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

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