Fx + aa − 2abx ; cos.ζ +bbxx

and thus henceforth. can be resolved into all its simple Therefore since in this way the proposed fraction M N fractions, all are contained in one or other of these forms : either A

( a +bx )

n

or

A+ Bx

( aa − 2abxcos.ζ +bbxx )

n

and now it is required to integrate the individual terms multiplied by dx; the value of the . function sought is the sum of all these integrals Xdx = Mdx N

∫

∫

COROLLARY 1 57. Hence for an integration of all the forms of this kind Mdx the whole calculation is N

reduced to the integration of the two forms of this kind :

∫ ( aAdx + bx )

n

and

( A+ Bx )dx

∫ ( aa −2abxcos.ζ +bbxx )

n

,

while for n there are written successively the numbers 1, 2, 3, 4, etc.


page 31

COROLLARIUM 2 58. And indeed the integral of the first form above (§ 53) is now obtained, from which it is apparent to be : Adx = A l a + bx + Const . ) a + bx b (

∫ ∫ ( aAdx + bx ) ∫ ( aAdx + bx )

2

=

−A b( a + bx )

3

=

−A 2 2b( a + bx )

+ Const . + Const .

et generally

∫ ( aAdx +bx )

n

=

−A n −1 1 − n b ( ) ( a +bx )

+ Const .

COROLLARY 3 59. Hence in order that the proposition can be completed, nothing else is required, except that the integration of this form

( A+ Bx )dx


n

must be shown, indeed for the case n = 1, then for the cases n = 2, n = 3, n = 4, etc. SCHOLIUM 1 60. Unless we want to avoid imaginary quantities, the whole calculation now can be completed from what has been treated; for with the denominator N resolved into all its simple factors, they are either real or imaginary, the proposed fraction always can be resolved into partial fractions of this form aAdx or of this Adx n ; since the integrals of + bx

( a +bx )

dx can be obtained. But then as which can be shown, the integral of the whole form M N

here there might be some inconvenience the pairs of imaginary parts [i. e. complex conjugates] thus are joined together, in order that a real expression results, which nevertheless completes the nature of the integration absolutely. SCHOLION 2 61. Certainly here it is to be conceded that we postulate the resolution of an integral function into factors, even if the algebra so far has by no means led to that [conclusion], in order that this resolution actually can be put in place. But this is accustomed to be postulated in analysis everywhere, in order that, as the longer we progress, those things which are left behind, even if they have not been explored well enough, we assume as known ; evidently it suffices here that all the factors are able to be assigned by the method of approximation, however close they can be assigned. In a like manner in the

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1 Part I, Section I,Chapter I. page 32 integral calculus the longer we proceed, the integrals of all the formulas of the form Xdx, whatever function of x may be signified by the letter X, we will consider as known and by us to be the most outstanding, if we can prevail to reduce more obscure integrals to these forms ; this also disturbs nothing in practical use, since the values of such formulas Xdx are allowed to be assigned almost to any extend you wish, as we shall show in Translated and annotated by Ian Bruce.

∫

what follows. Otherwise about these integrals, the resolution of the denominator N into its factors is absolutely necessary, therefore in order that the individual factors are present in the expression of the integral ; there are just a few cases and these are especially easy, in which we are able to do without that resolution ; just as if this formula is proposed x n −1dx 1+ x n

it is at once apparent on putting x n = ν that it goes into

(

)

dν , the integral of n(1+ν )

which is 1n l (1 + ν ) = 1n l 1 + x n , where there was no need for the resolution of the factors.

Now cases of this kind are so evident in themselves, so that no particular explanation of these may be needed. PROBLEM 4

62. To find the integral of this formula

∫

( A+ Bx )dx

y = aa − 2abxcos.ζ +bbxx

SOLUTION Since the numerator depends on the two parts Adx+Bxdx, this latter term Bxdx can be removed in the following manner. Since there arises l ( aa − 2abxcos.ζ + bbxx ) =

−2 abdxcos .ζ + 2bbxdx

∫ aa −2abxcos.ζ +bbxx

this equation is to be multiplied by 2Bbb and taken away from the proposed equation ; for thus there is produced y − B l ( aa − 2abxcos.ζ + bbxx ) = 2bb

( A+

Bacos .ζ b

)dx


thus in order that only the formula to be integrated remains. For brevity the formula is put in place Bacos .ζ A+ b =C so that this becomes

∫ aa −2abxCdx cos .ζ + bbxx



which thus can be expressed thus

∫ aa sin . ζ +Cdx ( bx − acos.ζ ) 2

2

.

There is put in place bx − acos.ζ = av sin.ζ and hence dx = adv bsin.ζ from which our formula becomes Cadv sin .ζ :b

C dv ∫ aa sin . ζ (1+vv ) = ab sin .ζ ∫ (1+ vv ) . 2

But from the differential calculus we know that this becomes

∫ (1+dvvv ) = Arc.tang.v = Arc.tang.

bx − a cos .ζ a sin .ζ

,

from which on account of C=

our integral is now

Ab + Ba cos .ζ b

Ab + Ba cos .ζ bx − a cos .ζ Arc.tang. a sin .ζ abb sin .ζ

On account of which the integral of the proposed formula

( A+ Bx )dx

∫ aa−2abxcos.ζ +bbxx is equal to B l aa − 2abxcos.ζ 2bb (

Ab + Ba cos .ζ

+ bbxx ) + abb sin .ζ

bx − a cos .ζ

Arc.tang. a sin .ζ

,

which so that is complete, an arbitrary constant C is added above. bx − a cos .ζ

63. If to Arc.tang. a sin .ζ

COROLLARY 1 cos .ζ

we add Arc.tang. sin .ζ , which obviously is considered bx sin .ζ

agreeable to be the constant to be added, there is produced Arc.tang. a −bx cos .ζ and thus we have


Translated and annotated by Ian Bruce. ( A+ Bx )dx aa − 2 abxcos .ζ +bbxx

∫

Ab + Ba cos .ζ bx sin .ζ = 2Bbb l ( aa − 2abxcos.ζ + bbxx ) + abb sin .ζ Arc.tang. a −bx cos .ζ

with the constant C added. COROLLARY 2 64. If we wish, so that the integral thus vanishes on putting x = 0, the constant C may be B laa and thus there arises taken as equal to 2−bb

( A+ Bx )dx

∫ aa−2abxcos.ζ +bbxx Bl = bb

( aa − 2abxcos.ζ +bbxx ) a

Ab + Ba cos .ζ

+ abb sin .ζ

bx sin .ζ

Arc.tang. a −bx cos .ζ

Hence this integral depends on logarithms in part, and in part on the arc of circles or angles. COROLLARY 3 65. If the letter B vanishes, the part depending on logarithms vanishes and there is produced bx sin .ζ

= A Arc.tang. a −bx cos .ζ + C ∫ aa−2abxAdx cos .ζ + bbxx ab sin .ζ and thus is defined by an angle alone. COROLLARY 4 66. If the angle ζ is right and thus cos .ζ = 0 and sin .ζ = 1 , there is obtained

( A+ Bx )dx

∫ aa +bbxx

Bl = bb

( aa +bbxx ) a

A Arc.tang. bx + C ; + ab a

if the angle ζ is 60° and thus cos.ζ = 12 and sin .ζ = 23 , then there arises

( A+ Bx )dx

∫ aa−abx+bbxx = bbB l

( aa − abx +bbxx ) a

3 + 2 Ab + Ba Arc.tang. 2bx a −bx

But if ζ = 120o and thus cos.ζ = − 12 and sin .ζ =

abb 3 3 , then there arises 2


∫

( A+ Bx )dx

= Bl aa + abx +bbxx bb

( aa + abx +bbxx ) a

page 35

3 + 2 Ab − Ba Arc.tang. 2bx a + bx abb 3

SCHOLIUM 1 67. Here it comes about entirely worthy to be noted, that in the case ζ = 0 , in which the denominator aa − 2abx + bbxx becomes a square, the account of the angle disappears from the integral. Indeed on putting the angle ζ to be infinitely small then there becomes B l a −bx and the other part cos.ζ = 1 and sin .ζ = ζ , from which the logarithmic becomes bb a Ab + Ba Arc.tang. bxζ = ( Ab + Ba ) x abbζ a −bx ab( a −bx )

, because the arc of the infinitely small tangent itself is

bxζ

equal to a −bx , and thus this part becomes algebraic. On account of which there becomes

( A+ Bx )dx B a −bx ( Ab + Ba ) x ∫ ( a−bx )2 = bb l a + ab( a −bx ) + Const., the truth of which is obvious from the preceding; for there is

( A+ Bx ) − B + Ab + Ba 2 = 2 ( a −bx ) b( a −bx ) b( a −bx ) Now there becomes B la = B l a −bx , = B l ( a − bx ) − bb ∫ b(−aBdx bb a −bx ) bb

∫

( Ab + Ba )dx Ab + Ba Ab + Ba ( Ab + Ba ) x = − abb = , 2 bb( a −bx ) ab( a −bx ) b( a −bx )

if indeed each integration is thus determined, as in the case x = 0 the integrals vanish. SCHOLION 2 68. In a similar manner, which we have used here, if in the formula with the differential the sum of the powers of x in the numerator M shall be one degree less than fraction Mdx N

in the denominator N, also that boundary can be lifted. For let the equations be M = Ax n −1 + Bx n − 2 + Cx n −3 + etc.

and N = α x n + β x n −1 + γ x n − 2 + etc.

and there is put Mdx = dy . Now since there shall be N dN = nα x n −1dx + ( n − 1) dxβ x n − 2 + ( n − 2 ) dxγ x n −3 + etc.

then there will be

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1 Part I, Section I,Chapter I. AdN nα N

Translated and annotated by Ian Bruce. dx ⎛ Ax n −1 + ( n −1) Aβ x n − 2 + ( n − 2 ) Aγ x n −3 + etc.⎞ , ⎟ N ⎜⎝ nα nα ⎠

=

page 36

with which value then subtracted there remains ⎛⎛ ⎞ ( n −1) Aβ ⎞ ( n − 2 ) Aγ ⎞ ⎛ = dx dy − nAdN B − nα ⎟ x n − 2 + ⎜ C − nα ⎟ x n −3 + etc.⎟ N ⎜ ⎜⎝ αN ⎠ ⎝ ⎠ ⎝ ⎠ Whereby if for the sake of brevity there is put B−

( n −1) Aβ

there is obtained

nα

= B, C −

( n − 2 ) Aγ

= C, D −

nα

( n −3) Aδ nα

= D, etc.,

dx( Bx n−2 +Cx n−3 + Dx n −4 + etc.) y = nAα lN + = Mdx . n n −1 n−2 n −3 N

∫ α x +β x

+γ x

+δ x

∫

etc.

Therefore in this manner all the differential formulas of the fractions can be reduced by that, so that the sum of the powers of x in the numerator shall be less by two or more degrees than in the denominator. PROBLEM 5

69. The integral formula y=

( A+ Bx )dx


n +1

is to be reduced to another like formula, where the power of the denominator is less by one degree.

SOLUTION For the sake of brevity let the formula be aa − 2abxcos.ζ + bbxx = X and there is put

∫ Since on account of

( A+ Bx )dx X n +1

= y.

dX = −2abdxcos.ζ + 2bbdxx

let d . C + Dx = n X

− n( C + Dx ) dX X

n +1

1 + Ddx n X

2

and thus [for some constants C and D ], C + Dx = Xn

∫

2nb( C + Dx )( a cos .ζ −bx ) dx + X n +1

, ∫ Ddx X n



we have y + C + Dx = n X

∫

dx( A+ 2 nCab cos .ζ + x( B + 2nDab cos .ζ − 2nCbb ) − 2 nDbbxx ) X n+1

∫X

+ Ddx n .

Now in the first formula the letters C and D thus are to be defined, in order that the numerator becomes divisible by X ; hence [see * below] it is necessary for the numerator to be equal to −2nDXdx , [i.e. the numerator becomes −2nDdx ( aa − 2abx cos .ζ + bbxx ) = −2nDXdx ] from which we arrive at A + 2nCab cos .ζ = −2nDaa

and

B + 2nDab cos .ζ − 2nCbb = 4nDab cos .ζ

or B − 2nCbb = 2nDabcos.ζ and hence 2nCbb ; 2nDa = Bb−cos .ζ but from the former condition there is

2nDa =

− A− 2nCab cos .ζ a

from which equated there arises Ba + Abcos.ζ − 2nCabbsin 2 .ζ = 0

or C=

Ba + Abcos .ζ 2 nabbsin 2 .ζ

,

from which B − 2nCbb =

Ba sin .2 ζ − Ba − Ab cos .ζ a sin .2 ζ

thus so that there is found D=

=

− Ab − Ba cos .ζ . 2naab sin .2 ζ

Hence with the letters taken : C=

Ba + Ab cos .ζ 2nabb sin 2 .ζ

− Ab cos .ζ − Ba cos .2 ζ a sin .2 ζ

and D =

− Ab − Ba cos .ζ 2 naab sin 2 .ζ

then there becomes [*] dx y + C + Dx = −2nDdx + Ddx n n n = − ( 2 n − 1) D n X

and thus

∫ or

∫

( A+ Bx )dx X n +1

∫X

X

∫X

= −C −nDx − ( 2n − 1) D dxn , X

∫X

,


∫

page 38

Translated and annotated by Ian Bruce. ( A+ Bx )dx − Baa − Aab cos .ζ +( Abb + Bab cos .ζ ) x = 2 naabb sin .2 ζ X n X n +1

+

( 2n −1)( Ab + Ba cos .ζ ) 2 naab sin .2 ζ

∫ Xdx . n

∫X

Whereby, if the formula dxn should be agreed upon, then it is also possible to assign the integral

∫

( A+ Bx )dx

.

X n+1

COROLLARY 1 70. Therefore since on keeping X = aa − 2abx cos .ζ + bbxx there shall be

∫ dxX

bx sin .ζ

1 = absin Arc.tang . a −bx cos .ζ + Const ., .ζ

then there becomes

∫

( A+ Bx )dx X

2

=

− Baa − Aab cos .ζ +( Abb + Bab cos .ζ ) x Ab + Ba cos .ζ + 2aabb sin .2 ζ X 2 na 3bb sin .3 ζ

bx sin .ζ

Arc.tang. a −bx cos .ζ + Const .

And thus on putting B = 0 and A = 1 there becomes a cos .ζ +bx

bx sin .ζ

∫ Xdx = 2aab sin . ζ X + 2a b 1sin . ζ Arc.tang. a −bx cos .ζ + Const . 2

Hence the integral

∫

2

( A+ Bx )dx X2

3

3

does not involve logarithms.

