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.
 A: After the first step from @ClaudeLeibovici’s answer, you can do a crazy u-sub: $$u=\frac{x^2-1}{x^2+1}$$ so $$x=\sqrt{\frac{1+u}{1-u}}$$ By chain rule, $$\implies dx=\frac{1}{2\sqrt{\frac{1+u}{1-u}}}\cdot\frac{2}{(1-u)^2}du$$$$=\frac{du}{(1+u)^\tfrac12(1-u)^\tfrac32}$$ Also, $$\frac{1}{\sqrt{1+x^4}}= \frac{1}{\sqrt{1+\left(\frac{1+u}{1-u}\right)^2}}=\frac{1-u}{\sqrt{2(1+u^2)}}$$ so the integral simplifies to $$-\int u\cdot\left(\frac{1-u}{\sqrt{2(1+u^2)}}\right)\frac{du}{(1+u)^\tfrac12(1-u)^\tfrac32}$$$$=-\int\frac{udu}{\sqrt{2(1+u^2)} (1+u)^\tfrac12(1-u)^\tfrac12}$$$$=-\int\frac{udu}{\sqrt{2(1-u^4)}}=-\frac{1}{2\sqrt2}\arcsin(u^2)+C$$ Putting appropriate limits, you get $\displaystyle\frac{\pi}{4\sqrt2}$ so the final answer should be $3$.
A: 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}
A: First, getting rid of all these $2$'s,
$$I=\frac{24}{\pi}\int_0^\sqrt 2\frac{2-x^2}{(2+x^2)\sqrt{4+x^4}}\,dx=\frac{24}{\pi\sqrt 2}\int_0^1\frac{1-t^2}{\left(t^2+1\right) \sqrt{t^4+1}}\,dt$$ Now, I am afraid that whatever you could do, you will face elliptic integrals.
Using a CAS
$$\int\frac{1-t^2}{\left(t^2+1\right) \sqrt{t^4+1}}\,dt=\sqrt[4]{-1} \left(F\left(\left.i \sinh ^{-1}\left(\sqrt[4]{-1}
   t\right)\right|-1\right)-2 \Pi \left(-i;\left.i \sinh ^{-1}\left(\sqrt[4]{-1}
   t\right)\right|-1\right)\right)$$ Using the bounds
$$\int_0^1\frac{1-t^2}{\left(t^2+1\right) \sqrt{t^4+1}}\,dt=\frac {1+i}{\sqrt 2}\left(F\left(\left.i \sinh ^{-1}\left(\sqrt[4]{-1}\right)\right|-1\right)-2 \Pi
   \left(-i;\left.i \sinh ^{-1}\left(\sqrt[4]{-1}\right)\right|-1\right)\right)$$
Computed with high accuracy, this is
$$K=0.55536036726979578087698512375758671232682771117196,\cdots$$ which, by luck, is identified by the $ISC$ as the solution of
$$8K-\pi \sqrt 2=0$$ (have a look here, clicking  the  Integer Relation Algorithms button on the front page).
Just finish the calculation to find a very complicated result.
A: A linear fractional transformation
$$x = \frac{\sqrt{2}u}{u-1}$$
transforms the integral into an integral an algebraic function from $-\infty$ to $\infty$, and then you can use the residue theorem.
