Find $\int_0^1\frac{\ln^2(1-x)}{x}\ dx$ In solving $\displaystyle\int_0^\frac{\pi}{4}\dfrac{\ln(\sin x)\ln(\cos x)}{\sin x\cos x}\ dx,$ I have found that this is equal to $\dfrac{1}{16}\displaystyle\int_0^1\dfrac{\ln^2(1-x)}{x}\ dx.$ WolframAlpha says that the desired value is $\dfrac{\zeta(3)}{8},$ so I suspect a conversion to a series is necessary.
How do I prove $\displaystyle\int_0^1\dfrac{\ln^2(1-x)}{x}\ dx=\displaystyle\sum_{n=1}^\infty\dfrac{2}{n^3}$?
Note that the above integral can also be given as $\displaystyle\int_0^1\dfrac{\ln^2x}{1-x}\ dx$, which I know is equal to $\displaystyle\sum_{n=0}^\infty x^n\ln^2x.$
Also for reference, here is a picture of my original work to get to this point.

 A: I thought it might be instructive to present a way forward that exploits the Polylogarithm Functions.  To that end, we proceed.
Note that integrating by parts with $u=\log^2(1-x)$ and $v=\log(x)$, we have 
$$\begin{align}
\int_0^1 \frac{\log^2(1-x)}{x}\,dx=2\int_0^1 \frac{\log(1-x)\log(x)}{1-x}\,dx \tag 1
\end{align}$$
Integrating by parts the right-hand side of $(1)$ with $u=\log(1-x)$ and $v=\text{Li}_2(1-x)$ yields
$$\begin{align}
2\int_0^1 \frac{\log(1-x)\log(x)}{1-x}\,dx&=2\int_0^1 \frac{\text{Li}_2(1-x)}{1-x}\,dx\\\\
&=2\int_0^1 \frac{\text{Li}_2(x)}{x}\,dx\\\\
&=2\text{Li}_3(1)\\\\
&=2\zeta(3)
\end{align}$$
as expected!
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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*

*This one is $\ul{slightly\ different}$ of the straightforward @Dr. MV answer:
\begin{align}
\color{#f00}{{1 \over 16}\int_{0}^{1}{\ln^{2}\pars{1 - x} \over x}\,\dd x} &\,\,\,
\stackrel{x\ \mapsto\ \pars{1 - x}}{=}\,\,\,
{1 \over 16}\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x}\,\dd x
\end{align}

Integrating by Parts a few times ( the main purpose is to 'sit' a
$\ds{\ln\pars{1 - x}}$-factor in the integrand numerator ): 
\begin{align}
\color{#f00}{{1 \over 16}\int_{0}^{1}{\ln^{2}\pars{1 - x} \over x}\,\dd x} & =
{1 \over 16}\int_{0}^{1}\ln\pars{1 - x}
\bracks{2\ln\pars{x}\,{1 \over x}}\,\dd x = 
-\,{1 \over 8}\int_{0}^{1}\Li{2}'\pars{x}\ln\pars{x}\,\dd x
\\[5mm] & =
{1 \over 8}\int_{0}^{1}\Li{2}\pars{x}\,{1 \over x}\,\dd x =
{1 \over 8}\int_{0}^{1}\Li{3}'\pars{x}\,\dd x = {1 \over 8}\,\Li{3}\pars{1}
\\[5mm] & = \color{#f00}{{1 \over 8}\,\zeta\pars{3}}
\end{align}

*Another approach uses the Beta Function
$\ds{\mrm{B}\pars{\mu,\nu} =
\int_{0}^{1}x^{\mu - 1}\,\pars{1 - x}^{\nu - 1}\,\,\dd x =
{\Gamma\pars{\mu}\Gamma\pars{\nu} \over \Gamma\pars{\mu + \nu}}}$ with
$\ds{\Re\pars{\mu} > 0\,,\ \Re\pars{\nu} > 0}$. $\ds{\Gamma\,}$: Gamma
Function.
\begin{align}
&\color{#f00}{{1 \over 16}\int_{0}^{1}{\ln^{2}\pars{1 - x} \over x}\,\dd x} =
{1 \over 16}\,\lim_{\mu \to 0}\,\,\partiald[2]{}{\mu}
\int_{0}^{1}{\pars{1 - x}^{\mu} - 1 \over x}\,\dd x
\\[5mm] & =
{1 \over 16}\,\lim_{\mu \to 0}\,\,\partiald[2]{}{\mu}\bracks{\mu
\int_{0}^{1}\ln\pars{x}\pars{1 - x}^{\mu - 1}\,\dd x} =
{1 \over 16}\,\lim_{\mu \to 0 \atop \nu \to 0}\,\,{\partial^{3} \over \partial\mu^{2}\,\partial\nu}
\bracks{\mu\int_{0}^{1}x^{\nu}\pars{1 - x}^{\mu - 1}\,\dd x}
\\[5mm] & =
{1 \over 16}\,\lim_{\mu \to 0 \atop \nu \to 0}\,\,{\partial^{3} \over \partial\mu^{2}\,\partial\nu}\bracks{\mu\,{\Gamma\pars{\nu + 1}\Gamma\pars{\mu} \over \Gamma\pars{\mu + \nu + 1}}} =
{1 \over 16}\,\lim_{\mu \to 0 \atop \nu \to 0}\,\,{\partial^{3} \over \partial\mu^{2}\,\partial\nu}\bracks{\Gamma\pars{\nu + 1}\Gamma\pars{\mu + 1} \over \Gamma\pars{\mu + \nu + 1}}
\\[5mm] & = \color{#f00}{{1 \over 8}\,\zeta\pars{3}}
\end{align}