COROLLARY 2 71. Hence therefore since there becomes a cos .ζ +bx

∫ Xdx = 4aab sin . ζ X 3

2

2

+

3 4 aa sin .2 ζ

∫ Xdx + Const ., 2

on substituting that value, it becomes

∫

dx X3

=

− a cos .ζ +bx 4 aab sin .2 ζ X 2

+

and hence in turn it is concluded

3( − a cos .ζ +bx ) 2⋅4 a 4b sin .4 ζ X

+

1⋅3 2⋅4 a 5b sin .5 ζ

bx sin .ζ

Arc.tang . a −bx cos .ζ


∫ Xdx = 4

Translated and annotated by Ian Bruce. 5( − a cos .ζ +bx ) 3( − a cos .ζ +bx ) − a cos .ζ + bx + + 6aab sin .2 ζ X 3 4⋅6 a 4b sin .4 ζ X 2 2⋅4⋅6 a 6b sin .6 ζ X

+

1⋅3⋅5 2⋅4⋅6 a 7b sin .7 ζ

page 39

bx sin .ζ


COROLLARIUM 3 72. Thus by progressing further the integrals of all the formulas of this kind are obtained :

∫ dxX , ∫ Xdx , ∫ Xdx , ∫ Xdx 2

3

4

etc.,

the first of which is expressed by the arc of a circle only, while the remaining contain algebraic parts as well. SCHOLIUM 73. Moreover it is sufficient to know the integral dxn+1 , because the formula

∫

( A+ Bx )dx X n +1

∫X

is easily reduced to that ; for thus it is possible to represent this by 1 2bb

∫

2Abbdx + 2 Bbbxdx − 2 Babdxcos .ζ + 2 Babdxcos .ζ X n+1

,

which on account of 2bbxdx − 2abdxcos.ζ = dX , goes into this form 1 2bb

+ 21b ∫ ∫ BdX X

( Ab + Ba )cos.ζ dx

n +1

X n +1

.

But

∫ XdX

n +1

= − 1n , nX

from which there is obtained

∫

( A+ Bx )dx X n +1

=

− B + Ab + Bacos .ζ b 2nbbX n

from which there is only the need to know the integral

∫ Xdx

n +1

∫ XdX

n +1

,

, but which we have shown.

And these are all the aids we require by which all the fractional formulas M dx are to be N integrated. Provided M and N are integral [i. e. polynomial] functions of x. On account of which in general the integration of all the formulas of this kind Vdx , where V is some

∫

rational function of x, is in the rule; from which it is to be noted, unless the integration should be algebraic, it can always be shown in terms of either logarithms or angles.

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1 Part I, Section I,Chapter I. page 40 Therefore nothing different remains, unless that we should illustrate the method with a few examples. Translated and annotated by Ian Bruce.

EXAMPLE 1 (A+ Bx )dx

74. With the proposed differential formula α + β x +γ xx , to define the integral of this. Since in the numerator of the variable x there is fewer dimensions than in the denominator, this fraction includes no integral parts. Hence the nature of the denominator must be assessed carefully, whether it should have two real simple factors or not, and in the first case, if the factors are equal ; from which we have three cases to be explained. 1. The denominator has both factors equal and let it be equal to (a + bx)2 and the fraction

(A+ Bx )dx (a +bx ) 2

is resolved into these two : Ab − Ba b (a + bx ) 2

+ b(aB+bx ) ,

from which there is produced

∫

(A+ Bx )dx = bbBa(a−+Ab + B l (a + bx) + Const. ; bx ) bb (a +bx )2

if the integral is thus to be determined, so that it vanishes on putting x = 0, then there is found :

∫

(A+ Bx )dx (a + bx ) 2

( Ab − Ba ) x

B l a + bx . = ab(a +bx ) + bb a

II. If the denominator has two unequal factors and let this be the proposed formula : A+ Bx dx ; and this fraction is resolved into these partial fractions : (a + bx )(f + gx ) Ab − Ba ⋅ dx bf -ag a +bx

Ag − Bf

+ ag-bf ⋅ f dx , + gx

from which there is obtained the integral sought

( A+ Bx )dx

Ag − Bf

f + gx f

Ag − Bf g (ag-bf )

= m − n,

− Ba l a + bx + l ∫ (a +bx)(f + gx) = bAb (bf -ag ) a g (ag-bf )

+ Const .

Putting Ab − Ba b (bf -ag )

= m + n and

in order that the integral becomes ml

there becomes :

( a +bx )( f + gx ) af

+ nl

f ( a + bx ) ; a ( f + gx )

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1 Part I, Section I,Chapter I. Translated and annotated by Ian Bruce. B ( bf − ag ) 2 Abg − Bag − Bbf 2m = = B and 2n = bg ( bf − ag ) bg bg ( bf − ag )

hence there becomes :

( A+ Bx )dx

∫ (a +bx)(f + gx) = 2Bbg l

( a +bx )( f + gx ) af

+

page 41 ;

2 Abg − B ( ag +bf ) f ( a + bx ) l . 2bg (bf -ag ) a ( f + gx )

III. Let the simple factors of the denominator both be imaginary, in which case there is had aa − 2abxcos.ζ + bbxx ; now as we have treated this case above [§ 64], it becomes

( A+ Bx )dx



Ab + Ba cos .ζ

+ abb sin .ζ

bx sin .ζ


COROLLARIUM 1 75. In the second case, in which f = a and g = −b , there will be

( A+ Bx )dx

∫ aa −bbxx

B l aa −bbxx + A l a +bx = 2−bb aa 2 ab a −bx

hence separately there follows :

= A l a +bx + C , ∫ aaAdx −bbxx 2 ab a −bx and bbxx = B l = − B l aa −aa ∫ aaBxdx −bbxx 2bb bb

a

( aa −bbxx )

+ C.

COROLLARY 2 76. In the third case, if we put cos.ζ = 0 , then we have

∫

( A+ Bx )dx

Bl = bb aa +bbxx

( aa +bbxx ) a

A Arc.tang. bx + c + ab a

and hence singly:

= A Arc.tang. bx +c, ∫ aaAdx a + bbxx ab and

∫

Bxdx aa +bbxx

Bl = bb

( aa +bbxx ) a

+c



EXAMPLE 2 m −1 77. With the proposed differential formula x dx n , if indeed the exponent m – 1 is less

1+ x

than n, to define the integral.

In the final chapter of the Institutionum Calculi Differentialis we have found the m −1 is to be resolved, on taking π for the simple fractions, into which this fraction x dx n

1+ x

measure of two right angles, to be contained in this general form : 2 sin .(

2 k −1)π n

sin .(

(

2 k −1)π n

− 2 cos .

n 1− 2 x cos .

( x −cos .( ) + xx )

m( 2 k −1)π n

2 k −1)π n

( 2 k −1 π n

),

where for k it is agreed to substitute all the numbers 1, 2, 3 etc., until 2k - 1 begins to surpass the number n . [See the notes at the end of this translation.] Hence this form is multiplied by dx and since it is to be compared with our general form

( A+ Bx )dx aa − 2 abxcos.ζ +bbxx

then a = 1, b = 1, ζ =

and

( 2k −1)π n

( 2k − 1) π sin . m ( 2k − 1) π + 2 cos . ( 2k − 1) π cos . m ( 2k − 1) π A = n2 sin . n n n n n

or

( m − 1)( 2k − 1) π A = n2 cos . n B = − n2 cos .

from which there arises Ab + Ba cos .ζ = n2 sin .

and

m ( 2k − 1) π , n

( 2k − 1) π sin . m ( 2k − 1) π n

n

and therefore the integral of this part is : − n2 cos .

m ( 2k − 1) π ⎛ ( 2k − 1) π + xx ⎞ l ⎜1 − 2 x cos . ⎟ n n ⎝ ⎠

( 2k −1)π

x sin . m ( 2k − 1) π n + n2 sin . Arc.tang. . ( 2k −1)π n 1− x cos .

n



And if n is an odd number, in addition to the agreed fraction ± dx

n(1+ x )

, the integral of

which is ± 1n l (1 + x) , where the upper sign prevails, if m is odd, now the lower, if m is

∫ 1+ x

m −1 even. On account of which the integral sought x dx can be expressed in the following n

manner − n2 cos . mπ l n

x sin .π

)

(

n 1 − 2 x cos . π + xx + n2 sin . mπ Arc.tang. n n 1− x cos .π

n

x sin .3π

)

− n2 cos . 3mπ l n

(


(

− n2 cos . 7 mπ l n

(

n 1 − 2 x cos . 3π + xx + n2 sin . 3mπ Arc.tang. n n 1− x cos .3π n

x sin .5π

)


x sin .7π

)

n 1 − 2 x cos . 7π + xx + n2 sin . 7 mπ Arc.tang. n n 1− x cos .7π n

etc. according to the odd numbers being less than to n itself ; and thus the whole integral is obtained, if n should be an even number, but if n should be an odd number, the part ± 1n l (1 + x) is agreed upon above according to whether m should be either an odd or even number; from which if m = 1, + 1n l (1 + x) is agreed above. COROLLARY 1 78. We take m = 1, in order that the form dx n is obtained, and for the various cases of n

∫ 1+ x

we arrive at : I. 1dx = l (1 + x ) +x

∫ II. ∫ dx = Arc.tang.x 1+ x III. ∫ dx = − 23 cos . π3 l (1 − 2 x cos . π3 + xx ) 1+ x 2

3

+ 23 sin . π3 Arc.tang.

x sin . π3 1− x cos . π3

+ 13 l (1 + x )



∫ 1+dxx

IV.

4

(1 − 2 x cos . π4 + xx ) + 42 sin . π4 Arc.tang.1−xxsincos. . x sin . − 24 cos . 34π l (1 − 2 x cos . 34π + xx ) + 24 sin . 34π Arc.tang. 1− x cos . π

= − 24 cos . π4 l

4

π

4

3π 4 3π 4

V.

∫

dx 1+ x5

= − 52 cos . π5 l

(

1 − 2 x cos . π5 + xx

(1 − 2 x cos . 35π + xx )

− 52 cos . 35π l

VI.

∫

)

+ 52 sin . π5 Arc.tang .


+ 52 sin . 35π Arc.tang.

x sin . 35π 1− x cos . 35π

+ 15 l (1 + x )

(1 − 2 x cos . π6 + xx ) + 62 sin . π6 Arc.tang.1−xxsincos. . x sin . − 62 cos . 36π l (1 − 2 x cos . 35π + xx ) + 62 sin . 36π Arc.tang . 1− x cos . x sin . − 62 cos . 56π l (1 − 2 x cos . 55π + xx ) + 62 sin . 56π Arc.tang. 1− x cos . π

dx = − 2 cos . π l 6 6 1+ x 6

6

π

6

3π 6 3π 6 5π 6 5π 6

COROLLARY 2 79. In place of the sines and cosines the values are to be substituted, where this can be conveniently done, and we obtain :

∫ 1+dxx

3

= − 13 l

(1 − x + xx ) +

1 Arc.tang . x 3 + 1 l 1 + x ) 2− x 3 ( 3

or

∫ 1+dxx

3

= − 13 l

1+ x

(1− x + xx )

+ 1 Arc.tang . x2− 3x 3

Then on account of sin . π4 = cos . π4 = 1 = sin . 34π = − cos . 34π there becomes 2

∫

dx 1+ x 4

=+ 1 l

(1+ x (1− x

) 2 + xx )

+ 1 Arc.tang.1x− xx2 , 2 2

∫

dx 1+ x 6

=+ 1 l

(1+ x (1− x

) 3 + xx )

+ 16 Arc.tang.

2 2

then 2 3

2 + xx

3 + xx

3 x (1− xx ) 1− 4 xx + x 4



EXAMPLE 3 m −1 80. With the proposed differential formula x dx n , if indeed the exponent m − 1 is less

1+ x

than n, to define the integral itself. m −1 The part of the fraction of the function x n is required to come from some factor

1+ x

contained in this form :

2 sin . 2 knπ sin . 2 mkn π − 2 cos . 2 mkn π ( x − cos . 2 knπ ) n(1− 2 x cos . 2 knπ + xx )

( A+ Bx )dx

which on comparison with our form aa − 2abxcos.ζ +bbxx gives a = 1, b = 1, ζ = 2knπ , A = n2 sin . 2kπ sin . 2mkπ + n2 cos . 2kπ cos . 2mkπ , B = − n2 cos . 2mkπ n n n n n and hence Ab + Ba cos .ζ = n2 sin . 2kπ sin . 2mkπ . n n From which the integral hence arising is equal to − n2 cos . 2mkπ l n

)

(

x sin . 2 kπ

n , 1 − 2 x cos . 2kπ + xx + n2 sin . 2mkπ Arc.tang. 2 n n 1− x cos . kπ n

where all the successive values for k ,1, 2, 3 etc. must be substituted, as long as 2k is less than n. These parts of the integral arising, − 1n l (1 − x ) and ∓ 1n l (1 + x ) , are to be added above from the fraction n(11− x) and, if n is an even number, from the fraction ∓1 , where the upper sign prevails if m is even, and truly the lower, if m is odd. [A n (1+ x )

correction to the first edition has been added here, by the editor of the O. O. edition] On account of which the integral

∫ x1+ xdx can be expressed in this way : m −1

n



− 1n l (1 − x ) x sin . 2π

)


(


(1 − 2x cos . 4nπ + xx ) +

2 sin . 4mπ n n


(1 − 2x cos . 6nπ + xx ) +

2 sin . 6mπ n n

n 1 − 2 x cos . π + xx + n2 sin . 2mπ Arc.tang. n n 1− x cos . 2π

n x sin . 4π n Arc.tang. 1− x cos . 4π n x sin .6π n Arc.tang. 1− x cos .6π n

etc. COROLLARY 81. If m = 1 and for n successively the numbers 1, 2, 3 etc. are substituted, in order that we arrive at the following integrations :

I. 1dx = −l (1 − x ) −x

∫ (1+ x ) II. ∫ 1−dxxx = − 12 l (1 − x ) + 12 l (1 + x ) = 12 l (1− x ) III.

∫ 1−dxx

3

= − 13 l (1 − x ) − 23 cos . 23π l

(1 − 2 x cos . 23π + xx )

+ 23 sin . 23π Arc.tang. IV.

∫

dx 1− x 4

= − 14 l (1 − x ) − 24 cos . 24π l

x sin . 23π 1− x cos . 23π

(1 − 2 x cos . 24π + xx )


x sin . 24π 1− x cos . 24π

+ 24 l (1 + x ) V.

