Evaluating $\int_0^1 \frac{\ln x\text{Li}_2(1-x)}{1+x}dx$ Trying to compute How can I evaluate $\int _0^1\frac{\text{Li}_2\left(-x\right)\ln \left(1-x\right)}{1+x}\:dx$
i came across the integral
$\displaystyle J= \int_0^1 \frac{\ln x\text{Li}_2(1-x)}{1+x}dx$
Probably
$\displaystyle J=4\text{Li}_4\left(\frac{1}{2}\right)-\frac{33}{16}\zeta(4)-\zeta(2)\ln^2 2+\frac{1}{6}\ln^4 2$
Is it possible to compute it using (generalised) harmonic series?
NB:
Probably i can compute it using only integration techniques.
 A: By using $\text{Li}_2(1-x)=\zeta(2)-\ln x\ln(1-x)-\text{Li}_2(x)$ we have
$$\int_0^1\frac{\ln x\text{Li}_2(1-x)}{1+x}dx=\zeta(2)\underbrace{\int_0^1\frac{\ln x}{1+x}dx}_{I_1}-\underbrace{\int_0^1\frac{\ln^2x\ln(1-x)}{1+x}dx}_{I_2}-\underbrace{\int_0^1\frac{\ln x\text{Li}_2(x)}{1+x}dx}_{I_3}$$
where
$$I_1=\int_0^1\frac{\ln x}{1+x}dx=-\eta(2)=-\frac12\zeta(2)$$
$$I_2=\int_0^1\frac{\ln^2x\ln(1-x)}{1+x}dx=\sum_{n=1}^\infty (-1)^{n-1}\int_0^1 x^{n-1}\ln^2x \ln(1-x)dx$$
$$=\sum_{n=1}^\infty (-1)^{n-1}\left(2\frac{\zeta(3)-H_n^{(3)}}{n}+2\frac{\zeta(2)-H_n^{(2)}}{n^2}-\frac{2H_n}{n^3}\right)$$
$$I_3=\int_0^1\frac{\ln x\text{Li}_2(x)}{1+x}dx=\sum_{n=1}^\infty (-1)^{n-1}\int_0^1 x^{n-1}\ln x\text{Li}_2(x)dx$$
$$=\sum_{n=1}^\infty (-1)^{n-1}\left(\frac{H_n^{(2)}}{n^2}+\frac{2H_n}{n^3}-\frac{2\zeta(2)}{n^2}\right)$$
Collecting all integrals we have
$$\int_0^1\frac{\ln x\text{Li}_2(1-x)}{1+x}dx=-\frac54\zeta(4)-2\zeta(3)\ln2-2\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n}-\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^2}$$
and I am sure you are well familiar with these two sums.
