TITLE: How to find $\int_{0}^{1}\dfrac{\ln^2{x}\ln^2{(1-x)}}{2-x}dx$
QUESTION [12 upvotes]: How to find 

$$
I=\int_{0}^{1}{\ln^{2}\left(x\right)\ln^{2}\left(1 - x\right) \over 2 - x}
\,{\rm d}x
$$

My idea: 
Let $x=1-t$, then

$$
I
=\int_{0}^{1}{\ln^{2}\left(1 - x\right)\ln^{2}\left(x\right) \over 1 + x}\,{\rm d}x
$$

I can't proceed any further. I can't remember that Math SE has this similar problem already posted.
this integral problem is from This Euler Sums  :
$$\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2}}}} \left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right)\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{{{k^2}}}} } \right) =  - \frac{{23}}{{1440}}{\pi ^4}\ln 2 + \frac{1}{{18}}{\pi ^2}{\ln ^3}2 - \frac{1}{{18}}{\ln ^5}2 - 4L{i_4}\left( {\frac{1}{2}} \right)\ln 2 - 8L{i_5}\left( {\frac{1}{2}} \right) + \frac{1}{{12}}{\pi ^2}\zeta \left( 3 \right) + \frac{{27}}{4}\zeta \left( 5 \right)$$
when I deal this series,then we must solve this integral .
Thank you.

REPLY [4 votes]: Here a closer form for this integral
$$
I=−\frac{13}{2}\zeta_2\zeta_3 +\frac{2}{3}\zeta_2\ln^3 2   +\frac{15}{2}\zeta_5+\zeta_4\ln2-\frac{1}{15}\ln^5 2+8\operatorname{Li}_5\left(\frac{1}{2}\right)
$$