Revisiting Ahmed Integral in $(0,\infty)$ A recent post in MSE:
Evaluate $\int_0^{\infty } \frac{\tan ^{-1}\left(a^2+x^2\right)}{\left(x^2+1\right)\sqrt{a^2+x^2}} \, dx$
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.

\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}
