TITLE: A challenging logarithmic integral $\int_0^1 \frac{\log(1+x)\log(1-x)}{1+x}dx$
QUESTION [23 upvotes]: While playing around with Mathematica, I found that
$$\int_0^1 \frac{\log(1+x)\log(1-x)}{1+x}dx = \frac{1}{3}\log^3(2)-\frac{\pi^2}{12}\log(2)+\frac{\zeta(3)}{8}$$
Please help me prove this result.

REPLY [2 votes]: It is easy to prove that $$\int\frac{\log(1-x)}{x+1}dx=\log(2)\log(x+1)-\operatorname{Li}_2\left(\frac{x+1}{2}\right)$$

The proof is straightforward  if we set $1+x=y$ and rest it follows by definition of dilogarithm and logarithmic integrals.
Next we do integration by parts giving us $$\log(1+x)\left(\log(2)\log(1+x)-\operatorname{Li}_2\left(\frac{x+1}{2}\right)\right)\Bigg|_0^1-\int_0^1\frac{dx}{1+x}\left(\log(2)\log(1+x)-\operatorname{Li}_2\left(\frac{x+1}{2}\right)\right)=\log^3(2)-\frac{\pi^2}{6}\log(2)-\frac{\log^3(2)}{3}+\int_0^1\frac{\operatorname{Li}_2\left(\frac{x+1}{2}\right)}{x+1}dx=\frac{\log^3(2)}{2}-\frac{\pi^2}{6}\log(2)+\zeta(3)-\hbox{Li}_3\left(\frac{1}{2}\right)$$ Now using the  known value of $\hbox{Li}_3\left(\frac{1}{2}\right)=\frac{\log^3(2)}{6}-\frac{\pi^2}{12}\log(2)+\frac{7}{8}\zeta(3)$ and simplifying gives us  the desires closed form
for it.
$$\frac{\log^3(2)}{3} -\frac{\pi^2}{12}\log(2)+\frac{\zeta(3)}{8}$$.