How to find this integral $\int_{0}^{1}\frac{x}{1-x^4}\arctan{\frac{x-x^5}{1+x^6}}\,dx$ Find the integral value
$$
I=\int_{0}^{1}
{x \over 1 - x^{4}}\,\arctan\left(x - x^{5} \over 1 + x^{6}\right)\,{\rm d}x
$$ 
My good friends gave me this problem, and I can't solve it. Using computer I found closed form 
$$
I=\int_{0}^{1}\left[{x \over 1 - x^{4}}\,
\arctan\left(x - x^{5} \over 1 + x^{6}\right)\right]\,{\rm d}x
={\pi \over 8}\,
\left[\left(1 + \,\sqrt{\,5\,}\, \over 2\right)^{3} - {\ln\left(5\right) \over 2}\right]
$$
 A: Using Lucian's suggestion, rewrite the integral as
$$\mathcal{I}=\int_0^1\frac{x\left(\arctan x-\arctan x^5\right)dx}{1-x^4}dx.$$
Its calculation using complex-analytic methods is essentially one of my very first posts on MSE.
I briefly sketch the idea so that your question could be considered as answered. For more details, see the link above.


*

*Integrate by parts using that $\int\frac{xdx}{1-x^4}=\frac14\ln\frac{1+x^2}{1-x^2}$ and rewrite $\mathcal{I}$ as
\begin{align*}
\mathcal{I}
=\frac14\int_0^1\frac{(1+x^2)(1-3x^2+x^4)}{x^8-x^6+x^4-x^2+1}\ln\left(\frac{1-x^2}{1+x^2}\right)dx.
\end{align*}

*Use the symmetry w.r.t. the change of variable $x\rightarrow\frac1x$ and parity to further rewrite this as
$$\mathcal{I}
=\Re\int_{\mathbb{R}+i0} f(z)\,dz,\qquad f(z)=\frac{1}{16}\frac{(1+z^2)(1-3z^2+z^4)}{z^8-z^6+z^4-z^2+1}\ln\left(\frac{1-z^2}{1+z^2}\right).$$

*Pull the contour of integration to $i\infty$. The residues at simple poles are easily computable and the $2\pi$-jump on the logarithmic branch cut $[i,i\infty)$ produces an integral of a rational function.