A: Here is an approach that makes use of an Euler sum.
We will first find a Maclaurin series expansion for $\ln^2 (1 - x)$. As 
$$\ln (1 - x) = - \sum_{n = 1}^\infty \frac{x^n}{n},$$
we have
\begin{align*}
\ln^2 (1 - x) &= \left (- \sum_{n = 1}^\infty \frac{x^n}{n} \right ) \cdot \left (- \sum_{n = 1}^\infty \frac{x^n}{n} \right ).
\end{align*}
Shifting the summation index $n \mapsto n + 1$ gives
\begin{align*}
\ln^2 (1 - x) &= x^2 \left (- \sum_{n = 0}^\infty \frac{x^n}{n + 1} \right ) \cdot \left (- \sum_{n = 0}^\infty \frac{x^n}{n + 1} \right )\\
&= \sum_{n = 0}^\infty \sum_{k = 0}^n \frac{x^{n + 2}}{(k + 1)(n - k + 1)},
\end{align*}
where the last line has been obtained by applying the Cauchy product.
Shifting the summation indices as follows: $n \mapsto n - 2, k \mapsto k - 1$ gives
\begin{align*}
\ln^2 (1 - x) &= \sum_{n = 2}^\infty \sum_{k = 1}^{n - 1} \frac{x^n}{k(n - k)}\\
&= \sum_{n = 2}^\infty \sum_{k = 1}^{n - 1} \left (\frac{1}{nk} + \frac{1}{n(n - k)} \right ) x^n\\
&= 2 \sum_{n = 2}^\infty \frac{x^n}{n} \sum_{k = 1}^{n - 1} \frac{1}{k}\\
&= 2 \sum_{n = 2}^\infty \frac{H_{n - 1} x^n}{n},
\end{align*}
where $H_n$ is the $n$th harmonic number.
Now evaluating the integral. From the above Maclaurin series expansion for $\ln^2 (1 - x)$ the integral can be written as
\begin{align*}
\int_0^1 \frac{\ln^2 (1 - x)}{x} \, dx &= 2 \sum_{n = 2}^\infty \frac{H_{n - 1}}{n} \int_0^1 x^{n - 1} \, dx = 2 \sum_{n = 2}^\infty \frac{H_{n - 1}}{n^2}.
\end{align*}
From properties of the harmonic numbers we have
$$H_n = H_{n - 1} + \frac{1}{n},$$
thus
\begin{align*}
\int_0^1 \frac{\ln^2 (1 - x)}{x} \, dx &= 2 \sum_{n = 2}^\infty \frac{H_n}{n^2} - 2 \sum_{n = 2}^\infty \frac{1}{n^3} = 2 \sum_{n = 1}^\infty \frac{H_n}{n^2} - 2 \sum_{n = 1}^\infty \frac{1}{n^3}.
\end{align*}
Each sum can be readily found. They are:
$$\sum_{n = 1}^\infty \frac{1}{n^3} = \zeta (3) \quad \text{and} \quad \sum_{n = 1}^\infty \frac{H_n}{n^2} = 2 \zeta (3).$$
A proof of the result for the second sum containing the harmonic number can, for example, be found here. Thus
$$\int_0^1 \frac{\ln^2 (1 - x)}{x} \, dx = 4 \zeta (3) - 2 \zeta (3) = 2 \zeta (3),$$
as required.
A: We have $$\int_{0}^{1}\frac{\log^{2}\left(1-x\right)}{x}dx\stackrel{x\rightarrow1-x}{=}\int_{0}^{1}\frac{\log^{2}\left(x\right)}{1-x}dx$$ $$\stackrel{DCT}{=}
 \sum_{k\geq0}\int_{0}^{1}\log^{2}\left(x\right)x^{k}dx\stackrel{IBP}{=}
 2\sum_{k\geq0}\frac{1}{\left(k+1\right)^{3}}=\color{red}{2\zeta\left(3\right)}.$$
A: Everything you did is good. Indeed, it suffices to show that $I_n:=\int_0^1 x^n \ln^2 x = \frac{2}{(n+1)^3}$ to conclude. The interversion $\int / \Sigma$ is possible as everything here is positive.
To compute $I_n$, I tried an integration by parts (using that a primitive of $\ln^2 x$ is $x(\ln^2 x-2\ln x)$) to obtain the relation $$I_n = \frac{-2}{n+1}\int_0^1 x^n\ln x.$$
Let us call this latter integral $J_n$. Once again with an integration by parts, you can show that $J_n = \frac{-1}{(n+1)^2}$, and thus conclude.
