How to find this integral $\int_{0}^{1}\frac{x}{1-x^4}\arctan{\frac{x-x^5}{1+x^6}}\,dx$

Question

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 $$ I=\int_{0}^{1}\left[{x \over 1 - x^{4}}\, \arctan\left(x - x^{5} \over 1 + x^{6}\right)\right]\,{\rm d}x ={\pi \over 8}\, \left[\left(1 + \,\sqrt{\,5\,}\, \over 2\right)^{3} - {\ln\left(5\right) \over 2}\right] $$

Answer 1

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.

1. 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*}

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

3. 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.