<s>[INST] Find $\int_0^\infty \sqrt{x} \left( \tan^{-1} \left(\frac{x+a}{c} \right)- \tan^{-1} \left(\frac{x-a}{c} \right) \right)dx$ How to compute the integral
\begin{align}
\int_{0}^{\infty}\sqrt{x}\left[\tan^{-1}\left(\frac{x+a}{c} \right)- \tan^{-1} \left(\frac{x-a}{c} \right) \right]\mathrm{d}x
\end{align}
for $c,a>0$.
If the it the integration is to difficult can we give a good upper bound? 
 [/INST] With the short hand $b=\sqrt{a^2+c^2}$ and the substitution $x=t^2$, the integral is
\begin{align}
I=& \ 2\int_0^\infty t^2\left( \tan^{-1} \frac{t^2+a}{c} - \tan^{-1} \frac{t^2-a}{c} \right)dt\\
\overset{ibp}=& \ \frac{4c}3 \int_0^\infty
\frac{2at^2-b^2}{t^4-2at^2+b^2}+ \frac{2at^2+b^2}{t^4+2at^2+b^2}\ dt\\
=& \ \frac{4c}3 \bigg(\frac{(2a-b)\pi}{2\sqrt{2(b-a)}}+ \frac{(2a+b)\pi}{2\sqrt{2(b+a)}} \bigg)\\
= &\ \frac{2\pi}3\left(2a\sqrt{b+c}-b \sqrt{b-c} \right)
\end{align}
</s>