<s>[INST] Evaluate $\int_{-\frac{1}{\sqrt 3}}^{\frac{1}{\sqrt 3}} \frac{x^4}{1-x^4} \cos^{-1} (\frac{2x}{1+x^2}) dx$ Note that the integral can(not) be simplified as
$$
2\int_0^{1/\sqrt 3} \frac{x^4}{1-x^4} \cos^{-1} \left(\frac{2x}{1+x^2}\right)dx
$$
Since $\cos^{-1}$ is not an even function. Let $x=\tan y$
$$
\implies
\int_0^{\pi/6}\frac{x^4}{1-x^4} \left(\frac{\pi}{2} -2y\right) \sec^2(y)\ dy
$$
So how do I solve the original problem?
 [/INST] Hint:
$$\int_{-a}^a\dfrac{x^4}{1-x^4}\cos^{-1}\dfrac{2x}{1+x^2}\ dx$$
$$=\int_{-a}^0\dfrac{x^4}{1-x^4}\cos^{-1}\dfrac{2x}{1+x^2}\ dx+\int_0^a\dfrac{x^4}{1-x^4}\cos^{-1}\dfrac{2x}{1+x^2}dx=I_1+I_2$$
For the first integral set $x=-y,dx=-dy$
$$I_1=\int_{-a}^0\dfrac{x^4}{1-x^4}\cos^{-1}\dfrac{2x}{1+x^2}\ dx=-\int_a^0\dfrac{y^4}{1-y^4}\cos^{-1}\dfrac{(-2y)}{1+y^2}dy$$
$$=\int_0^a\dfrac{y^4}{1-y^4}\cos^{-1}\dfrac{(-2y)}{1+y^2}dy$$
Using How do I prove that $\arccos(x) + \arccos(-x)=\pi$ when $x \in [-1,1]$?,
$$I_1=\int_0^a\dfrac{y^4}{1-y^4}\left(\pi-\cos^{-1}\dfrac{2y}{1+y^2}\right)dy=\pi\int_0^a\dfrac{y^4}{1-y^4}dy-I_2$$
Hope you can take it from here!
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