Evaluate int_0^1 log^2(1-x) log^2(x) dx

Evaluate ${\int }_{0}^{1}{\mathrm{log}}^{2}\left(1-x\right){\mathrm{log}}^{2}\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx$
I have no idea where to even start. WolframAlpha cant compute it either.
${\int }_{0}^{1}{\mathrm{log}}^{2}\left(1-x\right){\mathrm{log}}^{2}\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx$
I think it can be done with series, but I am not sure, can someone help a little so I can get a start??
Thanks!
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Alaina Mcintosh
Using the series of ${\mathrm{ln}}^{2}\left(1-x\right)$

Then integrate $f\left(z\right)=\frac{\left(\gamma +{\psi }_{0}\left(-z\right){\right)}^{2}}{\left(z+1\right)\left(z+2{\right)}^{3}}$ along an infinitely large square. The integral vanishes which implies the sum of its residues is zero. At the positive integers,
$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\mathrm{R}\mathrm{e}\mathrm{s}\left(f,n\right)& =\sum _{n=1}^{\mathrm{\infty }}\underset{z=n}{Res}\left[\frac{1}{\left(z+1\right)\left(z+2{\right)}^{3}\left(z-n{\right)}^{2}}+\frac{2{H}_{n}}{\left(z+1\right)\left(z+2{\right)}^{3}\left(z-n\right)}\right]\\ & =\sum _{n=1}^{\mathrm{\infty }}\frac{2{H}_{n}}{\left(n+1\right)\left(n+2{\right)}^{3}}-\sum _{n=1}^{\mathrm{\infty }}\frac{4n+5}{\left(n+1{\right)}^{2}\left(n+2{\right)}^{4}}\\ & =\sum _{n=1}^{\mathrm{\infty }}\frac{2{H}_{n}}{\left(n+1\right)\left(n+2{\right)}^{3}}+\frac{{\pi }^{4}}{30}+2\zeta \left(3\right)-\frac{91}{16}\end{array}$
At z=0,
$\begin{array}{rl}\mathrm{R}\mathrm{e}\mathrm{s}\left(f,0\right)& =\underset{z=0}{Res}\frac{1}{{z}^{2}\left(z+1\right)\left(z+2{\right)}^{3}}\\ & =-\frac{5}{16}\end{array}$
At z=−2,
$\begin{array}{rl}\mathrm{R}\mathrm{e}\mathrm{s}\left(f,-2\right)& =\frac{1}{2}\underset{z\to -2}{lim}\frac{{\mathrm{d}}^{2}}{\mathrm{d}{z}^{2}}\frac{\left(\gamma +{\psi }_{0}\left(-z\right){\right)}^{2}}{z+1}\\ & =-\frac{{\pi }^{4}}{36}+2\zeta \left(3\right)+\frac{2{\pi }^{2}}{3}-6\end{array}$
Therefore,