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

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  • $\int\frac{dx}{\sqrt{x^2-a^2}}=\int log\mid x+\sqrt{x^2-a^2}\mid+c.$
Given:$I=\int\frac{x-1}{\sqrt{x^2-1}}dx.$
 
Separating the terms we can write as,
 
$I=\int\frac{x}{\sqrt{x^2-1}}dx-\int\frac{dx}{\sqrt{x^2-1}}.$
 
Put $x^2-1=t.$
 
On differentiating we get,
 
2xdx=dt.$\Rightarrow x dx=\frac{dt}{2}.$
 
Substituting this we get,
 
$I=\frac{1}{2}\int\frac{dt}{\sqrt t}-\int\frac{dx}{\sqrt {x^2-1}}.$
 
On integrating we get,
 
$\;\;\;=\frac{1}{2}\bigg(\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}\bigg)-log\mid x+\sqrt{x^2-1}\mid+c.$
 
Substituting back for t we get,
 
$\;\;\;=\frac{1}{2}\bigg(\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\bigg)-log\mid x+\sqrt{x^2-1}\mid+c.$
 
$\;\;\;=[\sqrt{x^2-1}-log\mid x+\sqrt{x^2-1}\mid]+c.$

 

answered Feb 4, 2013 by sreemathi.v
 
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