<s>[INST] 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.
 [/INST] 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.
</s>