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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function $\Large \frac{\large e^{\tan^{-1}x}}{\large 1+x^2}$

This question has appeared in model paper 2012

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Toolbox:
  • Method of substitution :
  • Given f(x)dx can be transformed into another form by changing independent variable x to t by substituting x=g(t).
  • Consider $I=\int f(x)dx.$
  • Put x=g(t) so that $\frac{dx}{dt}=g'(t).$
  • dx=g'(t)dt.
  • Thus $I=\int f(g(t).g'(t))dt.$
Given $I=\int \frac{e^{\tan^{-1}x}}{1+x^2}dx $.
 
Let $\tan^{-1}x=t.$
 
$\frac{1}{1+x^2}dx=dt.$
 
Now substituting for x and dx we get,
 
$I=\int e^tdt.$
 
On integrating we get,
 
$e^t+c.$
 
Substituting back for t we get,
 
$e^{\tan^{-1}x}+c.$
 
Hence $ \int {e^{\tan{-1}x}}{1+x^2}=e^{\tan{-1}x}+c.$
 
 
 

 

answered Jan 29, 2013 by sreemathi.v
 
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