∫

dx 1− x5

− 52 cos . 45π l VI.

∫

dx 1− x 6

(1 − 2 x cos . 25π + xx ) + 52 sin . 25π Arc.tang.1−xxsincos. . (1 − 2 x cos . 45π + xx ) + 52 sin . 45π Arc.tang.1−xxsincos. .

2π 5 2π 5

= − 15 l (1 − x ) − 52 cos . 25π l

4π 5 4π 5

= − 16 l (1 − x ) − 62 cos . 26π l − 62 cos . 46π l + 16 l (1 + x )

(

1 − 2 x cos . 26π + xx

(1 − 2 x cos . 45π + xx )

)


+ 62 sin . 46π Arc.tang .

x sin . 46π 1− x cos . 46π

x sin . 26π 1− x cos . 26π



EXAMPLE 4

82. With the proposed differential formula

(x

m −1

)

+ x n − m−1 dx 1+ x

n

on n > m − 1 arising, to define

the integral of this. From Example 2 it is apparent that any part of the integral in general, on taking i for some odd number not greater than n,

− n2 cos . imπ l n

x sin .iπ

)

(

n , 1 − 2 x cos . iπ + xx + n2 sin . imπ Arc.tang. n n 1− x cos .iπ n

i ( n − m )π − n2 cos . l n

x sin .iπ

i ( n − m )π n . 1 − 2 x cos . iπ + xx + n2 sin . Arc.tang. n n 1− x cos .iπ

)

(

n

Now there is cos .

i ( n − m )π = cos . iπ − imnπ = − cos . imnπ n

(

)

and

i ( n − m )π = sin . iπ − imnπ = + sin . imnπ , n from which the logarithmic parts cancel each other, and the part of the integral in general is equal to :

(

sin .

)

x sin .iπ n . 4 sin . imπ Arc.tang. n n 1− x cos .iπ n

Hence there is put in place the angle of convenience πn = ω and there becomes :

∫

(x

m −1

)

+ x n − m−1 dx 1+ x

n

.ω = + n4 sin .mω Arc.tang. 1−x xsin cos ω .3ω + n4 sin .3mω Arc.tang. 1−x xsin cos 3ω .5ω + n4 sin .5mω Arc.tang. 1−x xsin cos 5ω

⋅ ⋅ ⋅ .iω + n4 sin .imω Arc.tang. 1−x xsin cos iω

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1 Part I, Section I,Chapter I. page 48 on taking for i the greatest odd number not greater than n. If the number n is itself odd, the part arising from the position i = n vanishes on account of sin. mn = 0. Hence it is to be noted here that the whole integral is to expressed by pure angles. Translated and annotated by Ian Bruce.

COROLLARY 83. In a like manner the following integral can be elicited, where only in this manner the following integral is elicited, where logarithms alone are left, on retaining πn = ω :

∫

(x

m −1

)

− x n −m−1 dx 1+ x n

(1 − 2 x cos .ω + xx )

= − n4 cos .mω l − n4 cos .3mω l

(1 − 2 x cos .3ω + xx )

− n4 cos .5mω l

(1 − 2 x cos .5ω + xx )

⋅ ⋅ ⋅ − n4 cos .imω l

(1 − 2 x cos .iω + xx ) ,

clearly as long as the odd number i does not exceed the exponent n. EXAMPLE 5

84. With the proposed differential formula

(x

m −1

)

− x n −m−1 dx 1+ x n

with n > m − 1 arising, to define

the integral of this. From example 3 it can be deduced that any part of the integral, if indeed for the sake of gravity we put πn = ω , can be written as − n2 cos .2mkω l

.2 mkω (1 − 2 x cos .2kω + xx ) + n2 sin .2mkω Arc.tang. 1−xxsin cos .2 mkω

+ n2 cos .2k ( n − m ) ω l But there is and

.2mkω (1 − 2 x cos .2kω + xx ) + n2 sin .2k ( n − m ) ω Arc.tang. 1−xxsin cos .2mkω

cos .2k ( n − m ) ω = cos .( 2kπ − 2kmω ) = cos .2kmω sin .2k ( n − m ) ω = sin .( 2kπ − 2kmω ) = − sin .2kmω ,

from which the general part goes into 4 sin .2kmω Arc.tang. x sin .2kω , n 1− x cos .2mω


page 49

from which hence this integration is deduced :

∫

(x

m −1

)

− x n −m−1 dx 1+ x

n

.2ω = + n4 sin .2mω Arc.tang. 1−x xsin cos 2ω .4ω + n4 sin .4mω Arc.tang. 1−x xsin cos 4ω .6ω , + n4 sin .6mω Arc.tang. 1−x xsin cos 6ω

with the even numbers increasing all the while, as long as they do not exceed the exponent n. COROLLARY 85. Also from the same place this integral is completed with πn = ω in place :

∫

(x

m −1

)

+ x n −m−1 dx 1− x n

= − n2 l (1 − x ) − n4 cos .2mω l

(1 − 2 x cos .2ω + xx )

− n4 cos .4mω l

(1 − 2 x cos .4ω + xx )

− n4 cos .6mω l

(1 − 2 x cos .6ω + xx ) ,

where also the even numbers are not to be continued beyond the term n. EXAMPLE 6 86. With the proposed differential dy = 3 dx 4 , to find the integral of this.

(

x (1+ x ) 1− x

)

The fractional function attached to dx following the factors of the denominator is 1 2 x 3 (1+ x ) (1− x )(1+ xx )

which is resolved into these simple fractions dy 1 − 1 +1− 1 − 9 + 1 + 1+ x = , x 3 x 2 x 4(1+ x )2 8(1+ x ) 8(1− x ) 4(1+ xx ) dx

from which there is elicited by integration: y = − 1 2 + 1x + lx + 1 − 98 l (1 + x ) − 81 l (1 − x ) 4(1+ x ) 2x + 18 l (1 + xx ) + 14 Arc.tang.x, which expression can be changed into this form :

(

)

xx + 1 Arc.tang .x, y = C + −2+ 2 x +5 xx − l 1+xx + 18 l 11+− xx 4 4 xx(1+ x )


page 50

SCHOLIUM 87. Therefore by this the chapter has been made clear, so that nothing further from this dy

kind can be desired. Hence as often as a function y of x of this kind is sought, so that dx is equal to some rational function of x, so the integration presents no difficulty, except perhaps in the elucidation of the particular factor of the denominator where the teaching of algebra has not been sufficient; then indeed the difficulty must be attributed to the defects of the algebra, not the method of integration which we have treated here. Then dy

also chiefly it is agreed to be noted always, since dx is put equal to a rational function of x, the function of y, unless it is algebraic, does not involve other transcending functions in addition to logarithms and angles ; where indeed here it is to be observed always that hyperbolic logarithms are to be understood, since the differential of lx in not equal to 1x , unless the hyperbolic logarithm is taken ; but the reduction of these to common logarithms is very simple, thus so that hence the application of the calculus to the practice is not bothered by any impediment. Whereby we may progress to these other cases, in dy

which the formula dx is equal to an irrational function of x, where indeed at first it must be noted, whenever this function by a suitable substitution can be reduced to rationals, the case to be returned to this chapter. Just as if there should be 3 dy = 1+ x 3− xx dx,

1+ x

6

it is clear on putting x = z , from which there arises dx = 6 z 5dz , to become

(1+ z − z ) ⋅ 6 z 5dz 3

dy =

4

1+ zz

and thus dy = −6 z 7 + 6z 6 + 6 z 5 − 6 z 4 + 6zz − 6 + 1+6zz , dz

from which on integrating y = − 34 z 8 + 76 z 7 + z 6 − 65 z 5 + 2 z 3 − 6 z + 6Arc.tang .z

and on restoring the value : 6

y = − 34 x 3 x + 76 x 6 x + x − 65 x5 + 2 x − 6 6 x + 6Arc.tang .6 x + C.



APPENDICES FROM INTRODUCTIO IN ANALYSIN INFINITORUM CH.2 & INSTITUTIONES CALCULI DIFFERENTIALIS CH.18,Part 2. In Ch. 9 § 145 of the Introductio..., Euler proceeds as follows : The quadratic p − qz + rzz in the variable z has imaginary roots if 4 pr > qq or

q 2 pr

< 1. Since

the sines and cosines of angles are less than 1, then the formula p − qz + rzz has simple imaginary factors, if

q should be the sine or cosine of some angle. Hence, let 2 pr q 2 pr

= cos .ϕ , or q = 2 pr ⋅ cos .ϕ

and the trinomial p − qz + zz contains the imaginary simple factors. But lest the irrationality should produce any trouble, I assume this form pp − 2 pqz cos .ϕ + qqzz , and the simple imaginary factors of this are these : qz − p cos .ϕ + −1 ⋅ sin .ϕ and qz − p cos .ϕ − −1 ⋅ sin .ϕ .

(

)

(

)

Where indeed it is apparent, if cos .ϕ = ±1 , then both the factors are equal and real since sin .ϕ = 0 . [In modern terms, we would say that qz − peiϕ

and qz − pe−iϕ are the

(

)

n factors, and powers of z can be expressed in the form z n = r n cos .ϕ ± −1 ⋅ sin .ϕ ]. In

§ 147, Euler assumes DeMoivre's Theorem :

( cos .ϕ ± and hence

(

−1 ⋅ sin .ϕ

)

n

= cos .nϕ ± −1 ⋅ sin .nϕ ,

)

p

z n = r n cos .nϕ ± −1 ⋅ sin .nϕ , where r = q

In § 150 Euler examines the function a n + z n , and the form of the trinomial factors of p

this factor pp − 2 pqz cos .ϕ + qqzz is to be determined. On putting r = q into the equation, there are produced these two equations : a n = r n cos .nϕ , and 0 = r n sin .nϕ . From the latter equation we must have, for some integer k, nϕ = 2kπ , or nϕ = ( 2k + 1) π ; but in the former equation, we must choose nϕ = ( 2k + 1) π , in order that we have a n − r n = 0 , and hence r = a = q . Hence in the above equation, p = a, q = 1, and ϕ = ( 2k + 1) πn . p


page 52

Hence the quadratic factor with conjugate roots of the function a n + z n can be shown to be aa − 2az cos . 2kn+1 π + zz , where k is less than n. Euler goes on to look at numerous cases involving this factorization. The partial fraction reduction quoted above in Examples 2 and 3 comes from the Institutionum...., Part 2, Ch.18, which we now address at length, we take up the story at the end of § 406 to Example 1 of §417 in translation : P , then the values of the factors of the If the rational fraction is expressed by Q denominator can be found from the resolution of the equation Q = 0 , and here we make use of a method involving the differential calculus in order that any simple factors of the denominator can thus be defined. P to have a factor f + gx , 407. Therefore we arrange the denominator Q of the fraction Q

thus so that Q = ( f + gx ) S , and indeed here, neither does S contain in addition the same factor f + gx . Let the simple fraction arising from this factor be equal to f +Agx and the P. complement has a form of this kind VS , thus so that f +Agx + VS = Q P− A = P − f +Agx = P − AS , and thus V = P − AS . Hence there arises : VS = Q f + gx ( f + gx ) S ( f + gx ) S ( f + gx ) Therefore, as V is an integral function of x, it is necessary that P − AS should be divisible

−f

by f + gx ; and therefore, if there is put f + gx = 0 or x = g , then the expression P − AS vanishes, [as f + gx is a factor of P − AS or x =

−f g

is a root of this polynomial].

−f Hence on making x = g , since P − AS = 0 , then A = PS , [for this value of x] as now we ( f + gx ) P , if there is Q have found above. But since S = f + gx , then there arises A = Q −f

put f + gx = 0 or x = g everywhere. Now since in this case both the numerator

( f + gx ) P as well as the denominator Q vanish here, by means of that, which we are about to set out in the investigation of fractions of this kind, then if indeed there is put −f

x= g , A=

( f + gx ) dP + Pgdx . dQ

But in this case on account of ( f + gx ) dP = 0 then A =

Pgdx . dQ



−f

[One can see that Euler is thinking of the ratio becoming zero on zero when x = g , and so the ratio is to the differential of the numerator to the differential of the denominator. One can see also that the problem is essentially an application of L'Hôpital's Rule : as is well known, in the neighbourhood of a common simple zero x = a of well-behaved functions r and s of x, or a common root of polynomials, the values of the functions at a are represented by their differentials, and it is an easy matter to show that r ( a ) r' ( a ) = x → a s ( a ) s' ( a )

lim

, where the rule can be extended to higher derivatives if necessary .]

P has a simple factor 408. Therefore if the denominator Q of the proposed fraction Q Pgdx

f + gx , from that there arises the simple fraction f +Agx with A = being present, dQ −f

after everywhere here the value g is put in place of x arising from the equation f + gx = 0 must be substituted. Hence, if Q is not expressed in factors, this division can often be omitted without any bother, particularly if the denominator Q has indefinite exponents, since the value of A can be obtained from the formula

Pgdx dQ

. But if the

denominator now was expressed in factors, thus in order that hence the value of S should be at once apparent, then the other expression is to be preferred, in which we found −f A = PS on putting everywhere equally x = g . And thus for finding the value of A itself

in whatever case that formula can be used, which is seen to be the more convenient and expedient. But we illustrate the use of the new formula with some examples. EXAMPLE 1

Let this fraction be proposed

x9 1+ x17

, a simple fraction of which is required to be defined

from the factor of the denominator 1 + x .

