Question on definite integral How to compute the following:
$$ \int_{0}^{\infty} \frac{\tan^{-1}(ax)}{x(1+x^2)}dx $$
If you could, please, solve it completely, for, I am afraid, a hint won't do for me.

Let $\displaystyle I(a) = \int_{0}^{\infty}\frac{\tan^{-1}(ax)}{x.(1+x^2)}dx$
Now Diff. both side w.r.to $a$ , We Get
$\displaystyle \frac{dI(a)}{da} = \int_{0}^{\infty}\frac{1}{(1+a^2x^2).x.(1+x^2)}.x.dx$
$\displaystyle \frac{dI(a)}{da} = \int_{0}^{\infty}\frac{1}{(1+x^2).(1+a^2x^2)}dx$
$\displaystyle \frac{dI(a)}{da} = \frac{1}{(a^2-1)}\int_{0}^{\infty}\left(\frac{a^2}{1+a^2x^2}-\frac{1}{1+x^2}\right)dx$
$\displaystyle \frac{dI(a)}{da}= \frac{a}{(a^2-1)}.\frac{\pi}{2}-\frac{1}{(a^2-1)}.\frac{\pi}{2}$
$\displaystyle \frac{dI(a)}{da} = \frac{\pi}{2}.\frac{1}{a+1}$
Now $\displaystyle \int \frac{dI(a)}{da}da = \frac{\pi}{2}\int\frac{1}{a+1}da$
$\displaystyle I(a) =\frac{\pi}{2}.\ln |a+1|+C$
