In an answer to a question found [here](https://math.stackexchange.com/questions/3513905/computing-int-01-frac-ln1-x2x-operatornameli-2-left-frac1-x2-ri/3513994#3513994) @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. 

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> 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)?
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