Since here Q = 1 + x17 , even if 1 + x is agreed to be a factor of this, yet, if as the first method demands, we wish to divide by that, there is produced S = 1 − x + xx − x3 + ... + x16

Therefore it is more convenient to use the new formula A =

Pgdx dQ

g = 1 and P= x9, on account of dQ = 17 x16dx there becomes A =

putting x = –1, from which there arises A = − factor of the denominator 1 + x is

−1 . 17(1+ x )

1 17

; and thus since f= 1, x9 1 16 = 17 x 17 x 7

and on

, and the simple fraction arising from the


page 54

EXAMPLE 2

For the proposed fraction

m

x 1− x 2 n

, to find the simple fraction arising from the factor 1 − x

of the denominator. On account of the proposed factor 1 − x there is f = 1 and g = −1 . Then the

denominator 1 − x gives dQ = −2nx 2n −1dx , from which because of P = x m there is 2n

obtained A =

Pgdx − xm = −2nx 2 n−1 dQ

becomes A =

1 1 , thus so that the simple fraction is 2n 2 n(1− x )

. And on putting from the equation 1 − x , x = 1 there

EXAMPLE 3

For the proposed fraction 1 − x of the denominator.

m

x 1− 4 x k +3 x n

, to find the simple fraction arising from the factor

Hence here there becomes f = 1, g = − 1, P = x m ,and Q = 1 − 4 xk + 3 x n and − xm dQ k −1 n −1 = − 4 kx + 3 nx ; from which there arises A = and on putting dx −4 kx k −1 + 3nx n−1 1 x = 1 then A = . The simple fraction arising from this simple factor of the 4 k − 3n 1 . denominator 1 − x is now ( 4k −3n )(1− x )

P to have the square factor 409. Now we put the denominator Q of the fraction Q

( f + gx )2 and the simple fractions hence arising are

A 2 ( f + gx )

+ fB . + gx

Let Q = ( f + gx ) S and the complement = VS , thus in order that 2

V S

P− =Q

A 2 ( f + gx )

P − AS −B( f + gx ) S − fB and V = 2 + gx

( f + gx )

Now because V is an integral function, it is necessary that P − AS − B ( f + gx ) S is divisible by ( f + gx ) ; and since S does not contain a further factor f +gx, 2

2 − A − B ( f + gx ) is divisible by ( f + gx ) , and thus with the also this expression P S f

factor f + gx = 0 or x = − g not only this expression, but also the differential of this f

− Bgdx . Therefore on putting x = − , there arises from the first vanishes , d . P S g

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1 Part I, Section I,Chapter I. page 55 1 = d . P ; from equation A = PS , and from the second equation there now arises B gdx S Translated and annotated by Ian Bruce.

which the values of the fractions sought

A 2 + f ( gx )

+ fB can be found . + gx

EXAMPLE

From the proposed fraction,

m

x 1− 4 x 3 + 3 x 4

the denominator of which has the factor (1 – x ) , 2

to find all the simple fractions thus arising. Since here there is f = 1, g = −1, P = x m and Q = 1 − 4 x3 + 3x 4 then [on factorising the denominator], S = 1 + 2 x + 3 xx , then P S

m

= 1+ 2 xx +3 xx and d . PS =

mx m−1dx + 2( m −1) x m dx +3( m − 2 ) x m+1dx

(1+ 2 x +3 xx )

2

.

Hence on putting x = 1 there arises A = 16 and B = − 1⋅ 6m36−8 = 4−183m ; from which the fractions sought are 4−3m . 1 2 + 18(1− x )

6(1− x )

P have three equal simple fractions, or 410. Let the denominator Q of the fraction Q

Q = ( f + gx ) S and let the simple fractions arising from the cube of the factor ( f + gx ) 3

3

be these A 3 + f ( gx )

+

B 2 + f ( gx )

+

C

( f + gx )

;

P is now the complement of these fractions to be established for the proposed fraction Q

equal to VS and it follows that

V=

P − AS −BS ( f + gx ) −CS ( f + gx )

( f + gx )

2

3

2 Whereby this expression PS − A − B ( f + gx ) − C ( f + gx ) = 0 is divisible by ( f + gx )3 ; −f

from which on putting f + gx = 0 or x = g not only this expression, but also the first −f

and second differentials are able to become equal to 0. Clearly on putting x = g


Translated and annotated by Ian Bruce. P S

2 − A − B ( f + gx ) − C ( f + gx ) = 0

d . PS − Bgdx − 2Cgdx ( f + gx ) = 0 dd . PS − 2Cg 2dx 2 = 0. Hence from the first equation there becomes A = PS . Now from the second 1 d.P . B = gdx S

And from the third there is defined C=

1 dd . P . S 2 g 2 dx 2

P has the factor f + gx n , 411. Hence generally, if the denominator Q of the fraction Q ( )

thus in order that Q = ( f + gx ) S ,with these simple fractions arising from the factor n

( f + gx )n

: A

( f + gx )

n

+

B

( f + gx )

n −1

+

C

( f + gx )

n−2

+

D

( f + gx )

n −3

+

E

( f + gx )

n−4

+ etc.,

until finally the fraction is arrived at, of which the denominator is f + gx, if the method is put in place as before, then this expression is found: P − A - B f + gx − C f + gx 2 − D f + gx 3 − E f + gx 4 − etc. ( ) ( ) ( ) ( ) S

which must be divisible by ( f + gx ) ; hence this expression as well as the individual n

−f

differentials of this as far as the degree n − 1 must vanish when we set x = g −f

From which equations it can be concluded on putting x = g everywhere :

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1 Part I, Section I,Chapter I. Translated and annotated by Ian Bruce. A = PS 1 d. P B = 1gdx S

C=

1 1⋅2 g 2 dx 2

D=

1 1⋅2⋅3 g 3 dx3

E=

1 1⋅2⋅3⋅4 g 4 dx 4

page 57

dd . PS d 3 . PS d 4 . PS

etc. Where indeed it is to be observed it is required to take the differentials themselves of P S

−f

beforehand, as in place of x there is to be put g ;

412. Hence in this way these numbers A ,B,C ,D etc can be expressed more easily than in the manner treated in the Introductio, and the values of these can be found more readily on many occasions too by this new method. Which comparison by which, in the previous manner, we can define the values of the letters A ,B,C ,D etc , can be put in −f

place more easily : Put x = g For the variable x , the remainder is put in place P − AS = P, A= P S f + gx B= S

P − BS f + gx

= Q,

C=Q S

Q−CS f + gx

= R,

D= R S

R −DS f + gx

= S,

E=S S

and thus henceforth.

P

P should not have all real simple factors, 413. But if the denominator Q of the fraction Q

then two of the imaginary factors are to be taken together, the product of which is real. Hence let a factor of the denominator Q be ff − 2 fgx cos ϕ + ggxx , which on being put equal to zero gives this two-fold imaginary value : f

f

x = g cos ϕ ± sin ϕ ; g −1 from which there arises xn =

fn gn

cos nϕ ±

fn sin nϕ . g n −1

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1 Part I, Section I,Chapter I. page 58 We put the denominator Q = ( ff − 2 fgx cos ϕ + ggxx ) S and in addition S is not to be Translated and annotated by Ian Bruce.

divisible by ff − 2 fgx cos ϕ + ggxx . Let the fraction originating from this denominator be A +ax ff − 2 fgx cos ϕ + ggxx

P is equal to V ; then and the complement for the proposed fraction Q S P − ( A + ax ) S

V = ff − 2 fgx cos ϕ + ggxx , from which P − ( A + ax ) S and in addition also PS − A − ax is divisible by

ff − 2 fgx cos ϕ + ggxx . Hence PS − A − ax vanishes, if there is put

ff − 2 fgx cos ϕ + ggxx = 0 , that is, if there is put either f

f

f

f

x = g cos ϕ + sin ϕ g −1 or x = g cos ϕ − sin ϕ g −1 414. Because P and S are integral [i. e. polynomial] functions of x, the substitution can be made separately into each in turn ; and because for any power of x, for example x n , this binomial must be substituted xn = in the first place we can put

fn gn

fn gn

cos nϕ ±

fn sin nϕ , g n −1

cos nϕ for x n everywhere, and with this done P is

fn changed into P and S into S . Then in place of x n everywhere there is put n sin nϕ and g

with this done P becomes p and S becomes s ; where it is to be noted that before these substitutions each function P and S must be expanded out completely, thus in order that, if perhaps factors are to be involved, then these are to be removed. From these values f

f

P , p, S, s found it is evident, if there is put x = g cos ϕ ± sin ϕ , the function P is g −1 changed into P ± Hence since

p −1

and the function S changed into S ± s

P − A − ax or P − A + ax S ( ) S

P±

p −1

−1

must vanish in each case, then

(

af af ⎛ ⎞ sin ϕ ⎟ S ± s = ⎜ A + g cos ϕ ± g −1 −1 ⎝ ⎠

)

from which on account of the sign condition these two equations arise :


Translated and annotated by Ian Bruce. af S af s P = AS + g cos ϕ − g sin ϕ af s

af S

p = As + g cos ϕ + g sin ϕ , from which on eliminating A there arises

Sp- sP = and thus there becomes a=

(

af S2 + s2

g

) sin ϕ

g ( Sp-sP )

(

)

f S2 + s2 sin ϕ

Then on eliminating sin ϕ there becomes

)(

(

;

.

)

af

SP + sp = S2 + s2 A + g cos ϕ . Hence A=

SP+sp ( Sp-sP ) cos ϕ − S2 + s2 S2 + s2 sin ϕ

(

)

.

Q

415. Since now there shall be S = ff − 2 fgx cos ϕ + ggxx , because on putting ff − 2 fgx cos ϕ + ggxx = 0 both the numerator as well as the denominator vanish, in this case there arises dQ:dx

S = 2 ggx − 2 fgcosϕ Now we may put, if

fn gn

dQ

cos nϕ is substituted everywhere, the function dx to become Q ,

but if there is put in place everywhere f

fn gn

sin nϕ , that goes into q ; and it is evident, if q

dQ

f

sin ϕ , the function dx becomes Q ± there is put x = g cos ϕ ± g −1 −1 From which the function S is changed into Q±q: −1 . ±2 fgsinϕ : −1

Therefore since S = S ± s with the same value for x put in place, there is obtained : −1

Q±

q 2 fgS =± sin ϕ − 2 fg s sin ϕ . −1 −1

Hence there becomes Q s = 2 fg− sin and S = 2 fg sin ϕ ϕ q



With these values in place there arises

a= and

A=

2 gg ( pq+PQ ) Q2 +q2

2 fg ( Pq-pQ ) sin ϕ 2 fg ( pq+PQ) cos ϕ . − Q2 +q2 Q2 +q2

416. Hence therefore a suitable ratio is obtained and here a simple fraction can be formed from whatever the factor of the second power, since the denominator of the proposed denominator is retained in the computation, since we avoid a division, in which the value of the letter S must be defined and which often is more than a little troublesome. P has such a factor Therefore if the denominator Q of the fraction Q

ff − 2 fgx cos ϕ + ggxx = 0 , a simple fraction arising from this factor can be defined in the following manner, so that we may produce : +ax = ff − 2 fgxAcos . ϕ + ggxx fn

f

There is put x = g cos ϕ and for whatever the power of x n there is written n cos nϕ ; g dQ

with which done P is changed into P and the function dx into Q . Then likewise there fn

f

dQ

is put x = g sin ϕ and any power of this x n = n sin nϕ and P is changed into p and dx g into q . And in this manner with the values of the letters P , Q, p and q found, A and a can thus be found, in order that there becomes

A=

2 fg ( Pq-pQ ) sin ϕ Q2 +q2

a=

−

2 fg ( pq+PQ ) cos ϕ Q2 +q2

,

2 gg ( pq+PQ ) . Q2 +q2

Hence the fraction arising from the factor ff − 2 fgx cos ϕ + ggxx of the denominator Q is given by 2 fg ( Pq-pQ ) sin ϕ + 2 g ( PQ+pq)( gx − f cos ϕ )

( Q +q )( ff −2 fgx cos ϕ + ggxx ) 2

2

.


page 61

EXAMPLE 1

If this fraction is proposed

xm a +bx n

, the denominator of which a + bx n has this factor

ff − 2 fgx cos ϕ + ggxx , to find the simple fraction agreeing with this factor. dQ

Because here P = x m and Q = a + bx n , then dx = nbx n and thus there arises

P=

fm gm

Q=

nbf n−1 cos ( n − 1) ϕ , g n−1

cos mϕ ,

p=

fn gn

q=

nbf n−1 g n−1

sin mϕ , sin ( n − 1)ϕ .

From these there becomes

Q2 + q2 =

n 2b 2 f ( 2 n −1 g( )

2 n −1)

,

PQ - pq =

nbf m+ n−1 sin ( n − m − 1) ϕ , g m+ n−1

PQ + pq =

nbf m+ n−1 cos ( n − m − 1) ϕ . g m+ n−1

and

On account of which the simple fraction sought is 2 g n−m ( f sin ϕ ⋅sin( n − m −1)ϕ + gx cos ( n − m −1)ϕ − f cos ϕ ⋅cos( n − m −1)ϕ ) nbf n−m−1 ( ff − 2 fgx cos ϕ + ggxx )

or

2 g n−m ( gx cos ( n − m −1)ϕ − f cos ( n − m )ϕ ) nbf n− m−1 ( ff − 2 fgx cos ϕ + ggxx )

.

EXAMPLE 2 If this fraction should be proposed, m 1 n , the denominator of which has the factor x

( a +bx )

ff − 2 fgx cos ϕ + ggxx ; to find the simple fraction arising from this. dQ As there is p= 1 and Q = ax m + bx m + n , then dx = max m −1 + (m + n)bx m + n − 1

and thus on putting x n =

fn cos nϕ , on account of P = x 0 , P = 1 ,and gn

( m + n )bf maf m−1 m −1 cos ( m − 1) ϕ + g g m+ n−1

m + n −1

Q=

fn and [on putting x n = n sin nϕ ] p = 0 and g

cos ( m + n − 1) ϕ

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1 Part I, Section I,Chapter I. m −1

q=

maf g m−1

page 62

Translated and annotated by Ian Bruce. ( m + n )bf m+n−1 sin ( m − 1) ϕ + sin ( m + n − 1) ϕ . g m+ n−1

Hence

A 2 + q2 =

m2 a 2 f ( g m−1

2 m −1)

2 m( m + n ) abf 2 m+ n−2 ( m + n ) b2 f ( cos n ϕ + 2 m + n −1) g 2 m+ n −2 g(

2 m + n −1)

2

+

.

For if now ff − 2 fgx cos ϕ + ggxx is a divisor of a + bx n , then a+

bf n bf n bbf 2 n . n cos nϕ = 0 and n sin nϕ = 0 , on account of which aa = g g g 2n

Hence there becomes

A 2 + q2 =

( m+ n ) g

2

bbf

2( m + n −1)

2( m + n −1)

−

m( 2n + m )aaf g

2( m −1)

2( m −1)

=

nnaaf ( 2 m −1 g( )

2 m −1)

=

nnbbf ( 2 m + n −1) g(

2 m + n −1)

.

Then there now becomes

( m + n )bf maf m−1 sin ( m − 1) ϕ + g m−1 g m+ n−1

m + n −1

Pq - pQ =

sin ( m + n − 1) ϕ

=

bf m+ n−1 g m+ n−1

=

bf m+ n−1 n cos nϕ ⋅ sin ( m − 1) ϕ + ( m + n ) sin nϕ ⋅ cos ( m − 1) ϕ g m+ n−1

( ( m + n ) sin ( m + n − 1)ϕ − m cos nϕ ⋅ sin ( m − 1)ϕ ) (

)

and

Pq + pQ =

bf m+ n−1 g m+ n−1

( ( m + n ) cos ( m + n − 1)ϕ − m cos nϕ ⋅ cos ( m − 1)ϕ ) .

