A definite integral involving logarithmic functions: $\int_{0}^{1} \frac{\ln(x) \ln (1+x)}{1-x} \, \mathrm{d} x$

EDIT:

I modified my answer so that it no longer involves dealing with polylogarithms.

There is also a very nice evaluation HERE.

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The generating function of the alternating harmonic numbers (i.e., $\bar{H}_{n} = \sum_{k=1}^{n} \frac{(-1)^{k-1}}{k})$ is $$\frac{\ln(1+x)}{1-x} = \sum_{n=1}^{\infty} \bar{H}_{n} x^{n}. $$

This can be derived using the Cauchy product.

(This was mentioned in a partial answer that was deleted shortly after it was posted.)

Using this generating function, we find that

$$ \begin{align} \int_{0}^{1} \frac{\ln (x) \ln(1+x)}{1-x} \, dx &= \int_{0}^{1} \ln(x) \sum_{n=1}^{\infty} \bar{H}_{n} \, x^{n} \, dx \\ &= \sum_{n=1}^{\infty} \bar{H}_{n} \int_{0}^{1} \ln(x) x^{n} \, dx \\ &= -\sum_{n=1}^{\infty} \frac{\bar{H}_{n}}{(n+1)^{2}} \\&= -\sum _{n=1}^{\infty} \frac{\bar{H}_{n+1}}{(n+1)^{2}} + \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(n+1)^{3}} \\&= \left( -\sum_{n=1}^{\infty}\frac{\bar{H}_{n}}{n^{2}} +1 \right) + \left( \frac{ 3\zeta(3)}{4} -1 \right) \\ &= -\sum_{n=1}^{\infty}\frac{\bar{H}_{n}}{n^{2}} + \frac{3\zeta(3)}{4}. \end{align}$$

In general, $$2 \sum_{n=1}^{\infty} \frac{\bar{H}_{n}}{n^{q}} = 2 \zeta(q) \ln(2)-q \zeta(q+1) + 2 \eta(q+1) + \sum_{k=1}^{q} \eta(k)\eta(q-k+1) , $$ where $\eta(z)$ is the Dirichlet eta function.

(See Theorem 7.1 in this paper for details about how to derive this formula using contour integration.)

Therefore, $$\sum_{n=1}^{\infty} \frac{\bar{H}_{n}}{n^{2}} = \zeta(2) \ln(2) - \zeta(3) + \eta(3) + \eta(1) \eta(2) = \frac{\pi^{2}}{4} \ln(2) - \frac{\zeta(3)}{4}, $$ and

$$\int_{0}^{1} \frac{\ln (x) \ln(1+x)}{1-x} \, dx = \frac{\zeta(3)}{4} - \frac{\pi^{2}}{4} \log(2) + \frac{3\zeta(3)}{4} = - \frac{\pi^{2}}{4} \ln(2) + \zeta(3).$$