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Integrate the functions\[\frac{3x^2}{x^6+1}\]

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  • $(i)\;\int \frac{dx}{x^2+1}=\tan^{-1}x+c.$
Given $I=\int\frac{3x^2}{x^6+1}dx.$
 
Let $x^6=(x^3)^2.$
 
$\;\;\;=\int\frac{3x^2}{(x^3)^2+1}.$
 
Let $x^3=t.$
 
On differentiating we get,
 
$3x^2dx=dt.$
 
Substituting this we get,
 
$I=\int\frac{dt}{(t^3)^2+1}.$
 
On integrating we get,
 
$\;\;\;=\tan^{-1}t+c.$
 
substituting for t we get,
 
$\frac{3x^2}{x^6+1}=\tan^{-1}(x^3)+c.$
 

 

answered Feb 3, 2013 by sreemathi.v
 
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