Evaluating $ \frac{24}{\pi}\int_0^\sqrt 2\frac{2-x^2}{(2+x^2)\sqrt{4+x^4}}\,\mathrm dx$ I recently came across a problem on definite integration and couldn't solve it despite my efforts. It goes as
$$
\frac{24}{\pi}\int_0^\sqrt 2\frac{2-x^2}{(2+x^2)\sqrt{4+x^4}}\,\mathrm dx
$$
The $2-x^2$ term and the upper limit along with the $2+x^2$ term at the bottom motivated me to substitute $x=2^{1/2}\tan a$. However subsequent steps proceeded to a stage I couldnt simplify.
I tried other substitutions but was unable to proceed
Any help to solve this problem would be appreciated. Please feel free to share your first thoughts and intuition as well.

Integrate as follows
\begin{align}&\ \frac{24}{\pi}\int_0^\sqrt 2\frac{2-x^2}{(2+x^2)\sqrt{4+x^4}}\,dx\\
=&\ \frac{24}{\pi}\int_0^\sqrt2\frac{\frac{2-x^2}{(2+x^2)^2}}{\sqrt{1-\frac{4x^2}{(2+x^2)^2}}}\,
\overset{y=\frac{2x}{2+x^2}}{dx}
=\frac{12}{\pi}\int_0^{\frac1{\sqrt2}}\frac{dy}{\sqrt{1-y^2}}
 = \frac{12}{\pi}\cdot\frac\pi{4}=3\\
\end{align}
