<s>[INST] How to compute $\int_0^1\frac{\ln(x)}{1+x^5}dx$? 
Let $\phi$ denote the golden ratio $\phi=\frac{1+\sqrt5}{2}$.
  How can I prove this sum?
  $$\sum_{n=0}^{\infty}(-1)^n\left[\frac{\phi}{(5n+1)^2}+\frac{\phi^{-1}}{(5n+2)^2}-\frac{\phi^{-1}}{(5n+3)^2}-\frac{\phi}{(5n+4)^2}\right]=\left(\frac{2\pi}{5}\right)^2$$

My try:
Change  the sums into an integrals:
$\sum_{n=0}^{\infty}\frac{(-1)^n}{(5n+1)^2}=\int_0^1\frac{-\ln(x)}{1+x^5}dx$ 
Can somebody give a hint how on to integrate this integral.
Try substitution by letting $u=\ln(x)$ is not working and integration by part is making it more complicated than before. What kind of substitution should I be using?
 [/INST] \begin{align*}
\sum_{n=0}^{\infty}(-1)^{n}\left[\frac{\phi}{(5n+1)^{2}}-\frac{\phi}{(5n+4)^{2}}\right] & =\frac{2}{125}\left(5+2\sqrt{5}\right)\pi^{2}\\
\sum_{n=0}^{\infty}(-1)^{n}\left[\frac{\phi^{-1}}{(5n+2)^{2}}-\frac{\phi^{-1}}{(5n+3)^{2}}\right] & =\frac{2}{125}\left(5-2\sqrt{5}\right)\pi^{2}
\end{align*}
</s>