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.


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.