Either since ff − 2 fg cos ϕ + ggxx is also a divisor of ax m −1 + bx m + n −1 , then af m−1 bf m+ n−1 cos ( m + n − 1)ϕ = 0 m −1 cos ( m − 1) ϕ + g g m+ n−1

and af m−1 bf m+ n−1 sin ( m + n − 1) ϕ = 0 , m −1 sin ( m − 1) ϕ + g g m+ n−1

from which there becomes

Q=

nbf m+ n−1 cos ( m + n − 1) ϕ and g m+n−1

q=

nbf m+ n−1 sin ( m + n − 1) ϕ g m+ n−1

or

Q=

− naf m−1 cos ( m − 1)ϕ g m−1

and q =

From which there results the fraction sought

− naf m−1 g m−1

sin ( m − 1) ϕ

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1 Part I, Section I,Chapter I. Translated and annotated by Ian Bruce. 2 g m ( f cos mϕ − gx cos ( m −1)ϕ ) naf m−1 ( ff − 2 fgx cos ϕ + ggxx )

page 63

.

Which formula follows from the first example, if m is put negative, from which there was no need to have put in place this particular example. EXAMPLE 3

If the denominator of this fraction

xm should have a factor of the form a + bx n + cx 2 n

ff − 2 fgx cos ϕ + ggxx , to find the simple fraction arising from this factor.

If ff − 2 fgx cos ϕ + ggxx is a factor of the denominator a + bx n + cx 2n , then, as we have shown above, a+

bf n cf 2 n bf n cf 2 n cos n ϕ + cos 2 n ϕ = 0 and sin n ϕ + sin 2nϕ = 0 gn g 2n gn g 2n

Therefore since there shall be P = x m and Q = a + bx n + cx 2n , then dQ = nbx n −1 + 2ncx 2n −1 , dx

from which there is produced

P= Q= q=

fm cos mϕ and gm

p=

fm sin mϕ , gm

nbf n−1 2ncf 2 n−1 cos ( 2n − 1) ϕ , n −1 cos ( n − 1) ϕ + g g 2 n−1

2 ncf 2 n−1 nbf n−1 sin ( 2n − 1) ϕ . n −1 sin ( n − 1) ϕ + g g 2 n−1

On account of which we have A 2 +q2 =

n2 f ( ) 2 n −1 g( ) 2 n −1

⎛ bb + 4bcf n cos nϕ + 4ccf 2 n ⎞ ⎜ ⎟ gn g 2n ⎠ ⎝

But from the two former equations there is f 2n ⎛ 2bcf n ccf 2 n cos nϕ + 2 n ⎞⎟ = aa 2 n ⎜ bb + n g ⎝ g g ⎠

and thus 4bcf n 2 g 2 n aa 2ccf 2 n cos n ϕ = − 2 bb − ; gn f 2n g 2n



from which value substituted there, there becomes n 2 f 2 n − 2 2aag 2 n 2ccf 2 n A 2 + q2 = 2 n−2 ⎛⎜ 2 n − bb + 2 n ⎞⎟ g g ⎝ f ⎠ or 2

2

A +q =

(

n 2 2aag 4 n −bbf 2 n g 2 n + 2ccf 4 n ffg

4 n−2

).

Then there becomes

Pq - pQ =

nbf m+ n−1 2 ncf m+ 2 n−1 sin n − m − 1 ϕ + sin ( 2n − m − 1) ϕ , ( ) g m+ n−1 g m+ 2 n−1

PQ + pq =

2 ncf m+ 2 n−1 nbf m+ n−1 cos n − m − 1 + cos ( 2n − m − 1) ϕ . ϕ ( ) g m+ n−1 g m+2 n−1

From which values found the simple fraction sought becomes 2 fg ( Pq-pQ ) sin ϕ + 2 g ( PQ+pq)( gx − f cos ϕ )

( Q +q )( ff −2 fgx cos ϕ + ggxx ) 2

2

.

417. But these values are more easily expressed if we determine these factors themselves from the denominator. Therefore let the denominator of the proposed fraction be a + bx n ; of which if the trinomial factor is put in place ff − 2 fgx cos ϕ + ggxx , there will be, as we have shown in the Introductio, a+

bf n bf n sin nϕ = 0 n cosnϕ = 0 and g gn

therefore since sin nϕ = 0 , either nϕ = (2k - 1)π or nϕ = 2kπ ; in the first case then cos nϕ = −1 , in the latter cos nϕ = +1 . Hence if a and b are positive quantities, in the first case only the condition is had, in which there arises a = 1

bf n and hence gn

1

f = a n and g = b n . But we may retain in place of these irrational quantities the letters f and g or we may put rather a = f n and b = g n , thus in order that the factors of this function are to be found : f n + g n xn . (2 k -1)π , where k can be designated to some positive number, but n indeed numbers for k greater than 2kn-1 are not to be taken, than which return 2kn-1 less n n n

Therefore since ϕ =

than unity ; hence the factors of the proposed fraction f + g x are the following :



ff − 2 fgxcos πn + ggxx

ff − 2 fgxcos 3nπ + ggxx ff − 2 fgxcos 5nπ + ggxx etc, where it is to be noted that if n is an odd number, then this one binomial factor is present, f + gx but if n is an even number then no binomial factor is present. EXAMPLE 1

To resolve this fraction

xm into its simple fractions. f + g n xn n

Since a factor of each trinomial denominator is contained in this form ff − 2 fgxcos

( 2k −1)π n

+ ggxx ,

( 2k −1)π , in example 1 in the preceding paragraph, a = f n , b = g n , and ϕ = n from which there shall be sin ( n − m − 1) ϕ = sin (m + 1)ϕ = sin

( m+1)( 2k −1)π

and cos ( n − m − 1) ϕ = − cos (m + 1)ϕ = − cos

n

,

( m+1)( 2k −1)π n

.

Hence from this factor this simple fraction arises 2 f sin .(

2 k −1)π n

sin .(

nf

m +1)( 2 k −1)π n

n − m −1

g

m

− 2 cos .(

( gx− f cos .( ) + ggxx )

m +1)( 2 k −1)π n

( ff −2 fgx cos .

2 k −1)π n

)

( 2 k −1 π n

On account of which the proposed fraction can be resolved into these simple parts :

( gx − f cos .πn ) nf g ( ff − 2 fgx cos .πn + ggxx ) 3 m +1 π 3 m +1 π 2 f sin .3nπ sin . ( n ) − 2 cos . ( n ) ( gx − f cos . 3nπ ) + nf n−m−1 g m ( ff − 2 fgx cos .3nπ + ggxx ) 5 m +1 π 5 m +1 π 2 f sin .5nπ sin . ( n ) − 2 cos . ( n ) ( gx − f cos .5nπ ) + nf n−m−1 g m ( ff − 2 fgx cos .5nπ + ggxx ) 2 f sin .πn sin .(

m +1)π n n − m −1 m

− 2 cos .(

etc.

m +1)π n


page 66

Therefore if n were an even number, in this way all the simple fractions arise; but if n should be an odd number, on account of the binomial factor f +gx this must be added to the resulting fractions above ±1 − − n m 1 nf g m ( f + gx )

where the sign + prevails, if m is an even number, and conversely for the – sign. If m should be a number greater than n, then to these fractions the above integral parts of this kind must be added Ax m − n + Bx m − 2n + Cx m −3n + Dx m − 4n + etc.,

as long as the exponents remain positive, and there becomes Ag n = 1 hence

A = 1n

Af n + Bg n = 0

B=−

fn g 2n

Bf n + Cg n = 0

C=+

f 2n g 3n

Cf n + Dg n = 0

D=−

f 3n g 4n

etc.

g

etc.


page 67

LIBER PRIOR PARS PRIMA SEU METHODUS INVESTIGANDI FUNCTIONES UNIUS VARIABILIS EX DATA RELATIONE QUACUNQUE DIFFERENTIALIUM PRIMI GRADUS

SECTIO PRIMA DE

INTEGRATIONE FORMULARUM DIFFERENTIALIUM.



CAPUT I

DE INTEGRATIONE FORMULARUM DIFFERENTIALIUM RATIONALIUM DEFINITIO

40. Formula differentialis rationalis est, quando variabilis x, cuius functio quaeritur, differentiale dx multiplicatur in functionem rationalem ipsius x, seu si X designet functionem rationalem ipsius x, haec formula differentialis Xdx dicitur rationalis. COROLLARIUM 1 41. In hoc ergo capite eiusmodi functio ipsius x quaeritur, quae si ponatur y, ut aequetur functioni rationali ipsius x, seu posita tali functione = X ut sit

dy dx

dy dx

= X.

COROLLARIUM 2 42. Hinc quaeritur eiusmodi functio ipsius x, cuius differentiale sit = Xdx; huius ergo integrale, quod ita indicari solet Xdx , praebebit functionem quaesitam.

∫

COROLLARlUM 3 43. Quodsi P fuerit eiusmodi functio ipsius x, ut eius differentiale dPsit = Xdx, quoniam. quantitatis P + C idem est differentiale, formulae propositae Xdx integrale completum est P + C. SCHOLION 1 44. Ad libri promi partem priorem huiusmodi referuntur quaestiones, quibus functiones solius variabilis x ex data differentialium primi gradus relatione quaeruntur. Scilicet si

functio quaesita = y et

dy dx

= p, id praestari oportet, ut proposita aequatione quacunque

inter ternas quantitates x, y et p inde indoles functionis y seu aequatio inter x et y, elisa littera p, inveniatur. Quaestio autem sic in genere proposita vires Analyseos adeo superare videtur, ut eius solutio nunquam expectari queat. In casibus igitur simplicioribus vires nostrae sunt exercendae, inter quos primum occurrit casus, quo p functioni cuipiam ipsius x, puta X, aequatur, ut sit

dy dx

= X seu dy = Xdx

∫

ideoque integrale y = Xdx requiratur, in quo primam sectionem collocamus. Verum et hic casus pro varia indole functionis X latissime patet ac plurimis difficultatibus implicatur, unde in hoc capite eiusmodi tantum quaestiones evolvere instituimus, in quibus ista functio X est rationalis, deinceps ad functiones irrationales



atque adeo transcendentes progressuri. Hinc ista pars commode in duas sectiones subdividitur, in quarum altera integratio formularum simplicium, quibus p =

dy dx

functioni

tantum ipsius x aequatur, est tradenda, in altera autem rationem integrandi doceri conveniet, cum proposita fuerit aequatio quaecunque ipsarum x, y et p. Et cum in his duabus sectionibus ac potissimum priore a Geometris plurimum sit elaboratum, eae fere maximam partem totius operis complebunt. SCHOLION 2 45. Prima autem integrationis principia ex ipso calculo differentiali sunt petenda, perinde ac principia divisionis ex multiplicatione et principia extractionis radicum ex ratione evectionis ad potestates sumi solent. Cum igitur, si quantitas differentianda ex pluribus partibus constet ut P + Q – R, eius differentiale sit dP + dQ – dR, ita vicissim, si formula differentialis ex pluribus partibus constet ut Pdx +Qdx – Rdx, integrale erit

∫ Pdx + ∫ Qdx – ∫ Rdx singulis scilicet partibus seorsim integrandis. Deinde cum quantitatis aP differentiale sit adP, formulae differentialis aPdx integrale erit a Pdx scilicet per quam quantitatem

∫

constantem formula differentialis multiplicatur, per eandem integrale multiplicari debet. Ita si formula differentialis sit aPdx + bQdx + cRdx, quaecunque functiones ipsius x litteris P, Q, R designentur, integrale erit a Pdx + b Qdx + c Rdx ,

∫

∫

∫

ita ut integratio tantum in singulis formulis Pdx, Qdx et Rdx sit instituenda, hocque facto insuper adiici debet constans arbitraria C, ut integrale completum obtineatur. PROBLEMA 1 46. Invenire functionem ipsius x, ut eius differentiale sit = ax n dx , seu integrare formulam differentialem ax n dx . SOLUTIO m

Cum potestatis x differentiale sit mx

∫ mx

m −1

dx , erit vicissim

m −1

dx = m x m −1dx = x m

∫

ideoque

∫x fiat m – 1 = n seu m = n +1; erit

m −1

dx = 1 x m ; m



∫x

n

dx = 1 x n +1 et a x n dx = a x n +1 .

∫

n +1

n +1

Unde formulae differentialis propositae ax n dx integrale completum erit a x n +1 + C , n +1

cuius ratio vel inde patet, quod eius differentiaIe revera fit = ax n dx . Atque haec integratio semper locum habet, quicunque numerus exponenti n tribuatur, sive positivus sive negativus, sive integer sive fractus, sive etiam irrationalis. Unicus casus hinc excipitur, quo est exponens n = –1 seu haec formula integranda adx proponitur. Verum in calculo differentiali iam ostendimus, si lx denotet logarithmum x

hyperbolicum ipsius x, fore eius differentiale = dx , unde vicissim concludimus esse

x adx = lx et x adx Quare adiecta constante arbitraria erit formulae x

∫

dx x

∫

= alx . integrale completum

= alx + C= lxa+ C, quod etiam pro C ponendo 1c ita exprimitur: lcxa. COROLLARIUM 1

47. Formulae ergo differentialis ax n dx integrale semper est algebraicum solo excepto casu, quo n = –1 et integrale per logarithmos exprimitur, qui ad functiones transcendentes sunt referendi. Est scilicet

∫ adxx = alx + C = lcx

n

.

