<s>[INST] 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.
 [/INST] 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}
</s>