Consider
$$\int _\frac{ -1}{\sqrt{3}} ^ \frac {1}{\sqrt {3}}\frac {x^4}{1-x^4}\arccos \left(\frac{2x}{1+x^2}\right) \mathrm{d}x.$$
Every condition like the limits, the value which arccosine is taking is making me put $x=\tan (t) $. Thus, on simplifying we get $$\frac {\sin^4 (t)}{(1-\sin^2 (t))(1-2\sin^2 (t)}\arccos (\sin(2t)),$$ but from there I don't know what to do. Another approach was writing the first part as $-1+\frac {1 }{2}\left(\frac {1}{1-x^2}+\frac {1}{1+x^2}\right) $ but still I have no hope that this helps . Thus I am hoping for a better approach . Thanks!
Hint:
$$\int_a^bf(x) \ dx=\int_a^bf(a+b-x)\ dx$$
$\arccos x+\arccos(-x)=\pi$