COROLLARIUM 2 48. Si exponens n numeros positivos denotet, sequentes integrationes utpote maxime obviae probe sunt tenendae

∫ adx = ax + C, ∫ axdx = a2 xx + C , ∫ ax dx = a3 x + C, 3 4 4 5 5 6 ∫ ax dx = a4 x + C, ∫ ax dx = a5 x + C, ∫ ax dx = a6 x + C. 2

3

COROLLARIUM 3 49. Si n sit numerus negativus, posito n = – m fit

∫ adx x m

= a x1− m + C = 1− m

unde hi casus simpliciores notentur

−a

( m −l ) x m −1

+C ,


∫ ∫

Translated and annotated by Ian Bruce. adx = − a + C, adx = − a + C, adx = − a + C, 2 xx x 3 x3 x2 x3 x4 adx = − a + C, adx = − a + C, etc. 4 x4 5 x5 x5 x6

∫ ∫

∫

COROLLARIUM 4 50. Quin etiam si n denotet numeros fractos, integralia hinc obtinentur. Sit primo n = m ; erit 3

∫ adx

x m = 2a x x m + C, m+2

unde casus notentur

∫ adx x = 23a x x + C, ∫ axdx x = 25a x x + C, 3 3 4 ∫ axxdx x = 27a x x + C, ∫ ax dx x = 29a x x + C. 2

COROLLARIUM 5

51. Ponatur etiam n = − m et habebitur 2 adx = 2a −2a +C = +C, m m x 2−m x ( m − 2) xm−2

∫

unde hi casus notentur

∫ adxx = 2a

∫ xadxx = −2xa + C , ∫ xxadxx = 3−x2ax + C , ∫ xadxx = 5 x−2ax + C.

x + C,

3

2

COROLLARIUM 6

μ 52. Si in genere ponamus n = , fiet ν

∫

μ

ax ν dx = ν a x μ +ν

μ +ν ν

+C

seu per radicalia ν

∫ adx sin autem ponatur n =

−μ

ν

habebitur

ν x μ = ν a x μ +ν + C ; μ +ν

page 71


∫

= νν−aμ x

adx μ xν

ν −μ ν

page 72

+C

seu per radicalia

∫ adxx = νν−aμ ν

ν ν −μ

x

μ

+C .

SCHOLION 1 53. Quanquam in hoc capite functiones tantum rationales tractare institueram, tamen istae irrationalitates tam sponte se obtulerunt, ut perinde ac rationales tractari possint. Caeterum hinc quoque formulae magis complicatae integrari possunt, si pro x functiones alius cuiuspiam variabilis z statuantur. Veluti si ponamus x = f +gz, erit dx = gdz; quare si pro a scribamus ag , habebitur

∫ adz ( f + gz )

n

=

( n +1) g ( a

f + gz )

n +1

+C ,

casu autem singulari, quo n = –1, = a l ( f + gz ) + C . ∫ ( fadz + gz ) g Tum si sit n = – m, fiet

∫ ( f adz + gz )

=

m

−a

( m −1) g ( f + gz )

m −1

+C.

μ

At posito n = ν prodit

∫

μ

adz ( f + gz ) ν =

μ

ν a ( f + gz ) ν +1 + C ; (ν + μ ) g

μ

posito autem n = − ν obtinetur

∫ ( f + gz ) adz

μ ν

=

ν a( f + gz ) μ

(ν − μ ) g ( f + gz )ν

+C .

SCHOLION 2 54. Caeterum hic insignis proprietas annotari meretur. Cum hic quaeratur functio y, ut dy

sit dy = ax n dx , si ponamus dx = p , haec habebitur relatio p = ax n , ex qua functio y investigari debet. Quoniam igitur est y = na+1 x n +1 + C ,



ob ax n = p erit quoque px

y = n +1 + C sicque casum habemus, ubi relatio differentialium per aequationem quandam inter x, y et

p proponitur cuique iam novimus satisfieri per aequationem y = na+1 x n +1 + C . Verum px

haec non ampius erit integrale completum pro relatione in aequatione y = n +1 + C contenta, sed tantum particulare, quoniam integrale non involvit novam constantem, quae in relatione differentiali non insit. Integrale autem completum est

y = naD x n +1 + C +1 novam constantem D involvens; hinc fit dy = aDx n = p dx

ideoque

px

y= n +1 + C.

Etsi hoc non ad praesens institutum pertinet, tamen notasse iuvabit.

PROBLEMA 2 55. Invenire functionem ipsius x, cuius differentiale sit = Xdx denotante X functionem quamcunque rationalem integram ipsius x, seu definire integrale

∫ Xdx .

SOLUTIO Cum X sit functio rationalis integra ipsius x, in hac forma contineatur necesse est

X = α + β x + γ x 2 + δ x3 + ε x 4 + etc., unde per problema praecedens integrale quaesitum est

∫ Xdx = C + α x + 12 β x

2

+ 13 γ x3 + 14 δ x 4 + 15 ε x5 + 15 ζ x6 + etc.,

Atque in genere si sit

X = α x λ + β x μ + γ xν + etc., erit

∫ Xdx = C + λα+1 x

λ +1

β

γ

+ β +1 x μ +1 + γ +1 xν +1 + etc.,

ubi exponentes α , β ,ν etc. etiam numeros tam negativos quam fractos significare possunt, dummodo notetur, si fuerit λ = −1 , fore ordinem transcendentium referendus.

∫ αxdx = α lx , qui est unicus casus ad


page 74

PROBLEMA 3 56. Si X denotet functionem quamcunque rationalem fractam ipsius x, methodum describere, cuius ope formulae Xdx integrale investigari conveniat.

SOLUTIO Sit igitur X = M , ita ut M et N futurae sint functiones integrae ipsius x, ac primo N dispiciatur, num summa potestas ipsius x in numeratore M tanta sit vel etiam maior quam in denominatore N, quo casu ex fractione M partes integrae per divisionem eliciantur; N quarum integratio cum nihil habeat difficultatis, totum negotium reducitur ad eiusmodi fractionem M , in cuius numeratore M summa potestas ipsius x minor sit quam in N denominatore N. Tum quaerantur omnes factores ipsius denominatoris N, tam simplices, si fuerint reales, quam duplices reales, vicem scilicet binorum simplicium imaginariorum gerentes; simulque videndum est, utrum hi factores omnes sint inaequales necne; pro factorum enim aequalitate alio modo resolutio fractionis M in fractiones simplices est instituenda, N quandoquidem ex singulis factoribus fractiones partiales nascuntur, quarum aggregatum fractioni propositae M aequatur. Scilicet ex factore simplici a + bx nascitur fractio N A ; a + bx

si bini sint aequales seu denominator N factorem habeat (a + bx)2 , hinc nascuntur fractiones A B ; 2 + a +bx (a + bx )

ex huiusmodi autem factore (a + bx)3 hae tres fractiones A (a + bx )3

+

B (a +bx ) 2

+ a +Cbx :

et ita porro. Factor autem duplex, cuius forma est aa − 2abxcos.ζ + bbxx , nisi alius ipsi fuerit aequalis, dabit fractionem partialem A+ Bx ; aa − 2 abxcos .ζ + bbxx si autem denominator N duos huiusmodi factores aequales involvat, inde nascuntur binae huiusmodi fractiones partiales A+ Bx C + Dx 2 + aa − 2abxcos .ζ +bbxx



Translated and annotated by Ian Bruce. 3

at si cubus adeo ( aa − 2abxcos.ζ + bbxx ) fuerit factor denominatoris N, ex eo oriuntur huiusmodi tres fractiones partiales A+ Bx


3

+

C + Dx

( aa − 2abxcos.ζ +bbxx )

2

E + Fx + aa − 2abx ; cos.ζ +bbxx

et ita porro. Cum igitur hoc modo fractio proposita M in omnes suas fractiones simplices fuerit N resoluta, omnes continebuntur in alterutra harum formarum vel A+ Bx n aa − abx 2 cos .ζ + bbxx ) (

A n vel a + ( bx )

ac singulos iam per dx multiplicatos integrari oportet; erit omnium horum integralium aggregatum valor functionis quaesitae Xdx = Mdx . N

∫

∫

COROLLARIUM 1 57. Pro integratione ergo omnium huiusmodi formularum Mdx totum negotium reducitur N

ad integrationem huiusmodi binarum formularum

∫ ( aAdx + bx )

et

n

( A+ Bx )dx


n

,

dum pro n successive scribuntur numeri 1, 2, 3, 4 etc. COROLLARIUM 2 58. Ac prioris quidem formae integrale iam supra (§ 53) est expeditum, unde patet fore

= A l a + bx ) + Const . ∫ aAdx + bx b ( = − A + Const . ∫ ( aAdx b( a + bx ) + bx ) 2

∫ ( aAdx + bx )

3

=

−A 2 2b( a + bx )

+ Const .

et generatim

∫ ( aAdx +bx )

n

=

−A n −1 ( n −1)b( a +bx )

+ Const .

COROLLARIUM 3


page 76

59. Ad propositum ergo absolvendum nihil aliud superest, nisi ut integratio huius formulae

( A+ Bx )dx


n

doceatur, primo quidem casu n = 1, tum vero casibus n = 2, n = 3, n = 4 etc. SCHOLION 1 60. Nisi vellemus imaginaria evitare, totum negotium ex iam traditis confici posset; denominatore enim N in omnes suos factores simplices resoluto, sive sint reales sive imaginarii, fractio proposita semper resolvi poterit in fractiones partiales huius formae Adx vel huius Adx ; quarum integralia cum sint in promtu, totius formae n a + bx

( a +bx )

M dx integrale habetur. Tum autem non parum molestum foret binas partes imaginarias N

ita coniungere, ut expressio realis resultaret, quod tamen rei natura absolute exigit. SCHOLION 2 61. Hic utique postulamus resolutionem cuiusque functionis integrae in factores nobis concedi, etiamsi algebra neutiquam adhuc eo sit perducta, ut haec resolutio actu institui possit. Hoc autem in Analysi ubique postulari solet, ut, quo longius progrediamur, ea, quae retro sunt relicta, etiamsi non satis fuerint explorata, tanquam cognita assumamus; sufficere scilicet hic potest omnes factores per methodum approximationum quantumvis prope assignari posse. Simili modo cum in calculo integrali longius processerimus, integralia omnium huiusmodi formularum Xdx, quaecunque functio ipsius x littera X significetur, tanquam cognita spectabimus plurimumque nobis praestinisse videbimur, si integralia magis abscondita ad eas formas reducere valuerimus; atque hoc etiam in usu practico nihil turbat, cum valores talium formularum Xdx quantumvis prope assignare

∫

liceat, uti in sequentibus ostendemus. Caeterum ad has integrationes resolutio denominatoris N in suos factores absolute est necessaria, propterea quod singuli hi factores in expressionem integralis ingrediuntur; paucissimi sunt casus iique maxime obvii, quibus ista resolutione carere possumus; veluti si proponatur haec n −1 formula x dx statim patet posito x n = ν eam abire in n

1+ x

(

)

dν , cuius integrale est n(1+ν )

1 l 1 + ν = 1 l 1 + x n , ubi resolutione in factores non fuerat opus. Verum huiusmodi ) n n (

casus per se tam sunt perspicui, ut eorum tractatio nulla peculiari explicatione indigeat. PROBLEMA 4

62. Invenire integrale huius formulae

∫

( A+ Bx )dx

y = aa − 2abxcos.ζ +bbxx



SOLUTIO Cum numerator duabus constet partibus Adx+Bxdx, haec posterior B xdx sequenti modi tolli poterit. Cum sit −2 abdxcos .ζ + 2bbxdx l ( aa − 2abxcos.ζ + bbxx ) = aa − 2 abxcos .ζ +bbxx

∫

multiplicetur haec aequatio per 2Bbb et a proposita auferatur; sic enim prodibit y − B l ( aa − 2abxcos.ζ 2bb

+ bbxx ) =

( A+

Bacos .ζ b

)dx


ita ut haec tantum formula integranda supersit. Ponatur brevitatis gratia formula

Bacos .ζ =C b

A+ ut habeatur haec

∫ aa −2abxCdx cos .ζ + bbxx quae ita exhiberi potest

∫ aa sin . ζ +Cdx ( bx − acos.ζ ) 2

2

.

Statuatur bx − acos.ζ = av sin .ζ hincque dx = adv bsin.ζ unde formula nostra erit Cadv sin .ζ :b

C dv ∫ aa sin . ζ (1+vv ) = ab sin .ζ ∫ (1+ vv ) . 2

Ex calculo autem differentiali novimus esse

∫ (1+dvvv ) = Arc.tang.v = Arc.tang.

bx − a cos .ζ , a sin .ζ

unde ob C=

erit nostrum integrale

Ab + Ba cos .ζ b

Ab + Ba cos .ζ bx − a cos .ζ Arc.tang. a sin .ζ abb sin .ζ



Quocirca formulae propositae

( A+ Bx )dx

∫ aa−2abxcos.ζ +bbxx integrale est B l aa − 2abxcos.ζ 2bb (

Ab + Ba cos .ζ

+ bbxx ) + abb sin .ζ

bx − a cos .ζ

Arc.tang. a sin .ζ

,

quod ut fiat completum, constans arbitraria C insuper addatur. COROLLARIUM 1

bx − a cos .ζ cos .ζ 63. Si ad Arc.tang. a sin .ζ addamus Arc.tang. sin .ζ , quippe qui in constante addenda bx sin .ζ contentus concipiatur, prodibit Arc.tang. a −bx cos .ζ sicque habebimus

( A+ Bx )dx

∫ aa −2abxcos.ζ +bbxx Ab + Ba cos .ζ bx sin .ζ = 2Bbb l ( aa − 2abxcos.ζ + bbxx ) + abb sin .ζ Arc.tang. a −bx cos .ζ

adiecta constante C. COROLLARIUM 2 64. Si velimus, ut integrale hoc evanescat posito x = 0, constans C sumi debet B laa sicque fiet = 2−bb

( A+ Bx )dx



Ab + Ba cos .ζ

+ abb sin .ζ

bx sin .ζ


Pendet ergo hoc integrale partim a logarithmis, partim ab arcubus circularibus seu angulis.

COROLLARIUM 3 65. Si littera B evanescat, pars a logarithmis pendens evanescit fitque bx sin .ζ

= A Arc.tang. a −bx cos .ζ + C ∫ aa−2abxAdx cos .ζ + bbxx ab sin .ζ sicque per solum angulum definitur.


page 79

COROLLARIUM 4 66. Si angulus ζ sit rectus ideoque cos .ζ = 0 et sin .ζ = 1 , habebitur

( A+ Bx )dx

∫ aa +bbxx

Bl = bb

( aa +bbxx ) a

A Arc.tang. bx + C ; + ab a

si angulus ζ sit 60° ideoque cos.ζ = 12 et sin .ζ = 23 , erit

( A+ Bx )dx

∫ aa−abx+bbxx = bbB l

( aa − abx +bbxx ) a

3 + 2 Ab + Ba Arc.tang. 2bx a −bx abb 3

At si ζ = 120 ideoque cos.ζ = − 12 et sin .ζ = 23 , erit o

( A+ Bx )dx

∫ aa+ abx+bbxx = bbB l

( aa + abx +bbxx ) a

3 + 2 Ab − Ba Arc.tang. 2bx a + bx abb 3

SCHOLION 1 67. Omnino hic notatu dignum evenit, quod casu ζ = 0 , quo denominator aa − 2abx + bbxx fit quadratum, ratio anguli ex integrali discedat. Posito enim angulo ζ B l a −bx et altera pars infinite parvo erit cos.ζ = 1 et sin .ζ = ζ , unde pars logarithmica fit bb a Ab + Ba Arc.tang. bxζ abbζ a −bx

( Ab + Ba ) x bxζ , quia arcus infinite parvi a −bx tangens ipsi est aequalis, ab( a −bx )

=

sicque haec pars fit algebraica. Quocirca erit

∫

( A+ Bx )dx B a −bx ( Ab + Ba ) x = bb l a + + Const ., 2 ab( a −bx ) ( a −bx )

cuius veritas ex praecedentibus est manifesta; est enim

( A+ Bx ) − B + Ab + Ba 2 = −bx ) b( a −bx )2 b a ( ( a −bx ) Iam vero est B la = B l a −bx , = B l ( a − bx ) − bb ∫ b(−aBdx bb a −bx ) bb

∫

( Ab + Ba )dx Ab + Ba Ab + Ba ( Ab + Ba ) x = − abb = , 2 bb( a −bx ) ab( a −bx ) b( a −bx )

siquidem utraque integratio ita determinetur, ut casu x = 0 integralia evanescant.



