How to find $\int_{0}^{1}\dfrac{\ln^2{x}\ln^2{(1-x)}}{2-x}dx$ 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.
 A: Just a possible approach to solving this. 
Let $1-x=e^{-z}$ Then $$I=\int_{0}^{\infty}\frac{\ln^2(1-e^{-z})z^2}{1+e^{-z}}e^{-z}dz=\sum_{k=0}^{\infty}(-1)^k\int_{0}^{\infty}z^2\ln^2(1-e^{-z})e^{-(k+1)z}dz\\ =\sum_{k=0}^{\infty}(-1)^k I_k$$ where $$I_k=\int_{0}^{\infty}z^2\ln^2(1-e^{-z})e^{-(k+1)z}dz$$ Perhaps these later integrals can be calculated.
$$I_k=\sum_{i,j=1}^\infty \frac{1}{ij}\int_{0}^{\infty}z^2e^{-(k+j+i+1)z}dz\\=\sum_{i,j=1}^\infty \frac{1}{ij}\frac{2}{(k+i+j+1)^3}$$ So, $$I=\sum_{i,j,k=1}^{\infty}(-1)^k\frac{2}{ij(i+j+k)^3}$$
A: Since:
$$[x^n]\log^2(1-x)=\sum_{k=1}^{n}\frac{1}{k(n-k)}=\frac{1}{n}\sum_{k=1}^{n-1}\left(\frac{1}{k}+\frac{1}{n-k}\right)=\frac{2H_{n-1}}{n},\tag{1}$$
we have:
$$ \begin{eqnarray*}I &=& \sum_{n=2}^{+\infty}\frac{2H_{n-1}}{n}\int_{0}^{1}\frac{x^n \log^2 x}{1+x}\,dx = \sum_{n=2}^{+\infty}\frac{2H_{n-1}}{n}\int_{0}^{+\infty}\frac{t^2 e^{-(n+1)t}}{1+e^{-t}}dt\\ &=& 4\sum_{n=2}^{+\infty}\frac{H_{n-1}}{n}\left(\frac{1}{(n+1)^3}-\frac{1}{(n+2)^3}+\frac{1}{(n+3)^3}-\ldots\right)\\&=&4\sum_{s=3}^{+\infty}\frac{(-1)^s}{s^3}\sum_{n=2}^{s-1}(-1)^{n-1}\frac{H_{n-1}}{n},\tag{2}\end{eqnarray*}$$
that is (in disguise) the same identity found by Samrat Mukhopadhyay.
A: Here a closer form for this integral
$$
I=−\frac{13}{2}\zeta_2\zeta_3 +\frac{2}{3}\zeta_2\ln^3 2   +\frac{15}{2}\zeta_5+\zeta_4\ln2-\frac{1}{15}\ln^5 2+8\operatorname{Li}_5\left(\frac{1}{2}\right)
$$
A: Using Mathematica, I found the answer below, which matches numerically.
I wonder if this can be simplified. Mathematica couldn't. Do you have a reason to expect there to be a simpler form?
$$
8 (\zeta (3)-3) \text{Hypergeometric2F1}^{(0,0,1,0)}(1,1,2,-1)+4 \zeta (3) \text{Hypergeometric2F1}^{(0,1,0,0)}(1,1,2,-1)+12 \text{Hypergeometric2F1}^{(0,0,2,0)}(1,1,2,-1)-4 \text{Hypergeometric2F1}^{(0,0,3,0)}(1,1,2,-1)+\text{Hypergeometric2F1}^{(0,0,4,0)}(1,1,2,-1)-12 \text{Hypergeometric2F1}^{(0,1,0,0)}(1,1,2,-1)+2 \left(6 \text{Hypergeometric2F1}^{(0,1,1,0)}(1,1,2,-1)-3 \text{Hypergeometric2F1}^{(0,1,2,0)}(1,1,2,-1)+\text{Hypergeometric2F1}^{(0,1,3,0)}(1,1,2,-1)+\text{Hypergeometric2F1}^{(0,2,0,0)}(1,1,2,-1)-\text{Hypergeometric2F1}^{(0,2,1,0)}(1,1,2,-1)\right)+\text{Hypergeometric2F1}^{(0,2,2,0)}(1,1,2,-1)-\frac{2}{3} \pi ^2 \left(-2 \text{Hypergeometric2F1}^{(0,0,1,0)}(1,1,2,-1)+\text{Hypergeometric2F1}^{(0,0,2,0)}(1,1,2,-1)-\text{Hypergeometric2F1}^{(0,1,0,0)}(1,1,2,-1)+\text{Hypergeometric2F1}^{(0,1,1,0)}(1,1,2,-1)+\log (4)\right)-4 \zeta (3) \log (4)-\frac{1}{90} \pi ^4 \log (2)+24 \log (2)
$$
