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Integrate the functions\[\frac{1}{\sqrt{(2-x)^2+1}}\]

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

 

answered Feb 4, 2013 by sreemathi.v
 
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