Q:

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:

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:

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!

