<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] Consider the function $f(x)=\frac12(3x^2-1)$ for $x\in(-1,1)$ and its periodic extension with period $2$. Since $f(x)=f(-x)$ we can write
$$f(x)=\sum_{k=0}^{\infty}a_n\cos n\pi x$$
$$\int_{-1}^1\frac12(3x^2-1)dx=0=2a_0$$
$$\begin{align}\int_{-1}^1\frac12(3x^2-1)\cos n\pi x\,dx&=\left[\frac1{n\pi}\frac12(3x^2-1)\sin n\pi x+\frac{3x}{n^2\pi^2}\cos n\pi x-\frac3{n^3\pi^3}\sin n\pi x\right]_{-1}^1\\
&=\frac{6(-1)^n}{n^2\pi^2}=a_n\end{align}$$
So
$$f(x)=\frac6{\pi^2}\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos n\pi x$$
Note that $\cos\left(n\pi\left(1-\frac15\right)\right)=(-1)^n\cos\frac{n\pi}5$. So
$$\begin{align}f\left(\frac45\right)&=\frac{23}{50}=\frac6{\pi^2}\sum_{n=1}^{\infty}\frac1{n^2}\cos\frac{n\pi}5\\
&=\frac6{\pi^2}\left(\frac{\sigma}2+\sum_{n=1}^{\infty}\frac1{25n^2}(-1)^n\right)\\
&=\frac6{\pi^2}\left(\frac{\sigma}2+\frac1{25}\left(-\frac12\right)\frac{\pi^2}6\right)\end{align}$$
Here $\sigma$ is the topical sum, where the coefficient of $1/n^2$ is $2\cos n\pi/5$ except where $n=5k$ where the $\cos 5k\pi/5=(-1)^k$ has been omitted. Thus
$$\sigma=2\frac{\pi^2}6\left(\frac{23}{50}+\frac1{50}\right)=\frac{4\pi^2}{25}$$
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