<s>[INST] What's the value of $\int_0^1\frac{1}{2y} \ln(y) \ln^2(1-y) \, dy$? I came across this integral while doing a different problem: 

$$ \int_0^1\frac{1}{2y} \ln (y)\ln^2(1-y) \, dy$$

I think we can evaluate this integral by differentiating the common integral representation of the beta function, but it seems to get a bit messy.
 [/INST] I offer up yet another approach, this one relying on the Maclaurin series expansion for $\ln^2 (1 - x)$. It is similar to that used by @Frank W.
As was shown here
$$\ln^2 (1 - x) = 2 \sum_{n = 2}^\infty \frac{H_{n - 1} x^n}{n}, \qquad |x| < 1.$$
Here $H_n$ denotes the Harmonic number. So for the integral we may write
\begin{align}
\int_0^1 \frac{1}{2x} \ln x \ln^2 (1 - x) \, dx &= \sum_{n = 2}^\infty \frac{H_{n - 1}}{n} \int_0^1 x^{n - 1} \ln x \, dx = -\sum_{n = 2}^\infty \frac{H_{n - 1}}{n^3},
\end{align}
after integrating by parts. From properties for the Harmonic number, since
$$H_n = H_{n - 1} + \frac{1}{n},$$
the infinite sum can be rewritten as
\begin{align}
\int_0^1 \frac{1}{2x} \ln x \ln^2 (1 - x) \, dx &= \sum_{n = 2}^\infty \frac{1}{n^4} - \sum_{n = 2}^\infty \frac{H_n}{n^3} = \sum_{n = 1}^\infty \frac{1}{n^4} - \sum_{n = 1}^\infty \frac{H_n}{n^3}.
\end{align}
Values for each of these sums are well known. For the first
$$\sum_{n = 1}^\infty \frac{1}{n^4} = \zeta (4) = \frac{\pi^4}{90}.$$
For the second sum
$$\sum_{n = 1}^\infty \frac{H_n}{n^3} = \frac{\pi^4}{72}.$$
(for various proofs of this result, see here) Thus
$$\int_0^1 \frac{1}{2x} \ln x \ln^2 (1 - x) \, dx = \frac{\pi^4}{90} - \frac{\pi^4}{72} = -\frac{\pi^4}{360}.$$
</s>