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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function $\;\int\;x\;\tan^{-x}$

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Toolbox:
  • (i)When there are two functions u and v and if they are of the form $\int u dv,$then we can solve it by the method of integration by parts\[\int udv=uv-\int vdu\]
  • (ii)$\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}+c.$
  • (iii)$\int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan^{-1}(x/a)dx.$
Given $I=\int x\tan^{-1}xdx.$
 
Clearly the given integral function is of the form $\int u dv$,so let us follow the method of integration by parts where \[\int udv=uv-\int vdu\]
 
Here let us take
 
$u=\tan^{-1}x$
 
On differentiating with respect to x we get
 
$du=\frac{1}{1+x^2}dx.$
 
and let dv=xdx.
 
On integrating we get
 
$v=\frac{x^2}{2}$.
 
Now substituting for u,v,du and dv we get,
 
$\int x\tan^{-1}xdx=\frac{x^2}{2}\tan^{-1}x-\int\frac{x^2}{2}.\frac{1}{1+x^2}dx.$
 
$\qquad\qquad=\frac{x^2}{2}\tan^{-1}x-\frac{1}{2}\int \frac{x^2}{1+x^2}dx.$----(1)
 
consider$\int\frac{x^2}{1+x^2}dx.$
 
Add and subtract 1 to the numerator
 
$\;\;\;=\int\frac{x^2+1-1}{1+x^2}.$
 
On separating the terms we get
 
$\;\;\;=\int\frac{1+x^2}{1+x^2}dx-\int\frac{1}{1+x^2}dx.$
 
$\;\;\;=\int dx-\int\frac{1}{1+x^2}dx.$
 
On integrating we get
 
$\;\;\;=x-\tan^{-1}x+c.$
 
Now substituting this in equ(1)
 
$\int x\tan^{-1}xdx=\frac{x^2}{2}\tan^{-1}x-\frac{1}{2}(x-\tan^{-1}x)+c.$
 
$\;\;\;\quad\qquad\;\;\;\;=\frac{x^2}{2}\tan^{-1}x-\frac{x}{2}-\frac{1}{2}\tan^{-1}x+c.$

 

answered Feb 8, 2013 by sreemathi.v
 
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