SCHOLION 2 68. Simili modo, quo hic usi sumus, si in formula differentiali fracta Mdx summa N

potestas ipsius x in numeratore M uno gradu minor sit quam in denominatore N, etiam is terminus tolli poterit. Sit enim

M = Ax n −1 + Bx n − 2 + Cx n −3 + etc. et

N = α x n + β x n −1 + γ x n − 2 + etc. = dy Cum iam sit ac ponatur Mdx N dN = nα x n −1dx + ( n − 1) dxβ x n − 2 + ( n − 2 ) dxγ x n −3 + etc.

erit AdN nα N

⎛ n −1 ( n −1) Aβ n − 2 ( n − 2 ) Aγ n −3 ⎞ = dx + nα x + etc.⎟ , Ax + nα x N ⎜⎝ ⎠

quo valore inde subtracto remanebit ⎛⎛ ⎞ ( n −1) Aβ ⎞ ( n − 2 ) Aγ ⎞ ⎛ dy − nAdN B − nα ⎟ x n − 2 + ⎜ C − nα ⎟ x n −3 + etc.⎟ = dx N ⎜ ⎜⎝ αN ⎠ ⎝ ⎠ ⎝ ⎠ Quare si brevitatis gratia ponatur

B− obtinebitur

( n −1) Aβ nα

= B, C −

( n − 2 ) Aγ nα

= C, D −

( n −3) Aδ nα

= D, etc.,

dx( Bx n−2 +Cx n−3 + Dx n−4 + etc.) y = nAα lN + = Mdx . n n −1 n−2 n −3 N

∫ α x +β x

+γ x

+δ x

etc.

∫

Hoc igitur modo omnes formulae differentiales fractae eo reduci possunt, ut summa potestas ipsius x in numeratore duobus pluribusve gradibus minor sit quam in denominatore. PROBLEMA 5

69. Formulam integralem y=

( A+ Bx )dx


n +1

ad aliam similem reducere, ubi potestas denominatoris sit unogradu inferior. SOLUTIO



Sit brevitatis gratia aa − 2abxcos.ζ + bbxx = X ac ponatur

∫ Cum ob

( A+ Bx )dx X n +1

= y.

dX = −2abdxcos.ζ + 2bbdxx

sit

− n( C + Dx ) dX Ddx + n X n +1 X

d . C + Dx = n X

ideoque C + Dx Xn

habebimus

y + C + Dx = n X

∫

=

∫

2nb( C + Dx )( a cos .ζ −bx ) dx X n +1

∫X

+ Ddx n ,

dx( A+ 2 nCab cos .ζ + x( B + 2nDab cos .ζ − 2nCbb ) − 2 nDbbxx ) X n+1

∫X

+ Ddx n .

Iam in formula priori litterae C et D ita definiantur, ut numerator per X fiat divisibilis; oportet ergo sit = −2nDXdx , unde nanciscimur

A + 2nCab cos .ζ = −2nDaa et

B + 2nDab cos .ζ − 2nCbb = 4nDab cos .ζ

seu B − 2nCbb = 2nDabcos.ζ hincque 2nCbb ; 2nDa = Bb−cos .ζ

at ex priori conditione est 2nDa =

− A− 2nCab cos .ζ a

quibus aequatis fit

Ba + Abcos.ζ − 2nCabbsin 2 .ζ = 0 seu

C=

Ba + Abcos .ζ 2 nabbsin 2 .ζ

,

unde

B − 2nCbb =

Ba sin .2 ζ − Ba − Ab cos .ζ a sin .2 ζ

ita ut reperiatur

D= Sumtis ergo litteris

C=

=

− Ab cos .ζ − Ba cos .2 ζ , a sin .2 ζ

− Ab − Ba cos .ζ . 2naab sin .2 ζ

Ba + Ab cos .ζ et 2nabb sin 2 .ζ

D=

− Ab − Ba cos .ζ 2naab sin 2 .ζ



erit dx y + C + Dx = −2nDdx + Ddx n n n = − ( 2 n − 1) D n

∫

X

ideoque

∫ sive

∫

( A+ Bx )dx X n +1

( A+ Bx )dx X

+

∫X

= −C −nDx − ( 2n − 1) D dxn ,

∫X

X

=

n +1

∫X

X

− Baa − Aab cos .ζ + ( Abb + Bab cos .ζ ) x 2 naabb sin .2 ζ X n

( 2n −1)( Ab + Ba cos .ζ ) 2 naab sin .2 ζ

∫ Xdx . n

( A+ Bx )dx Quare, si formula dxn constet, etiam integrale hoc assignari n +1

∫

∫X

X

poterit. COROLLARIUM 1 70. Cum igitur manente X = aa − 2abx cos .ζ + bbxx sit

∫ dxX erit

∫

( A+ Bx )dx X2

bx sin .ζ

1 = absin Arc.tang . a −bx cos .ζ + Const ., .ζ

= +

− Baa − Aab cos .ζ +( Abb + Bab cos .ζ ) x 2 aabb sin .2 ζ X

Ab + Ba cos .ζ 2 na 3bb sin .3 ζ

bx sin .ζ

Arc.tang. a −bx cos .ζ + Const .

Ideoque posito B = 0 et A = 1 fiet a cos .ζ +bx

bx sin .ζ

∫ Xdx = 2aab sin . ζ X + 2a b 1sin . ζ Arc.tang. a −bx cos .ζ + Const . 2

Integrale ergo

∫

2

( A+ Bx )dx X2

3

3

logarithmos non involvit.

COROLLARIUM 2

71. Hinc ergo cum sit

a cos .ζ +bx


erit ilIum valorem substituendo

2

2

+

3 4 aa sin .2 ζ

∫ Xdx + Const ., 2


∫

Translated and annotated by Ian Bruce. bx sin .ζ 1⋅3 dx = − a cos .ζ + bx + 3( − a cos .ζ +bx ) + Arc.tang. a −bx cos .ζ 3 2 2 4 4 5 5 4 aab sin . ζ X 2⋅4 a b sin . ζ X 2⋅4 a b sin . ζ X

page 83

hincque porro concluditur − a cos .ζ +bx


2

+

3

+

5( − a cos .ζ +bx ) 4⋅6 a b sin . ζ X 4

1⋅3⋅5 2⋅4⋅6 a 7b sin .7 ζ

4

2

+

3( − a cos .ζ +bx ) 2⋅4⋅6 a 6b sin .6 ζ X

bx sin .ζ

Arc.tang . a −bx cos .ζ

COROLLARIUM 3 72. Sic ulterius progrediendo omnium huiusmodi formularum integralia obtinebuntur

∫ dxX , ∫ Xdx , ∫ Xdx , ∫ Xdx 2

3

etc.,

4

quorum primum arcu circulari solo exprimitur, reliqua vero praeterea partes algebraicas continent. SCHOLION 73. Sufficit autem integralia

∫ Xdx

n +1

nosse, quia formula

∫

( A+ Bx )dx X n +1

facile eo reducitur; ita

enim repraesentari potest 1 2bb

∫

2Abbdx + 2 Bbbxdx − 2 Babdxcos .ζ + 2 Babdxcos .ζ X n +1

quae ob 2bbxdx − 2abdxcos.ζ = dX abit in hanc 1 2bb

∫

BdX X n +1

+ 21b

∫

( Ab + Ba )cos.ζ dx X n +1

.

At

∫ XdX

n +1

unde habebitur

∫

( A+ Bx )dx X n +1

=

unde tantum opus est nosse integralia

= − 1n ,

−B 2nbbX n

∫ XdX

n +1

nX

+

Ab + Bacos .ζ b

∫ Xdx

n +1

,

, quae modo exhibuimus. Atque haec sunt

dx integrandas, omnia subsidia, quibus indigemus ad omnes formulas fractas M N

Dummodo M et N sint functiones integrae ipsius x. Quocirca in genere integratio omnium huiusmodi formularum Vdx , ubi V est functio rationalis ipsius x quaecunque, est in

∫

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1 Part I, Section I,Chapter I. page 84 potestate; de quibus notandum est, nisi integralia fuerint algebraica, semper vel per logarithmos vel angulos exhibeti posse. Nihil aliud igitur superest, nisi ut hanc methodum aliquot exemplis illustremus. Translated and annotated by Ian Bruce.

EXEMPLUM 1

74. Proposita formula differentiali

(A+ Bx )dx α + β x +γ xx

definire eius integrale.

Cum in numeratore variabilis x pauciores habeat dimensiones quam in denominatore, haec fractio nullas partes integras complectitur. Hinc denominatoris indoles perpendatur, utrum habeat duos factores simplices reales necne, ac priori casu, num factores sint aequales; ex quo tres habebimus casus evolvendos. 1. Habeat denominator ambos factores aequales sitque = (a + bx)2 et fractio

(A+ Bx )dx (a +bx ) 2

resolvitur in has duas Ab − Ba b (a + bx ) 2

+ b(aB+bx ) ,

unde fit

∫

(A+ Bx )dx (a +bx )2

= bbBa(a−+Ab + B l (a + bx)+Const. ; bx ) bb

si integrale ita determinetur, ut evanescat posito x = 0, reperitur

∫

(A+ Bx )dx ( Ab − Ba ) x B a + bx = ab(a +bx ) + bb l a . (a + bx ) 2

II. Habeat denominator duos factores inaequales sitque proposita haec A+ Bx formula (a +bx dx et haec fractio resolvitur in has partiales )(f + gx ) Ab − Ba ⋅ dx + Ag − Bf ⋅ dx , bf -ag a +bx ag-bf f + gx

unde obtinetur integrale quaesitum

( A+ Bx )dx

Ag − Bf

f + gx f

Ag − Bf g (ag-bf )

= m − n,

− Ba l a + bx + l ∫ (a +bx)(f + gx) = bAb (bf -ag ) a g (ag-bf )

Ponatur Ab − Ba b (bf -ag )

= m + n et

ut integrale fiat ml

( a +bx )( f + gx ) af

+ nl

f ( a + bx ) a ( f + gx )

+ Const .



erit 2m =

erit ergo

B ( bf − ag ) bg ( bf − ag )

( A+ Bx )dx

∫ (a +bx)(f + gx) = 2Bbg l

B = bg

et 2n =

( a +bx )( f + gx ) af

+

2 Abg − Bag − Bbf bg ( bf − ag )

;

2 Abg − B ( ag +bf ) f ( a + bx ) l . 2bg (bf -ag ) a ( f + gx )

III. Sint denominatoris factores simplices ambo imaginarii, quo casu formam habebit aa − 2abxcos.ζ + bbxx ; qui casus cum supra [§ 64] iam sit tractatus, erit

( A+ Bx )dx

∫ aa−2abxcos.ζ +bbxx ( aa − 2abxcos.ζ +bbxx )

Bl = bb

a

75. Casu secundo, quo f = a et

Ab + Ba cos .ζ

+ abb sin .ζ

bx sin .ζ


COROLLARIUM 1 g = −b , erit

( A+ Bx )dx

∫ aa −bbxx

B l aa −bbxx + A l a +bx = 2−bb 2 ab a −bx aa

hinc seorsim sequitur

= A l a +bx + C ∫ aaAdx −bbxx 2 ab a −bx et bbxx = B l = − B l aa −aa ∫ aaBxdx bb −bbxx 2bb

a ( aa −bbxx )

+ C.

COROLLARIUM 2 76. Casu tertio, si ponamus cos.ζ = 0 , habemus

( A+ Bx )dx

∫ aa +bbxx

Bl = bb

( aa +bbxx ) a

A Arc.tang. bx + c + ab a

hincque singillatim

= A Arc.tang. bx +c ∫ aaAdx + bbxx ab a et

= Bl ∫ aaBxdx +bbxx bb

( aa +bbxx ) a

EXEMPLUM 2

+c

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1 Part I, Section I,Chapter I. Translated and annotated by Ian Bruce. page 86 m −1 , siquidem exponens m – 1 minor sit quam n, 77. Proposita formula differentiali x dx 1+ x n

integrale definire. In capite ultimo Institutionum Calculi Differentialis [L. EULERI Institutiones calculus differentialis cum eius usu in analysi finitorum ac doctrina serierum, Petropoli 1755] m −1 resolvitur, sumto π pro invenimus fractiones simplices, in quas haec fractio x dx n

1+ x

mensura duorum angulorum rectorum in hac forma generali contineri 2 sin .(

2 k −1)π n

sin .(

(

2 k −1)π n

− 2 cos .

n 1− 2 x cos .

( x −cos .( ) + xx )

m( 2 k −1)π n

2 k −1)π n

( 2 k −1 π n

),

ubi pro k successive omnes nnmeros 1, 2, 3 etc. substitui convenit, quoad 2k - 1 numerum n superare incipiat. Hac ergo forma in dx ducta et cum generali nostra

( A+ Bx )dx

aa − 2 abxcos .ζ +bbxx

comparata fit

a = 1, b = 1, ζ = et

seu

( 2k −1)π n

( 2k − 1) π sin . m ( 2k − 1) π + 2 cos . ( 2k − 1) π cos . m ( 2k − 1) π A = n2 sin . n n n n n

( m − 1)( 2k − 1) π A = n2 cos . n B = − n2 cos .

unde fit

Ab + Ba cos .ζ = n2 sin . ac propterea huius partis integrale erit

et

m ( 2k − 1) π , n

( 2k − 1) π sin . m ( 2k − 1) π n

n



− n2 cos .

m ( 2k − 1) π ⎛ ( 2k − 1) π + xx ⎞ l ⎜1 − 2 x cos . ⎟ n n ⎝ ⎠

( 2k −1)π

x sin . m ( 2k − 1) π n . + n2 sin . Arc.tang. n ( 2k −1)π

1− x cos .

