How to prove $\int_{0}^{1}{ dx\over 1+x^2}\cdot{\tan^{-1}\left(\sqrt{x^2+2}{x+1\over x^2-x+2}\right)\over \sqrt{x^2+2}}={13\pi^2\over 288}?$ Variation of Ahmed's integral

$$\int_{0}^{1}{\mathrm dx\over 1+x^2}\cdot{\tan^{-1}\left(\sqrt{x^2+2}\cdot{x+1\over x^2-x+2}\right)\over \sqrt{x^2+2}}={13\pi^2\over 288}\tag1$$

Making an attempt:
I can't think of what to do. It is too complicate.
 A: One may observe that, for $x \in [0,1]$,
$$
\arctan\left(\sqrt{x^2+2}\cdot{x+1\over x^2-x+2}\right)=\arctan\left({1\over \sqrt{x^2+2}}\right)+\arctan\left({x\over \sqrt{x^2+2}}\right)
$$ then
$$
I:=\int_0^1 \frac{1}{(1+x^2)\sqrt{x^2+2}}\arctan\left(\sqrt{x^2+2}\cdot{x+1\over x^2-x+2}\right)dx=I_1+I_2
$$with
$$
\begin{align}
I_1:&=\int_0^1 \frac{1}{(1+x^2)\sqrt{x^2+2}}\arctan\left({1\over \sqrt{x^2+2}}\right)dx
\\&=\int_0^1 \frac{1}{(1+x^2)\sqrt{x^2+2}}\left(\frac \pi2-\arctan\sqrt{x^2+2}\right)dx
\\&=\frac \pi2\int_0^1 \frac{1}{(1+x^2)\sqrt{x^2+2}}-\int_0^1 \frac{\arctan\sqrt{x^2+2}}{(1+x^2)\sqrt{x^2+2}}\:dx
\\&=\frac{\pi^2}{12}-\frac{5\pi^2}{96}
\\&=\frac{\pi^2}{32}
\end{align}
$$ where the latter integral is Ahmed's evaluation.
On the other hand we have
$$
\begin{align}
I_2:&=\int_0^1 \frac{1}{(1+x^2)\sqrt{x^2+2}}\cdot\arctan\left({x\over \sqrt{x^2+2}}\right)dx
\\&=\int_1^\infty \frac{u}{(1+u^2)\sqrt{1+2u^2}}\cdot\arctan\left({1\over \sqrt{1+2u^2}}\right)du \quad (u=1/x)
\\&=\frac12\int_1^\infty \frac{1}{(1+v)\sqrt{1+2v}}\cdot\arctan\left({1\over \sqrt{1+2v}}\right)dv \qquad (v=u^2)
\\&=-\int_1^\infty \left[\arctan\left({1\over \sqrt{1+2v}}\right)\right]'\arctan\left({1\over \sqrt{1+2v}}\right)dv
\\&=-\left[\frac12\cdot\arctan^2\left({1\over \sqrt{1+2v}}\right)\right]_1^\infty 
\\&=\frac{\pi^2}{72}.
\end{align}
$$
Finally,

$$
\int_0^1 \frac{1}{(1+x^2)\sqrt{x^2+2}}\arctan\left(\sqrt{x^2+2}\cdot{x+1\over x^2-x+2}\right)dx=\frac{\pi^2}{32}+\frac{\pi^2}{72}=\color{blue}{\frac{13\pi^2}{288}}
$$ 

as announced.
