$$\int_0^1 \frac{\operatorname{Li}_2(-x)\operatorname{Li}_2(x)}{x}\,\mathrm dx=?$$
where $$\operatorname{Li}_2(-x)=\sum_{k=1}^{\infty}\frac{(-x)^k}{k^2}$$ for $$|x|>1$$
actually my goal is to edit the problem i had while solving
$$\sum_{n=1}^{\infty} \frac{1}{n^4}\left(\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+2}-\cdots\right)=\sum_{n=1}^{\infty}\frac{1}{n^4}\int_0^1\frac{x^{n-1}}{1+x}\,\mathrm dx$$ $$=\int_0^1\frac{1}{x(1+x)}\left(\sum_{n=1}^{\infty}\frac{x^n}{n^4}\right)\,\mathrm dx$$ $$=\int_0^1\frac{\operatorname{Li}_4(x)}{x(1+x)}\,\mathrm dx=\int_0^1\frac{\operatorname{Li}_4(x)}{x}\,\mathrm dx-\int_0^1\frac{\operatorname{Li}_4(x)}{1+x}\,\mathrm dx$$
here $$\int_0^1\frac{\operatorname{Li}_4(x)}{x}\,\mathrm dx=\zeta(5)$$ but $$\int_0^1\frac{\operatorname{Li}_4(x)}{1+x}\,\mathrm dx=?$$
When I apply partial integration, the integral I asked comes out.