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?
 A: 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
$$\sigma=2\frac{\pi^2}6\left(\frac{23}{50}+\frac1{50}\right)=\frac{4\pi^2}{25}$$
A: 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

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

A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
\int_{0}^{1}{\ln\pars{x} \over 1 + x^{5}}\,\dd x &=
\int_{0}^{1}\ln\pars{x}\pars{{1 \over 1 + x^{5}} + {1 \over 1 - x^{5}}}\,\dd x -
\int_{0}^{1}{\ln\pars{x} \over 1 - x^{5}}\,\dd x
\\[3mm] & =
2\int_{0}^{1}{\ln\pars{x} \over 1 - x^{10}}\,\dd x -
\int_{0}^{1}{\ln\pars{x} \over 1 - x^{5}}\,\dd x
\\[3mm] & =
{1 \over 50}\int_{0}^{1}{\ln\pars{x} \over 1 - x}\,x^{-9/10}\,\dd x -
{1 \over 25}\int_{0}^{1}{\ln\pars{x} \over 1 - x}\,x^{-4/5}\,\dd x
\\[3mm] & =
-\,{1 \over 50}\lim_{\mu \to -9/10}\totald{}{\mu}
\overbrace{\int_{0}^{1}{1 - t^{\mu} \over 1 - x}\,\dd x}
^{\ds{\Psi\pars{\mu + 1} + \gamma}}\ +\
{1 \over 25}\lim_{\mu \to -4/5}\totald{}{\mu}
\overbrace{\int_{0}^{1}{1 - t^{\mu} \over 1 - x}\,\dd x}^{\ds{\Psi\pars{\mu + 1} + \gamma}}
\\[3mm] & =
\fbox{$\ds{{1 \over 50}\bracks{%
           2\Psi\,'\pars{{1 \over 5}} - \Psi\,'\pars{{1 \over 10}}}}$}
\approx -0.9780
\end{align}
$\Psi\pars{z}$ is the digamma function and $\gamma$ is the Euler-Mascheroni constant. The final integral which involves the $\Psi$ function is a 'standard' identity and it appears in, for example, Abramowitz and Stegun table. 
A: \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*}
