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# Integrate the rational functions$\frac{2}{(1-x)(1+x^2)}$

Toolbox:
• $(i)\;$Form of the rational function$\frac{px+q}{(x+a)(x^2+b)}$
• $\;$Form of the partial function$\frac{Ax+B}{(x+a)}+\frac{C}{(x^2+b)}$
• $(ii)\;\int\frac{dx}{(x+a)}=log|x+a|+c.$
Given:$I=\int\frac{2}{(1-x)(1+x^2)}dx.$

$\frac{2}{(1-x)(1+x^2)}=\frac{A}{(1-x)}+\frac{Bx+C}{(1+x^2)}$

$\Rightarrow 2=A(1+x^2)+(Bx+C)(1-x)$

Equating the coefficients of $x^2$

0=A-B-----(1)

Equating the coefficients of x,

0=B-C-------(2)

Equating the constant terms,

2=A+C-----(3)

A-B=0
B-C=0
_____________
A-C=0------(4)
A+C=2
A-C=0
_________________
2A=2$\Rightarrow A=1$.

Substitute for A in equ(1)

A-B=0.

1-B=0$\Rightarrow B=1.$

Substitute of B in equ (2)

B-C=0.

1-C=0$\Rightarrow C=1.$

Hence A=1,B=1,C=1.

Substituting these values we get in I

$\frac{2}{(1-x)(1+x^2)}=\frac{1}{(1-x)}+\frac{x+1}{(1+x^2)}$

On separating the terms,

$\qquad=\frac{1}{(1-x)}+\frac{x}{(1+x^2)}+\frac{1}{(1+x^2)}$

On integrating we get,

$I=\int\frac{dx}{(1-x)}+\int\frac{xdx}{(1+x^2)}+\int\frac{dx}{1+x^2}.$

Put 1-x=t

Therefore -dx=dt$\Rightarrow dx=-dt.$

Put $1+x^2=u.$

2xdx=du.

$xdx=\frac{du}{2}.$

Substituting this we get,

I=$-\int\frac{dt}{t}+\frac{1}{2}\int\frac{du}{u}+\int\frac{dx}{1+x^2}$

On integrating

$\;\;\;=-log t+\frac{1}{2}log u+\tan^{-1}x+c.$

Substituting for t and u

$\;\;\;=-log|1-x|+\frac{1}{2}log|1+x^2|+\tan^{-1}(x)+c.$

edited Feb 6, 2013