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.

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 $$ =\frac{24}{6^5}  \left(\frac{-\psi^4(1/6)}{24} - \frac{-\psi^4(1/3)}{24} \right) $$ 
$$=\frac{1}{6^5}  \left({\psi^4(1/3)} - {\psi^4(1/6)} \right) $$
$$= {\frac{16{{\pi}^{5}}}{243\sqrt[]{{3}}}+\frac{605\zeta(5)}{54}} $$
