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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the functions\[\frac{3x}{1+2x^4}\]

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  • $\int\frac{dx}{\sqrt{a^2+x^2}}=\frac{1}{a}\tan^{-1}\big(\frac{x}{a}\big)+c.$
Given:$I=\int \frac{3x}{1+2x^4}dx.$
 
Put $\sqrt 2x^2=t.$
 
On differentiating we get,
 
2\sqrt 2 xdx=dt.$\Rightarrow dx=\frac{dt}{2\sqrt 2}.$
 
On substituting t and dt we get,
 
Hence $I=\frac{3}{2\sqrt 2}\int \frac{dt}{1+t^2}dt.$
 
On integrating we get,
 
$\;\;\;=\frac{3}{2\sqrt 2}\tan^{-1}(t)+c.$
 
substituting for t we get,
 
$\;\;\;=\frac{3}{2\sqrt 2}\tan^{-1}\big(\sqrt 2x^2\big)+c.$

 

answered Feb 4, 2013 by sreemathi.v
 
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