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?

Let's define the sum we are looking for with $S$.
Furthermore, notice that ($k \in [1,4] $)
$$
S_k=\sum_{n=0}^{\infty}\frac{(-1)^n}{(5n+k)^2}=\sum_{n=0}^{\infty}\frac{1}{(5(2n)+k)^2}-\sum_{n=0}^{\infty}\frac{1}{(5(2n+1)+k)^2}=\\
\frac{1}{5^22^2}\left[\sum_{n=0}^{\infty}\frac{1}{(n+k/10)^2}-\sum_{n=0}^{\infty}\frac{1}{(n+(k+5)/10)^2}\right]=\\
\frac{1}{5^22^2}\left[\psi^{(1)}(k/10)-\psi^{(1)}((k+5)/10)\right]
$$
where we used the definition of the Trigamma function
Taking now for example $\Delta_{14}=S_1-S_4$
$$
\Delta_{14} =\frac{1}{5^22^2}\left[\psi^{(1)}(1/10)-\psi^{(1)}(6/10)\right]-\frac{1}{5^22^2}\left[\psi^{(1)}(4/10)-\psi^{(1)}(9/10)\right]=
\frac{1}{5^22^2}(\psi^{(1)}(1/10)+\psi^{(1)}(9/10))-\frac{1}{5^22^2}(\psi^{(1)}(4/10)+\psi^{(1)}(6/10))
$$
Trigamma reflection $\psi^{(1)}(1-z)+\psi^{(1)}(z)=\frac{\pi^2}{\sin^2(\pi z)}$ yields the massive simplification
$$
\Delta_{14}=\frac{\pi^2}{100}\left(\frac{1}{\sin^2(\pi/10)}-\frac{1}{\sin^2(4\pi/ 10)}\right)=\\
\frac{\pi^2}{125}(5+3\sqrt{5})
$$
Playing the same game with $\Delta_{23}=S_2-S_3$ we obtain
$$
\Delta_{23}=\frac{\pi^2}{125}(-5+3\sqrt{5})
$$
Doing the algebra yields

$$
S=\phi\Delta_{14}+\frac{\Delta_{23}}{\phi}=\frac{4 \pi^2}{25}\\
\bf{QED}
$$

