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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the rational functions\[\frac{2}{(1-x)(1+x^2)}\]

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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)
 
Add equ(1) and equ(2)
A-B=0
B-C=0
_____________
A-C=0------(4)
Add equ(4) and equ(3)
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
 
$\;\;\;=-log|1-x|+\frac{1}{2}log|1+x^2|+\tan^{-1}(x)+c.$

 

answered Feb 6, 2013 by sreemathi.v
edited Feb 6, 2013 by sreemathi.v
 
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