Ahmed's integral is:
$$
A = \int_0^1 \frac{\tan^{-1}\sqrt{2 + x^2}}
{(1 + x^2)\sqrt{2 + x^2}}\,dx.
$$
If we denote the unknown integral above by $B$, then:
\begin{gather*}
A + B =
\int_0^1 \frac{\tan^{-1}\sqrt{2 + x^2}
+ \tan^{-1}\left(\frac{1}{\sqrt{2 + x^2}}\right)}
{(1 + x^2)\sqrt{2 + x^2}}\,dx \\
= \frac{\pi}{2}\int_0^1 \frac{dx}{(1 + x^2)\sqrt{2 + x^2}}
= \frac{\pi}{2}\left[
\tan^{-1}\left(\frac{x}{\sqrt{2 + x^2}}\right)\right]_0^1
= \frac{\pi^2}{12}.
\end{gather*}
So the conjecture $B = \pi^2/32$ is equivalent to the
known theorem $A = 5\pi^2/96$.
On p.190ff. of Paul J. Nahin's book Inside Interesting
Integrals (with an introduction to contour integration): A
Collection of Sneaky Tricks, Sly Substitutions, and Numerous Other
Stupendously Clever, Awesomely Wicked, and Devilishly Seductive
Maneuvers for Computing Nearly 200 Perplexing Definite Integrals
From Physics, Engineering, and Mathematics (Plus 60 Challenge
Problems with Complete, Detailed Solutions (2014), there is an
ingenious and well-explained derivation of these two results, which
amounts to proving, in the same way, the above equation
$A + B = \pi^2/12$, along with the simultaneous equation
$A - B = \pi^2/48$.
Part of the highly condensed argument given by MathSE user Jack D'Aurizio
in answer to a recent question
is equivalent to the case $\tau = 1$ of the following argument,
for $0 < \tau \leqslant 1$, which I've expanded considerably for
the sake of those unused to this kind of calculation (i.e. me).
For all real $a$,
$$
\tan^{-1} a = \int_0^1 \frac{a}{1 + a^2u^2}\,du.
$$
Using Wolfram Alpha to evaluate the inner integral after
changing the order of integration:
\begin{gather*}
\int_0^\tau \frac{\tan^{-1}\sqrt{\frac{1 - t^2}{2}}}{1 + t^2}\,dt =
\int_0^\tau \int_0^1 \frac{\sqrt{\frac{1 - t^2}{2}}}
{(1 + t^2)\left(1 + \frac{1 - t^2}{2}u^2\right)}\,du\,dt \\
= \int_0^1\int_0^\tau \frac{\sqrt{\frac{1 - t^2}{2}}}
{(1 + t^2)\left(1 + \frac{1 - t^2}{2}u^2\right)}\,dt\,du \\
= \int_0^1 \left(
\frac{\tan^{-1}\left(\frac{\sqrt{2}\tau}{\sqrt{1 - \tau^2}}\right)}
{u^2 + 1}
- \frac{\tan^{-1}\left(\frac{\sqrt{2}\tau}{\sqrt{1 - \tau^2}\sqrt{u^2 + 2}}\right)}
{(u^2 + 1)\sqrt{u^2 + 2}}
\right)du.
\end{gather*}
But we have already seen the values of these integrals:
$$
\int_0^1 \frac{du}{u^2 + 1}
= \frac{\pi}{4}, \ \
\int_0^1 \frac{du}{(u^2 + 1)\sqrt{u^2 + 2}}
= \frac{\pi}{6}, \ \
\int_0^1 \frac{\tan^{-1}\left(\frac{1}{\sqrt{u^2 + 2}}\right)}
{(u^2 + 1)\sqrt{u^2 + 2}}\,du
= \frac{\pi^2}{32}.
$$
Hence, taking $\tau = 1$ and $\tau = \frac{1}{\sqrt{3}}$ respectively:
$$
\int_0^1
\frac{\tan^{-1}\sqrt{\frac{1 - t^2}{2}}}{1 + t^2}\,dt
= \frac{\pi}{2}\cdot\frac{\pi}{4} - \frac{\pi}{2}\cdot\frac{\pi}{6}
= \frac{\pi^2}{24},
$$
and:
$$
\int_0^{\frac{1}{\sqrt{3}}}
\frac{\tan^{-1}\sqrt{\frac{1 - t^2}{2}}}{1 + t^2}\,dt
= \frac{\pi}{4}\cdot\frac{\pi}{4} - \frac{\pi^2}{32}
= \frac{\pi^2}{32}.
$$
The familiar trigonometrical identities:
$$
\tan\left(\theta - \frac{\pi}{2}\right) = -\frac{1}{\tan\theta}, \ \
\tan3\theta = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}
$$
combine to give:
$$
\tan^{-1}\left(\frac{\sqrt{2}(1-3t^2)}{(5+t^2)\sqrt{1-t^2}}\right)
= 3\tan^{-1}\sqrt{\frac{1-t^2}{2}} - \frac{\pi}{2}.
$$
Therefore:
\begin{align*}
\int_0^1
\frac{\tan^{-1}
\left(\frac{\sqrt{2}(1-3t^2)}{(5+t^2)\sqrt{1-t^2}}\right)}
{1 + t^2}\,dt
& = 3\cdot\frac{\pi^2}{24} - \frac{\pi}{2}\cdot\frac{\pi}{4}
= 0, \\
\int_0^{\frac{1}{\sqrt{3}}}
\frac{\tan^{-1}
\left(\frac{\sqrt{2}(1-3t^2)}{(5+t^2)\sqrt{1-t^2}}\right)}
{1 + t^2}\,dt
& = 3\cdot\frac{\pi^2}{32} - \frac{\pi}{2}\cdot\frac{\pi}{6}
= \frac{\pi^2}{96}.
\end{align*}
Finally, making the change of variables $t = \tan\phi$, and
subtracting these two integrals, we get:
\begin{align*}
\int_{0}^{\frac{\pi}{4}}\tan^{-1}\!\!
\left(\frac{\sqrt{2}\cos3 \phi}
{\left(2\cos 2 \phi+ 3\right)\sqrt{\cos 2 \phi}}\right)d\phi
& = 0, \\
\int_{0}^{\frac{\pi}{6}}\tan^{-1}\!\!
\left(\frac{\sqrt{2}\cos3 \phi}
{\left(2\cos 2 \phi+ 3\right)\sqrt{\cos 2 \phi}}\right)d\phi
& = \frac{\pi^2}{96}, \\
\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\tan^{-1}\!\!
\left(\frac{\sqrt{2}\cos3 \phi}
{\left(2\cos 2 \phi+ 3\right)\sqrt{\cos 2 \phi}}\right)d\phi
& = -\frac{\pi^2}{96}
\end{align*}
