Prove that $\displaystyle\int_{0}^{\frac{1}{\sqrt{3}}}\frac{\arctan\left(\frac{1}{\sqrt{1-2x^2}}\right)}{1+x^2} \mathrm{dx}=\frac{13\pi^2}{288}$ I need to prove that,
$\displaystyle \tag*{} \int_{0}^{\frac{1}{\sqrt{3}}}\frac{\arctan\left(\frac{1}{\sqrt{1-2x^2}}\right)}{1+x^2} \mathrm{dx}=\frac{13\pi^2}{288}$
Here is what I tried:
I tried to solve the integral by Integrating by parts.
We have:
$\displaystyle \tag{1} \int f(x) g'(x) \mathrm{ dx} = f(x)g(x) - \int f'(x) g(x) \mathrm{ dx}$
I noticed ,
$\displaystyle \tag*{} \dfrac{1}{1+x^2} = \arctan '(x)$
$\displaystyle \tag*{} g'(x) = \dfrac{1}{1+x^2} \Leftrightarrow g(x) = \arctan(x)$
Now, I defined $f(x)$
$\displaystyle \tag*{} f(x) = \arctan \left( \dfrac{1}{\sqrt{1-2x^2}}\right )$
and
$\displaystyle \tag*{} f'(x) = \arctan'\left(\dfrac{1}{\sqrt{1-2x^2}} \right ) = \text{arccot}'(\sqrt{1-2x^2}) = \dfrac{x}{\sqrt{1-2x^2}(1-x^2)}$
Now, using $(1)$ and substituting the values of $f(x)$ and $g(x)$, My indefinite integral becomes:
$\displaystyle \tag*{} \arctan(x) \arctan \left (\dfrac{1}{\sqrt{1-2x^2}} \right ) - \int  \dfrac{x \arctan(x)}{\sqrt{1-2x^2}(1-x^2)} \mathrm{ dx}$
Now, we want to evaluate:
$\displaystyle \tag*{} \int \dfrac{x \arctan(x)}{\sqrt{1-2x^2}(1-x^2)} \mathrm{ dx}$
Now this is integral I am having trouble solving. Any hints or different methods would be greatly appreciated. Thank you.

Integrate as follows
\begin{align}
&\int_{0}^{\frac{1}{\sqrt{3}}}\frac{\tan^{-1}\frac{1}{\sqrt{1-2x^2}}}{1+x^2} {dx}\\
\overset{ibp}=&\ \frac{\pi^2}{18}-\int_{0}^{\frac{1}{\sqrt{3}}}\frac{x\tan^{-1} x}{(1-x^2)\sqrt{1-2x^2}}dx 
\>\>\>\>\>\>\>\sqrt2 x=\cos t\\
=& \ \frac{\pi^2}{18} -\int_{\cot^{-1}\sqrt2}^{\pi/2}
\frac{\tan^{-1}(\cos t/\sqrt2)\cos t}{2-\cos^2 t}dt\\
 =& \ \frac{\pi^2}{18} -\int_{\cot^{-1}\sqrt2}^{\pi/2}\int_0^1
\frac{\sqrt2\cos^2t}{(2-\cos^2 t)(2+y^2\cos^2t)}dy\ dt\\
 =& \ \frac{\pi^2}{18} -\int_0^1\frac1{1+y^2}\bigg(\frac\pi4-\frac\pi{2\sqrt{2+y^2}}+\frac{\cot^{-1} \sqrt{2+y^2}}{\sqrt{2+y^2}}\bigg)dy\\
=& \ \frac{11\pi^2}{144}- \int_0^1 \frac{\cot^{-1} \sqrt{2+y^2}}{\sqrt{2+y^2}}dy 
=\frac{11\pi^2}{144}-\frac{\pi^2}{32}= \frac{13\pi^2}{288}
\end{align}
where
\begin{align}
I=&\int_0^1 \frac{\cot^{-1} \sqrt{2+y^2}}{\sqrt{2+y^2}}dy 
=\int_0^1 \int_0^1 \frac1{(1+y^2)(y^2+x^2+2)}dx \ dy\\
=& \int_0^1 \int_0^1 \frac1{1+x^2}\bigg(\frac1{1+y^2}-\frac1{y^2+x^2+2}\bigg)dy \ dx
=\frac{\pi^2}{16}-I=\frac{\pi^2}{32}
\end{align}
