Hard Definite integral involving the Zeta function Prove that: $$\displaystyle \int_{0}^{1}\frac{1-x}{1-x^{6}}{\ln^4{x}} \ {dx} = \frac{16{{\pi}^{5}}}{243\sqrt[]{{3}}}+\frac{605\zeta(5)}{54} $$
I was able to simplify it a bit by substituting ${y = -\ln{x}}$ and some further mathematical manipulation but was not able to get the correct form.
 A: Note that $\frac{1-x}{1-x^6}=\sum_{k=0}^\infty (x^{6k}-x^{6k+1})$. And the integration $\int_0^1 x^n \ln{x}^4=\partial_n^4 \int_0^1 x^n dx=\frac{24}{(n+1)^5}$. We have
$$LHS = 24\sum_{k=0}^\infty \left(\frac{1}{(6k+1)^5}-\frac{1}{(6k+2)^5}\right)$$
Use the discrete Fourier, and denote $\xi=\exp(i\frac{\pi}{3})$, $\xi_i =\xi^i$. Then
$$LHS=24\sum_{i=1}^6a_i\sum_{k=1}^\infty \frac{\xi_i^k}{k^5}=24\sum_{i=1}^6 a_i Li_5(\xi_i).$$
where $a_1=a_5=\frac{1}{6}$,$a_2=-\frac{i}{2\sqrt{3}}=-a_4$,$a_6=0$,$a_3=-\frac{1}{3}$.Thus we use the summation formula for polylogarithm
$$Li_5(\xi_2)+(-1)^5 Li_5(\xi_4)=-\frac{(2\pi i)^5}{5!}B_5(\frac{1}{3})=\frac{4i\pi^5}{729}.$$
$B$ is the Bernoulli polynomial. Also, $$Li_5(\xi_1)+Li_5(\xi_5)=(1-\frac{2}{2^5}-\frac{3}{3^5}+\frac{6}{6^5})\sum_{k=1}^\infty\frac{1}{k^5}=\frac{25}{27}\sum_{k=1}^\infty\frac{1}{k^5}.$$
Also note that $L_5(-1)=-(1-\frac{2}{2^5})\zeta(5).$ We conclude that
$$24\sum_{i=1}^6 a_i Li_5(\xi_i)=24\frac{-i}{2\sqrt{3}}\cdot\frac{4i\pi^5}{729}+24(\frac{1}{6}\frac{25}{27}+\frac{1}{3}\cdot\frac{15}{16})\zeta(5)=\frac{16\pi^5}{243\sqrt{3}}+\frac{605}{54}\zeta(5)$$
A: The series
$$ 24\sum_{k=0}^\infty \left(\frac{1}{(6k+1)^5}-\frac{1}{(6k+2)^5}\right)$$
$$ =\frac{24}{6^5}\sum_{k=0}^\infty \left(\frac{1}{(k+1/6)^5}-\frac{1}{(k+2/6)^5}\right)$$ 
Can be evaluated by the polygamma function $$ =\frac{24}{6^5}  \left(\frac{-\psi^4(1/6)}{24} - \frac{-\psi^4(1/3)}{24} \right) $$ 
$$=\frac{1}{6^5}  \left({\psi^4(1/3)} - {\psi^4(1/6)} \right) $$
$$= {\frac{16{{\pi}^{5}}}{243\sqrt[]{{3}}}+\frac{605\zeta(5)}{54}} $$
