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

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Toolbox:
  • (i)Form of rational function\[\frac{px+q}{(x+a)(x+b)}\]
  • Form of partial fraction\[\frac{A}{(x+a)}+\frac{B}{(x+b)}\]
  • (ii)$\int \frac{dx}{x+a}=log|x+a|+c$
Given:$I=\int\frac{1-x^2}{x(1-2x)}dx.$
 
$\frac{1-x^2}{x(1-2x)}=\frac{A}{x}+\frac{B}{(1-2x)}$
 
$1-x^2=A(1-2x)+Bx.$
 
Equating the coefficient of x we get
 
-2A+B=0----(1)
 
Equating the constant term
 
1=A$\Rightarrow A=1.$
 
Substituting for A in equ(1) we get,
 
-2(1)+B=0.
 
Therefore B=2.
 
$I=\int\frac{dx}{x}+2\int\frac{dx}{(1-2x)}$
 
Let 1-2x=t
 
Therefore -2dx=dt$\Rightarrow \frac{-dt}{2}$
 
Now substituting this we get
 
$I=\frac{dx}{x}+2\int\frac{-dt/2}{t}$
 
$\;\;\;=\int\frac{dx}{x}-\frac{dt}{t}$
 
On integrating we get
 
$log|x|-log|t|+c$.
 
On substituting for t we get
 
$\int\frac{1-x^2}{x(1-2x)}=log|x|-log|1-2x|+c.$

 

answered Feb 19, 2013 by sreemathi.v
 
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