Prove that $\int_0^1\frac{\ln(1-x)\ln^2x}{x-1}dx=\frac{\pi^4}{180}$

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{1}{\ln\pars{1 - x}\ln^{2}\pars{x} \over x - 1}\,\dd x ={\pi^{4} \over 180}:\ {\large ?}}$

\begin{align}&\int_{0}^{1}{\ln\pars{1 - x}\ln^{2}\pars{x} \over x - 1}\,\dd x =-\lim_{\mu \to -1\,\,\,\,\atop \nu \to 0}\partiald{}{\mu}\partiald[2]{}{\nu} \int_{0}^{1}\pars{1 - x}^{\mu}x^{\nu}\,\dd x \\[5mm]&=-\lim_{\mu \to - 1\,\,\,\,\atop \nu \to 0} \partiald{}{\mu}\partiald[2]{}{\nu}{\rm B}\pars{\mu + 1,\nu + 1} \end{align} where $\ds{{\rm B}\pars{x,y}}$ is the Beta Function. Use expressions $\pars{23}$, $\pars{24}$, $\pars{25}$ and $\pars{26}$ from the above cited link.

The idea is to use the fact that $\dfrac{\ln(1-x)}{x-1} = \sum\limits_{n \geq 1} H_n x^n$ where $|x| < 1$ and $H_n$ is the $n$'th harmonic number. This reduces the integral to a linear combination of $\sum\limits_{n \geq 1} H_n/n^3$ and $\sum\limits_{n \geq 1} 1/n^4$. The latter term is $\zeta(4) = \pi^4/90$; the former term is more difficult to calculate, but can be calculated with residues as well; a proof can be found here here. It seems very difficult to do this using only elementary methods.

More explicitly:

$$ I = \int_0^1\frac {\log(1 - u)}{u - 1}\, (\log u)^2 \, du = \int_0^1 \sum_{n = 0}^\infty H_n u^n \, (\log u)^2 \, du = \sum_{n = 0}^\infty H_n \int_0^1 u^n\, (\log u)^2\, du = \sum_{n = 0}^\infty H_n\, \frac {2}{(n + 1)^3} = 2\sum_{n = 1}^\infty \frac {H_{n - 1}}{n^3} = 2\left(\sum_{n = 1}^\infty \frac {H_n}{n^3} - \sum_{n = 1}^\infty \frac {1}{n^4}\right) = 2\left(\frac54\zeta(4) - \zeta(4)\right) = \frac{\pi^4}{180}.$$ Here we used the fact that $\int_0^{1}dx\, x^n \ln^2 x = \int_0^{\infty} dx\, e^{-(n+1)x} x^2 = \frac{2!}{(n+1)^3}$. The penultimate equality is not so trivial to show (but see the comment by Tunk-Fey), but a proof is given in the link.

\begin{align} -\int^1_0\frac{\ln^2{x}\ln(1-x)}{1-x}dx &=-\int^1_0\frac{\ln{x}\ln(1-x)\ln(1-x)}{x}dx\tag1\\ &=\int^1_0\frac{\ln{x} \ \mathrm{Li}_2(x)}{1-x}dx-\int^1_0\underbrace{\frac{\ln(1-x) \ \mathrm{Li}_2(x)}{x}}_{-\mathrm{Li}_2(x)\mathrm{Li}_2'(x)}dx\tag2\\ &=-\frac{\pi^4}{120}+\frac{\pi^4}{72}\tag3\\ &=\frac{\pi^4}{180} \end{align} Explanation:
$(1)$ Substitute $x \mapsto 1-x$
$(2)$ Integrate by parts, with $u=\ln{x}\ln(1-x)$ and $v=\mathrm{Li}_2(x)$
$(3)$ The first integral is cleverly solved here (I take absolutely no credit for it). For the second, observe that \begin{align} \int^1_0\mathrm{Li}_2(x)\mathrm{Li}_2'(x)dx &=\left[\frac{1}{2}\mathrm{Li}_2(x)^2\right]^1_0\\ &=\frac{1}{2}\left(\frac{\pi^2}{6}\right)^2\\ &=\frac{\pi^4}{72} \end{align}