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Prove that: $$\int_{0}^{1}\frac{\tan^{-1}\sqrt{2+x^2}}{\left ( 1+x^2 \right )\sqrt{x^2+2}}\,dx=\frac{5\pi^2}{96}$$

Let us begin by the identity

$$\arctan x + \arctan \frac{1}{x} = \frac{\pi}{2} \quad \text{forall} \; x>0$$

Thus

\begin{align*} \mathcal{J} &= \int_{0}^{1} \frac{\arctan \sqrt{2+x^2}}{\left ( 1+x^2 \right ) \sqrt{2+x^2}} \, {\rm d}x \\
&= \frac{\pi}{2} \int_{0}^{1} \frac{{\rm d}x}{\left ( 1+x^2 \right ) \sqrt{2+x^2}} - \bigintsss_{0}^{1} \frac{\arctan \left ( \frac{1}{\sqrt{2+x^2}} \right )}{\left ( 1+x^2 \right )\sqrt{2+x^2}} \, {\rm d}x \\
&=\frac{\pi^2}{12} - \bigintsss_{0}^{1} \frac{\arctan \left ( \frac{1}{\sqrt{2+x^2}} \right )}{\left ( 1+x^2 \right )\sqrt{2+x^2}} \, {\rm d}x
\end{align*}

Now one of the $\arctan$ ‘s definition is

$$\arctan \frac{1}{a} = \int_{0}^{1} \frac{a}{x^2+a^2} \, {\rm d}x \Leftrightarrow \frac{1}{a} \arctan \frac{1}{a} = \int_{0}^{1} \frac{{\rm d}x}{x^2+a^2}$$

Hence

\begin{align*} \bigintsss_{0}^{1} \frac{\arctan \left ( \frac{1}{\sqrt{2+x^2}} \right )}{\left ( 1+x^2 \right )\sqrt{2+x^2}} \, {\rm d}x &= \int_{0}^{1} \int_{0}^{1} \frac{{\rm d}(x, y)}{\left ( 1+x^2 \right )\left ( 2+x^2+y^2 \right )} \\
&=\int_{0}^{1} \int_{0}^{1} \frac{1}{y^2+1} \bigg( \frac{1}{1+x^2} - \frac{1}{2+x^2+y^2} \bigg )\, {\rm d}(x, y) \\\\
&=\int_{0}^{1} \int_{0}^{1} \frac{{\rm d}(x, y)}{\left ( 1+x^2 \right ) \left ( 1+y^2 \right )} -\int_{0}^{1} \int_{0}^{1} \frac{{\rm d}(x, y)}{2+x^2+y^2} \\\\
&= \frac{1}{2} \int_{0}^{1} \int_{0}^{1} \frac{{\rm d}(x, y)}{\left ( 1+x^2 \right )\left ( 1+y^2 \right )} \\
&= \frac{1}{2}\left ( \int_{0}^{1} \frac{{\rm d}x}{1+x^2} \right )^2 \\
& = \frac{\pi^2}{32}
\end{align*}

and the result follows.