A recent post in MSE:
re-emphasizes that when Ahmed Integral is converted to a two-dimensional integral in $x,y$, then the sameness of domain the $(0,1)$ of $x$ and $y$ makes it do-able. Further, it asks for the evaluation of a slightly different Integral, which is challenging.
In this light, now the question here is: How to find $$\int_{0}^{\infty} \frac{\tan^{-1}\sqrt{2+x^2}}{(1+x^2)\sqrt{2+x^2}} dx?$$
Mind the upper limit that is $\infty$, there is square root sign in the argument of $\tan^{-1}$, and yes the integrand is the same as that of Ahmed integral.
Best Answer
\begin{align} &\int_{0}^{\infty} \frac{\tan^{-1}\sqrt{2+x^2}}{(1+x^2)\sqrt{2+x^2}}~ dx\\ =& \int_{0}^{\infty}\int_0^1 \frac{1}{(1+x^2)(1+y^2(2+x^2))}dy~dx\\ = & \int_{0}^{1}\int_0^\infty \frac{1}{1+y^2} \bigg( \frac{1}{1+x^2}-\frac{y^2}{1+y^2(2+x^2) }\bigg) dx~dy\\ = & \ \frac\pi2\int_{0}^1 \frac{1}{1+y^2} \bigg( 1-\frac{y}{\sqrt{1+2y^2} }\bigg)dy\\ =& \ \frac\pi2 \left(\tan^{-1}y-\tan^{-1}\sqrt{1+2y^2}\right)\bigg|_0^1 =\frac\pi2\cdot \frac\pi{6}=\frac{\pi^2}{12} \end{align}