What real tools excepting the ones provided here Closed-form of $\int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx $ would you like to recommend? I'm not against them, they might be great, but it seems they didn't lead anywhere for the version $\displaystyle \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx$. Perhaps we can find an approach that covers both cases, also
$$\int_0^1 \log(x) \left(\frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}}\right)^2 \,dx$$ that I would like to calculate.
Might we possibly expect a nice closed form as in the previous case? What do you propose?
EDIT: Thanks David, I had to modify it a bit to fix the convergence issue. Also, for the previous question there is already a 300 points bounty offered for a full solution with all steps clearly explained.
We have: $$\sum_{k=1}^{n-1}\frac{1}{k^2(n-k)^2}=\frac{1}{n^2}\sum_{k=1}^{n-1}\left(\frac{1}{k}+\frac{1}{n-k}\right)=\frac{2H_{n-1}^{(2)}}{n^2}+\frac{4H_{n-1}}{n^3}$$ so: $$ \text{Li}_2(x)^2 = \sum_{n\geq 2}\left(\frac{2H_{n-1}^{(2)}}{n^2}+\frac{4H_{n-1}}{n^3}\right) x^n\tag{1}$$ and since: $$ \int_{0}^{1}\frac{x^n \log x}{1-x^2}\,dx = -\sum_{m\geq 0}\frac{1}{(n+2m+1)^2}\tag{2}$$ we have: $$ \int_{0}^{1}\log(x)\left(\frac{\text{Li}_2(x)}{\sqrt{1-x^2}}\right)^2\,dx = -\sum_{n\geq 2}\left(\frac{2H_{n-1}^{(2)}}{n^2}+\frac{4H_{n-1}}{n^3}\right)\sum_{m\geq 0}\frac{1}{(n+2m+1)^2}\tag{3}$$ and the problem boils down to the computation of a complicated Euler sum.
In order to perform partial summation, it is useful to recall that: $$ \sum_{n=1}^{N}\frac{2H_{n-1}^{(2)}}{n^2}=\left(H_{N}^{(2)}\right)^2-H_{N}^{(4)},$$ $$\sum_{n=1}^{N}\frac{H_{n-1}}{n^3} = H_N^{(3)}H_{N-1}-\sum_{n=1}^{N-1}\frac{H_{n}^{(3)}}{n}.\tag{4}$$