TITLE: How find this integral $I=\int_{0}^{1}\int_{0}^{1}\frac{\ln{(1+xy)}}{1-xy}dxdy$
QUESTION [11 upvotes]: Find this integral
$$I=\int_{0}^{1}\int_{0}^{1}\dfrac{\ln{(1+xy)}}{1-xy}dxdy$$

My try: since
$$\dfrac{1}{1-xy}=\sum_{n=0}^{\infty}(xy)^n$$
so
$$I=\sum_{n=0}^{\infty}\int_{0}^{1}y^n\int_{0}^{1}x^n\ln{(1+xy)}dx$$
because
$$\int_{0}^{1}x^n\ln{(1+xy)}dx=\dfrac{x^{n+1}\ln{(1+xy)}}{n+1}|_{0}^{1}-\dfrac{1}{n+1}\int_{0}^{1}\dfrac{x^{n+1}y}{1+xy}dx=\dfrac{\ln{(1+y)}}{n+1}-I_{1}$$
where
$$I_{1}=y\int_{0}^{1}\dfrac{x^{n+1}}{1+xy}dx$$  even if $I_{1}$ can use the beta function,
But  then  follow I can't it.Thank you

REPLY [3 votes]: Though I'd actually prefer the solutions other users have already posted to the solution below, I thought it worth pointing out that there's really nothing stopping you from solving this integral by brute force.
The main non-trivial fact needed beforehand is the anti-derivative,
$$\int\mathrm{d}u\,\frac{\ln{(1+u)}}{1-u}=-\operatorname{Li}_2{\left(\frac{x+1}{2}\right)}-\ln{\left(\frac{1-x}{2}\right)}\ln{(1+x)}+constant,$$
which can be verified via differentiation.
Here's a sketch of the rest of the calculation:
$$\begin{align}
I
&=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{\ln{(1+xy)}}{1-xy}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{1}{x}\int_{0}^{x}\mathrm{d}u\,\frac{\ln{(1+u)}}{1-u}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{1}{x}\left[-\operatorname{Li}_2{\left(\frac{x+1}{2}\right)}-\ln{\left(\frac{1-x}{2}\right)}\ln{(1+x)}-\frac12\ln^2{2}+\frac12\zeta{(2)}\right]\\
&=-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(\frac{1-x}{2}\right)}\ln{(1+x)}}{x}-\int_{0}^{1}\mathrm{d}x\,\frac{\zeta{(2)}-\ln^2{2}-2\operatorname{Li}_2{\left(\frac{x+1}{2}\right)}}{2x}\\
&=\frac58 \zeta{(3)}+\frac12\zeta{(2)}\ln{2}-\frac{13}{8}\zeta{(3)}+\zeta{(2)}\ln{2}\\
&=\frac32\zeta{(2)}\ln{2}-\zeta{(3)}.
\end{align}$$