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PE 63. Let be a rational fraction (Pand Q being functions of x);

Q and let the greatest exponent of « in P be less, at least by unity than in Q. If it be not so, divide the numerator by the denominator, till this latter condition takes place. For exaniple, if the fraction

a

IC

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atbp3

xic

6 were

of which the at binti we shall have by division

b

Pr second part is such as we have supposed for

Q 64. This done, find the factors of Q, as if we had to resolve the equation Q=0; and if they are all of the first degree, real, and unequal, the proposed fraction will then be of the form

ar*-*+60*2 + &c.... ta: (x-1) (x-g) (x-h) &c. To find the fluent in this case, we must decompose the proposed

Ax Br C: fraction into these others.

it + + &c.; of which the in

x-g x—h tegral (art. 56) is Al (x-f) + Bl (x-8) +&c. (where l stands for the hyperbolic logarithm of the quantity within the parenthesis.) To this value we must add a constant C, and determine the com efficients A, B, C, &c. by first reducing them to the same denominator, and then transposing, and successively equating to zero the coefficient of each power of x, which process will give as many equam tions as there are unknown quantities. Ex. Required the fluent of

x (aa—xx)' x (a +x)(a

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or

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+

a

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aa

+

cx

Where lo

G

+

re

aa

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aa

نم

.

Ai B. Cå I decompose this fraction into the following ones : +

ats and reducing to the same denominator, transposing, and ordering the terms, I find

Aa? + Bax + Bxx -1 +Ca -A

-C Therefore A=

B=

1
C= ; and consequently
2aa'
2aa

x(aa-I) 1 1

1

of which the integral is Laa

2aai satu k _! (_)_!(a+i), lo 1 aa 2aa

2aa presents the correction C, the form of which is optional.

65. This method will always succeed when the factors of the denominator of the proposed fraction are all real and unequal; but it several among them were equal; if, for example, (r—a)represented any number m of these factors, we must then decompose the fraction into the following ones. BE

A'z+B'xm? +&c.... +R
+ +&C.... +
g

(-a)" and, after having determined the coefficients as before, we may pro

Ax-1 Be-2 ceed to integrate

* +

. + &c. or in general (t-a)" (-a)" shi (ma), by making ama=%.

Example. Let it be required to find the integral of (=+*+2); (send)* (*+1)**

(**+x2+2). Ac (Bx+C): (Dx+E);
I suppose

+
(-1)(2+1)= (1-1) (+1)
8
D

;

therefore
(x3 + x2 +2)
2.3

(5x+7)

ė

Now to inte*(1-1)” (x+1)* (1-1) (x+1)

make -l=%, which changes it into

(1-1) (4432) 3_4ż sz

4 of which the fluent is

-31(x-1) -1

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+1(7—3x).

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grate the fraction (7—31)

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and treating in the same manner the other fraction, we obtain for the
complete integral or fluent
1 1

3
Ql2-
+

(x-
r+1

1

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2

66. If in the denominator of the proposed fraction

P*

there

Q should be imaginary factors ; on representing one of them by x+a+bv --1, there will be another of the form rta- a-bn-1, and their product x? + 2ax + a + b* will be a real factor of Q. We must therefore find the coefficients 2a, a', b, of the factor of the second degree, which is always possible; and the real factor of the second degree 2+ + 2ax + a + b, or more concisely

? +mz+n, will be determined. Then we must suppose that (Ar+B):

is one of the partial factors of the proposed fraction, and + m2 +12 determine A and B, as before. After which, making r+am=, the fraction will become

A' 2+B) 2

zz +66% A'zz B'2

A'zz

A'
+
Now

1(xe +66'), and S.
zz +66 +22+60°
22 +6

zz +6b

Be

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arc tang

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+C; and consequently we shall by this means arrive at the fluent required.

Examples. ?

Az (Bz+C)ż

+ ; we shall find that A= (1+z) (1 + x2) 1+2

C--, this changes the proposed fraction into these 3 1

3

of which the integral is 2 1+2 12 1+zz

2 (1+z) – 2 (1+z2)+ arc tan z+C.

:

be proposed, which is reducible to X (1+x) '(1 +3+x1)

B

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1

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2

1+ 2z

2

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Again let

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(2x+3)
+

To integrate this last quantity, make
(1+x)? 'i+x+x.c
(34)

2 X-2-1, and it becomes.

of which the zz +

zz + 1

2z integral is 1 (zz+)-- arc lan

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zz +

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Consequently by substituting the value of z, we have for the com

1 1 plete integral lx—21 (1+x)+31(1+x+xx)+

arc tan

1+30 73 2.r

13

67. .The last case which remains to be considered, is that in which the denominator Q has one or more factors of this form (xx+ax +b)". Here we must suppose that the partial fraction proceeding from this

«2m-' + Bxem-? +&c.... +R); factor is (Ax?m-1

t, and then determine the (rx +ar+6) coefficients as before. After which, making x=2-4a, and substituting, the fraction will become of the form A'zzm-' + Bz2m-*+&c. +R' :

2, which may be thus decomposed
(22 +66')
A'22–1 Bizam-
% +

2+&c. (zz+b'b') (z2 +6°6) Now those terms in which the numerator contains any odd power of s, are integrable, in part algebraically, and in part by logarithms (57); and those in the numerator of which z is raised to any even power,

being of the form_Mz2t

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may be reduced to the form.

by (zz +6%)m? art (63); that is, we are able to integrate them, partly algebraically, and partly by circular arcs ; consequently we shall by combining these means obtain the integral of the proposed fraction.

68. The following example will serve to elucidate these different methods, as it contains them all. Let the fraction

c
(1+x) xx (xx+2) (xx+1)%

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+

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Ai
(Bx+C) = (Dx+

E) _ (FrP +Gx* +Hr+1);
+
xx +2

(rr+1) We shall find by reducing these fractions to the same denominator, that AS

6'

=,= н.

and that the proposed fraction=

3

1

2

+
1+2 2

12

+

6

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XX+2

+
(1–2):71
1. (x-1)& +(4x?–4+*+{):

4==; Thens /(1+)

*=

1

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}. (1)

1 2x

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- S

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xx- 1

2xx

1 72 6

XX+2 12° xx+2 6213 +112 1

arc tan 6/2

72 v!.

1

) +
(xx+1)*
xx

3
(xx+1) ( 8 (rr+1)

+

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Now to integrate the partial fractions - 1.

and

(rx+1)% we must have recourse to the method of reduction before (or+1) explained. Thus by formula (A), article 61, we find /

(1 +18) sat and again by formula (D) article 63, it appears (1 +37)"

=t. Totusi -= : + f arc tan s (1 +xx)'

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