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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the functions\[\frac{x^2}{1-x^6}\]

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  • $\int\frac{dx}{\sqrt{a^2+x^2}}=\frac{1}{2a}log\mid\frac{a+x}{a-x}\mid+c.$
Given:$I=\int \frac{x^2}{1-x^6}dx.$
 
Put $x^3=t.$
 
On differentiating we get,
 
$3x^2dx=dt.\Rightarrow x^2dx=\frac{dt}{3}.$
 
On substituting t and dt we get,
 
Hence $I=\frac{1}{3}\int \frac{dt}{1-t^2}dt.$
 
On integrating we get,
 
$\;\;\;=\frac{1}{3}.\frac{1}{2}log\mid\frac{1+t}{1-t}\mid+c.$
 
substituting back for t we get,
 
$\int\frac{x^2}{1-x^6}dx=\frac{1}{6}log\mid\frac{1+x^3}{1-x^3}\mid+c.$

 

answered Feb 4, 2013 by sreemathi.v
 
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