Compute $ \int_{0}^{1}\frac{\ln(x) \ln^2 (1-x)}{x} dx $ Compute
$$ \int_{0}^{1}\frac{\ln(x) \ln^2 (1-x)}{x} dx $$
I'm looking for some nice proofs at this problem. One idea would be to use Taylor expansion and then integrating term by term. What else can we do? Thanks.

In this answer I will make use of a Maclaurin series expansion for the term $\ln^2 (1 - x)$, which I show here to be
$$\ln^2 (1 - x) = 2 \sum_{n = 2}^\infty \frac{H_{n - 1} x^n}{n},$$
and the well-known Euler sum of
$$\sum_{n = 1}^\infty \frac{H_n}{n^3} = \frac{1}{2} \zeta^2 (2),$$
several proofs for which can be found here. 
From the above Maclaurin series expansion for $\ln^2 (1 - x)$ the integral can be written as
\begin{align*}
\int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = 2 \sum_{n = 2}^\infty \int_0^1 x^{n - 1} \ln x \, dx.
\end{align*}
The integral that appears to the right can be readily found by parts. The result is
$$\int_0^1 x^{n - 1} \ln x \, dx = -\frac{1}{n^2}.$$
Thus
$$\int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = -2 \sum_{n = 2}^\infty \frac{H_{n - 1}}{n^3}.$$
From properties of harmonic numbers, since
$$H_n = H_{n - 1} + \frac{1}{n},$$
the integral becomes
$$\int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = 2 \sum_{n = 2}^\infty \frac{1}{n^4} - 2 \sum_{n = 2}^\infty \frac{H_n}{n^3} = 2 \sum_{n = 1}^\infty \frac{1}{n^4} - 2 \sum_{n = 1}^\infty \frac{H_n}{n^3}.$$
As 
$$\sum_{n = 1}^\infty \frac{1}{n^4} = \zeta (4) \quad \text{and} \quad \sum_{n = 1}^\infty \frac{H_n}{n^3} = \frac{1}{2} \zeta^2 (2),$$
we have
$$\int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = 2 \zeta (4) - \zeta^2 (2) = - \frac{\pi^4}{180} = -\frac{1}{2} \zeta (4),$$
as expected.
