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Integrate the function $\;\int\;x\;\tan^{-x}$

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  • (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
On differentiating with respect to x we get
and let dv=xdx.
On integrating we get
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)
Add and subtract 1 to the numerator
On separating the terms we get
$\;\;\;=\int dx-\int\frac{1}{1+x^2}dx.$
On integrating we get
Now substituting this in equ(1)
$\int x\tan^{-1}xdx=\frac{x^2}{2}\tan^{-1}x-\frac{1}{2}(x-\tan^{-1}x)+c.$


answered Feb 8, 2013 by sreemathi.v