Prove that $\int_0^1\frac{1-x}{1-x^6}\ln^4x\,dx=\frac{16\sqrt{3}}{729}\pi^5+\frac{605}{54}\zeta(5)$ This integral comes from a well-known site (I am sorry, the site is classified due to regarding the OP.)

$$\int_0^1\frac{1-x}{1-x^6}\ln^4x\,dx$$

I can calculate the integral using the help of geometric series and I get the answer
\begin{align}
\sum_{n=0}^\infty\left(\frac{24}{(6n+1)^5}-\frac{24}{(6n+2)^5}\right)
&=\frac{1}{6^5}\left(\Psi^{(4)}\left(\frac{1}{3}\right)-\Psi^{(4)}\left(\frac{1}{6}\right)\right)\\
&=\frac{16\sqrt{3}}{729}\pi^5+\frac{605}{54}\zeta(5)
\end{align}
To be honest, I use Wolfram Alpha to calculate the sum of series. The problem is I don't think this is the correct way to calculate the integral because I use a machine to help me. I tried another way, I used partial fraction to decompose the integrand as
$$\frac{\ln^4x}{3(x+1)}+\frac{\ln^4x}{2(x^2+x+1)}-\frac{2x-1}{6(x^2-x+1)}\ln^4x$$
but none of them seemed easy to calculate. Could anyone here please help me to calculate the integral preferably (if possible) with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.

You can use the geometric series to evaluate. In fact,
\begin{eqnarray}
I&=&\int_0^1\frac{(1-x)\ln^4x}{1-x^6}dx\\
&=&\int_0^1\sum_{n=0}^\infty(1-x)x^{6n}\ln^4xdx\\
&=&\sum_{n=0}^\infty(1-x)x^{6n}\ln^4xdx\\
&=&\sum_{n=0}^\infty\left(\frac{24}{(6n+1)^5}-\frac{24}{(6n+2)^5}\right)\\
&=&\frac{1}{324}\left(\sum_{n=0}^\infty\frac{1}{(n+1/6)^5}-\sum_{n=0}^\infty\frac{1}{(n+1/3)^5}\right)\\
&=&\frac{1}{324}(\zeta(5,\frac{1}{6})-\zeta(5,\frac{1}{3}))\\
&=&\frac{16\pi^5}{243\sqrt3}+\frac{605}{54}\zeta(5).
\end{eqnarray}
The (tedious) calculation of $\zeta(5,\frac{1}{6})$ and $\zeta(5,\frac{1}{3})$ is derived from two results from this (in Pages 1628-1629). First we have
$$ \zeta(5,\frac{1}{6})-\zeta(5,\frac{5}{6})=\frac{44\pi^5}{\sqrt3}, \zeta(5,\frac{1}{6})+\zeta(5,\frac{5}{6})=(2^5-1)(2^5-1)\zeta(5), $$
from which we obtain
$$ \zeta(5,\frac{1}{6})=\frac{22\pi^5}{\sqrt3}+3751\zeta(5). $$
Since
$$ \zeta(5,\frac{1}{6})+\zeta(5,\frac{1}{3})+\zeta(5,\frac{1}{2})+\zeta(5,\frac{2}{3})+\zeta(5,\frac{5}{6})+\zeta(5)=6^5\zeta(5),\zeta(5,\frac{1}{2})=31\zeta(5) $$
we have
$$ \zeta(5,\frac{1}{3})+\zeta(5,\frac{2}{3})=242\zeta(5). $$
Also
$$ \zeta(5,\frac{1}{3})-\zeta(5,\frac{2}{3})=\sum_{n=-\infty}^\infty\frac1{(n+\frac{1}{3})^5}=\frac{4\pi^5}{3\sqrt3}.$$
From this, we obtain
$$\zeta(5,\frac{1}{3})=\frac{2\pi^5}{3\sqrt3}+121\zeta(5). $$