Ac si n sit numerus impar, praeterea accedit fractio ± dx

n(1+ x )

n

, cuius integraIe

est ± 1n l (1 + x) , ubi signum superius valet, si m impar, inferius vero, si m par. Quocirca integrale quaesitum

∫ x1+ xdx m −1

n

x sin .π

)

(

− n2 cos . mπ l n

sequenti modo exprimetur

n 1 − 2 x cos . π + xx + n2 sin . mπ Arc.tang. n n 1− x cos .π

n

)


(


(

− n2 cos . 7 mπ n

(

l

x sin .3π


)

x sin .5π


)

x sin .7π

n 1 − 2 x cos . 7π + xx + n2 sin . 7 mπ Arc.tang. n n 1− x cos .7π n

etc. secundum numeros impares ipso n minores; sicque totum obtinetur integrale, si n fuerit numerus par, sin autem n sit numerus impar, insuper accedit haec pars ± 1n l (1 + x) , prout

m sit numerus vel impar vel par; unde si m = 1, accedit insuper + 1n l (1 + x) . COROLLARIUM 1 78. Sumamus m = 1, ut habeatur forma dx n , et pro variis casibus

∫ 1+ x

ipsius n adipiscimur



I. 1dx = l (1 + x ) +x

∫ II. ∫ dx = Arc.tang.x 1+ x III. ∫ dx = − 23 cos . π3 l (1 − 2 x cos . π3 + xx ) 1+ x 2

3

+ 23 sin . π3 Arc.tang. IV.

V.

∫ ∫

dx 1+ x 4

dx 1+ x 5


( ) − 24 cos . 34π l (1 − 2 x cos . 34π + xx )

= − 24 cos . π4 l

1 − 2 x cos . π4 + xx

+ 13 l (1 + x )




x sin . 34π 1− x cos . 34π

(1 − 2 x cos . π5 + xx ) + 52 sin . π5 Arc.tang.1−xxsincos. . x sin . − 52 cos . 35π l (1 − 2 x cos . 35π + xx ) + 52 sin . 35π Arc.tang. 1− x cos . π

= − 52 cos . π5 l

5

π

5

3π 5 3π 5

VI.

∫

dx 1+ x 6

( ) − 62 cos . 36π l (1 − 2 x cos . 35π + xx ) − 62 cos . 56π l (1 − 2 x cos . 55π + xx )

= − 62 cos . π6 l

1 − 2 x cos . π6 + xx


+ 15 l (1 + x )



x sin . 36π 1− x cos . 36π


x sin . 56π 1− x cos . 56π

COROLLARIUM 2 79. Loco sinuum et cosinuum valores, ubi commode fieri potest, substituendo obtinemus

∫ 1+dxx

3

= − 13 l

(1 − x + xx ) +

1 Arc.tang . x 3 2− x 3

+ 13 l (1 + x )

seu

∫ 1+dxx

3

= − 13 l

1+ x (1− x + xx )

+ 1 Arc.tang . x2− 3x 3

Deinde ob sin . π4 = cos . π4 = 1 = sin . 34π = − cos . 34π fit 2

∫ tum vero

dx 1+ x 4

=+ 1 l 2 2

(1+ x (1− x

) 2 + xx ) 2 + xx

+ 1 Arc.tang.1x− xx2 , 2 2



∫

=+ 1 l

dx 1+ x 6

2 3

(1+ x (1− x

) 3 + xx ) 3 + xx

+ 16 Arc.tang.

3 x(1− xx ) 1− 4 xx + x 4

EXEMPLUM 3


x m−1dx , siquidem exponens m − 1 1+ x n

sit minor quam n,

eius integrale definire. m −1 Functionis fractae x n pars ex factore quocunque oriunda hac forma continetur

1+ x

2 sin . 2 knπ sin . 2 mkn π − 2 cos . 2 mkn π ( x − cos . 2 knπ ) n(1− 2 x cos . 2 knπ + xx )

( A+ Bx )dx

quae cum forma nostra aa − 2abxcos.ζ +bbxx comparata dat a = 1, b = 1, ζ = 2knπ , A = n2 sin . 2kπ sin . 2mkπ + n2 cos . 2kπ cos . 2mkπ , B = − n2 cos . 2mkπ n n n n n

hincque

Ab + Ba cos .ζ = n2 sin . 2kπ sin . 2mkπ . n n Ex quo integrale hinc oriundum erit − n2 cos . 2mkπ l n

)

(

x sin . 2 kπ

n , 1 − 2 x cos . 2kπ + xx + n2 sin . 2mkπ Arc.tang. 2 n n 1− x cos . kπ n

ubi pro k successive omnes numeri 1, 2, 3 etc. substitui debent, quamdiu 2k minor est quam n. Accedunt insuper hae ex fractione n(11− x) et, si n est numerus par, ex fractione n(1∓+1x ) oriundae integralis partes − 1n l (1 − x ) et ∓ 1n l (1 + x ) , ubi signum superius valet, si m est par, inferius vero, si m impar. Quocirca integrale

∫ x1+ xdx hoc modo exprimitur m −1

n



− 1n l (1 − x ) x sin . 2π

)


(


(1 − 2x cos . 4nπ + xx ) +

2 sin . 4mπ n n


(1 − 2x cos . 6nπ + xx ) +

2 sin . 6mπ n n

n 1 − 2 x cos . π + xx + n2 sin . 2mπ Arc.tang. n n 1− x cos . 2π

n x sin . 4π n Arc.tang. 1− x cos . 4π n x sin .6π n Arc.tang. 1− x cos .6π n

etc. COROLLARIUM 81. Sit m = 1 et pro n successive numeri 1, 2, 3 etc. substituantur, ut nanciscamur sequentes integrationes

I. 1dx = −l (1 − x ) −x

∫ (1+ x ) II. ∫ 1−dxxx = − 12 l (1 − x ) + 12 l (1 + x ) = 12 l (1− x ) III.

∫ 1−dxx

3

= − 13 l (1 − x ) − 23 cos . 23π l

(1 − 2 x cos . 23π + xx )

+ 23 sin . 23π Arc.tang. IV.

∫

dx 1− x 4

= − 14 l (1 − x ) − 24 cos . 24π l

x sin . 23π 1− x cos . 23π

(1 − 2 x cos . 24π + xx )


x sin . 24π 1− x cos . 24π

+ 24 l (1 + x ) V.

∫

dx 1− x5

− 52 cos . 45π l VI.

∫

dx 1− x 6

(1 − 2 x cos . 25π + xx ) + 52 sin . 25π Arc.tang.1−xxsincos. . (1 − 2 x cos . 45π + xx ) + 52 sin . 45π Arc.tang.1−xxsincos. .

2π 5 2π 5

= − 15 l (1 − x ) − 52 cos . 25π l

4π 5 4π 5

= − 16 l (1 − x ) − 62 cos . 26π l − 62 cos . 46π l + 16 l (1 + x )

(

1 − 2 x cos . 26π + xx

(1 − 2 x cos . 45π + xx )

)


+ 62 sin . 46π Arc.tang .

x sin . 46π 1− x cos . 46π

x sin . 26π 1− x cos . 26π



EXEMPLUM 4


(x

m −1

)

+ x n −m−1 dx 1+ x

n

existente n > m − 1 eius integrale

definire. Ex exemplo 2 patet integralis partem quamcunque In genere esse, sumto i pro numero quocunque impare non maiore quam n, − n2 cos . imπ l n − n2 cos .

x sin .iπ

)

(

n , 1 − 2 x cos . iπ + xx + n2 sin . imπ Arc.tang. n n 1− x cos .iπ

i ( n − m )π l n

n

)

(

1 − 2 x cos . iπ + xx + n2 sin . n

Verum est cos .

x sin .iπ i ( n − m )π n . Arc.tang. n 1− x cos .iπ n

i ( n − m )π = cos . iπ − imnπ = − cos . imnπ n

(

)

et

i ( n − m )π = sin . iπ − imnπ = + sin . imnπ , n unde partes logarithmicae se destruent, eritque pars integralis In genere

(

sin .

)

x sin .iπ n . 4 sin . imπ Arc.tang. n n 1− x cos .iπ n

Ponatur commoditatis ergo angulus πn = ω eritque

∫

(x

m −1

)

+ x n− m−1 dx 1+ x

n

.ω = + n4 sin .mω Arc.tang. 1−x xsin cos ω .3ω + n4 sin .3mω Arc.tang. 1−x xsin cos 3ω .5ω + n4 sin .5mω Arc.tang. 1−x xsin cos 5ω

⋅ ⋅ ⋅ .iω + n4 sin .imω Arc.tang. 1−x xsin cos iω

sumto pro i maximo numero impare exponentem n non excedente. Si ipse numerus n sit impar, pars ex positione i = n oriunda ob sin. mn = 0 evanescet. Notetur ergo hic totum integrale per meros angulos exprimi.


page 92

COROLLARIUM 83. Simili modo sequens integrale elicitur, ubi soli logarithmi relinquentur, manente πn = ω :

∫

(x

m −1

)

− x n−m−1 dx 1+ x

n

(1 − 2 x cos .ω + xx )

= − n4 cos .mω l − n4 cos .3mω l

(1 − 2 x cos .3ω + xx )

− n4 cos .5mω l

(1 − 2 x cos .5ω + xx )

⋅ ⋅ ⋅ − n4 cos .imω l

(1 − 2 x cos .iω + xx ) ,

donec scilicet numerus impar i non superet exponentem n. EXEMPLUM 5


(x

m −1

)

− x n − m−1 dx 1+ x

n

existente n > m − 1 eius integrale

definire. Ex exemplo 3 integralis pars quaecunque concluditur, siquidem brevitatis gratia πn = ω statuamus, − n2 cos .2mkω l

.2mkω (1 − 2 x cos .2kω + xx ) + n2 sin .2mkωArc.tang. 1−xxsin cos .2mkω

+ n2 cos .2k ( n − m ) ω l At est et

.2 mkω (1 − 2 x cos .2kω + xx ) + n2 sin .2k ( n − m ) ω Arc.tang. 1−xxsin cos .2 mkω

cos .2k ( n − m ) ω = cos .( 2kπ − 2kmω ) = cos .2kmω sin .2k ( n − m ) ω = sin .( 2kπ − 2kmω ) = − sin .2kmω ,

unde ista pars generalis abit in 4 sin .2kmω Arc.tang. x sin .2kω , n 1− x cos .2mω quare hinc ista integratio colligitur


∫

(x

Translated and annotated by Ian Bruce. m −1

−x

n − m −1

1+ x

n

)dx = + 4 sin .2mω Arc.tang. n

page 93

x sin .2ω 1− x cos 2ω

.4ω + n4 sin .4mω Arc.tang. 1−x xsin cos 4ω .6ω + n4 sin .6mω Arc.tang. 1−x xsin cos 6ω

numeris paribus tamdiu ascendendo, quoad exponentem n non superent. COROLLARIUM 85. Indidem etiam haec integratio absolvitur manente πn = ω

∫

(x

m −1

)

+ x n−m−1 dx 1− x

n

= − n2 l (1 − x ) − n4 cos .2mω l

(1 − 2 x cos .2ω + xx )

− n4 cos .4mω l

(1 − 2 x cos .4ω + xx )

− n4 cos .6mω l

(1 − 2 x cos .6ω + xx ) ,

ubi etiam numeri pares non ultra terminum n sunt continuandi. EXEMPLUM 6 86. Proposita formula differentiali dy = 3 dx 4 eius integrale invenire.

(

x (1+ x ) 1− x

)

Functio fracta per dx affecta secundum denominatoris factores est 1 2 x3 (1+ x ) (1− x )(1+ xx )

quae in has fractiones simplices resolvitur dy 1 − 1 +1− 1 − 9 + 1 + 1+ x = , x3 x 2 x 4(1+ x )2 8(1+ x ) 8(1− x ) 4(1+ xx ) dx

unde per integrationem elicitur y = − 1 2 + 1x + lx + 2x

1 − 9 l 1+ x − 1 l 1− x ( ) 8 ( ) 4(1+ x ) 8

+ 18 l (1 + xx ) + 14 Arc.tang.x, quae expressio in hanc formam transmutatur

(

)

xx + 1 Arc.tang .x, y = C + −2+ 2 x +5 xx − l 1+xx + 18 l 11+− xx 4 4 xx(1+ x )

SCHOLION 87. Hoc igitur caput ita pertractare licuit, ut nihil amplius in hoc genere desiderari possit. dy

Quoties ergo eiusmodi functio y ipsius x quaeritur, ut dx aequetur functioni rationali ipsius x, toties integratio nihil habet difficultatis, nisi forte ad denominatoris singulos

EULER'S INSTITUTIONUM CALCULI INTEGRALIS VOL. 1 Part I, Section I,Chapter I. page 94 factores eliciendos Algebrae praecepta non sufficiant; verum tum defectus ipsi Algebrae, non vero methodo integrandi, quam hic tractamns, est tribuendus. Deinde etiam Translated and annotated by Ian Bruce.

dy

potissimum notari convenit semper, cum dx functioni rationali ipsius x aequale ponitur, functionem y, nisi sit algebraica, alias quantitates transcendentes non involvere praeter logarithmos et angulos; ubi quidem observandum est hic perpetuo logarithmos hyperbolicos intelligi oportere, cum ipsius lx differentiale non sit = 1x , nisi logarithmus hyperbolicus sumatur; at horum reductio ad vulgares est facillima, ita ut hinc applicatio calculi ad praxin nulli impedimento sit obnoxia. Quare progrediamur ad eos casus, dy

quibus formula dx functioni irrationali ipsius x aequatur, ubi quidem primo notandum est, quoties ista functio per idoneam substitutionem ad rationalitatem perduci poterit, casum ad hoc caput revolvi. Veluti si fuerit 3 dy = 1+ x 3− xx dx,

1+ x 5

6

evidens est ponendo x = z , unde fit dx = 6 z dz , fore

(1+ z − z ) ⋅ 6 z5dz 3

dy =

4

1+ zz

ideoque dy = −6 z 7 + 6z 6 + 6 z 5 − 6 z 4 + 6zz − 6 + 1+6zz , dz

unde integrale y = − 34 z 8 + 76 z 7 + z 6 − 56 z 5 + 2 z 3 − 6 z + 6Arc.tang .z

et restituto valore 6

y = − 34 x 3 x + 76 x 6 x + x − 65 x5 + 2 x − 6 6 x + 6Arc.tang .6 x + C.