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

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  • $\int\frac{dx}{\sqrt{a^2+x^2}}=log\mid x+\sqrt{x^2+a^2}\mid+c.$
Given:$I=\int \frac{1}{\sqrt{1+4x^2}}dx.$
 
Let 2x=t.
 
On integrating we get,
 
2dx=dt.$\Rightarrow dx=\frac{dt}{2}.$
 
On substituting t and dt we get,
 
Hence $I=\frac{1}{2}\int\frac{1}{\sqrt{1+t^2}}dt.$
 
On integrating we get,
 
$\;\;\;=\frac{1}{2}log\mid t+\sqrt{t^2+1}\mid+c.$
 
Substituting for t we get,
 
$\int\frac{1}{\sqrt{1+4x^2}}=\frac{1}{2}log\mid2x+\sqrt{4x^2+1}\mid+c.$

 

answered Feb 4, 2013 by sreemathi.v
 
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