In an answer to a question found here @user97357329 implies the following integral $$\int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x)}{1 + x} \, dx,$$ can be found relatively easily.
So far what I have managed to come up with is the following. Since $$\sum_{n = 1}^\infty H^{(2)}_n x^n = \frac{\operatorname{Li}_2 (x)}{1 - x},$$ where $H^{(2)}_n = \sum_{k = 1}^n \frac{1}{k^2}$ denotes the 2nd order generalised harmonic number, replacing $x$ with $-x$ gives $$\sum_{n = 1}^\infty (-1)^n H^{(2)}_n x^n = \frac{\operatorname{Li}_2 (-x)}{1 + x}.$$ So the integral becomes \begin{align} \int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x)}{1 + x} \, dx &= \sum_{n = 1}^\infty (-1)^n H^{(2)}_n \int_0^1 x^n \ln (1 - x) \, dx\\ &= \sum_{n = 2}^\infty (-1)^{n - 1} H^{(2)}_{n - 1} \int_0^1 x^{n - 1} \ln (1 - x) \, dx\\ &= \sum_{n = 2}^\infty (-1)^n \frac{H^{(2)}_{n - 1} H_n}{n}, \end{align} where the result $\int_0^1 x^{n - 1} \ln (1 - x) \, dx = -\frac{H_n}{n}$ has been used. This gives a difficult non-linear Euler sum.
How can one find the value of the integral without using the value for the Euler sum just found or other difficult non-linear Euler sums (linear ones are fine)?