How find this integral $I=\int_{0}^{1}\int_{0}^{1}\frac{\ln{(1+xy)}}{1-xy}dxdy$ 
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
 A: You have

$$
\int_{0}^{1}\!\int_{0}^{1}\dfrac{\ln{(1+xy)}}{1-xy}\mathrm dx \: \mathrm dy = \frac{\pi^2}{4}\ln 2 -\zeta(3). \tag1
$$

To obtain $(1)$ one may write
$$
\begin{align}
I=\int_{0}^{1}\int_{0}^{1}\dfrac{\ln{(1+xy)}}{1-xy}dxdy &= \sum_{n=1}^{\infty}\int_{0}^{1}\int_{0}^{1} \frac{(-1)^{n-1}}{n}\dfrac{(xy)^n}{1-xy}dxdy \\
& =
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\int_{0}^{1}\int_{0}^{1} \dfrac{(xy)^n}{1-xy}dxdy \\
& =
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\Phi(1,2,n+1)\\
& =
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\left(\sum_{k=1}^{\infty}\frac{1}{k^2} -\sum_{k=1}^{n}\frac{1}{k^2}  \right)\\
& =\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\left(\frac{\pi^2}{6} \quad -\sum_{k=1}^{n}\frac{1}{k^2}\,\right) \\
& = \frac{\pi^2}{4}\ln 2 -\zeta(3)
\end{align}
$$ where we have used a result on double integrals from J. Guillera and J. Sondow (30): 
$$
\int_{0}^{1}\int_{0}^{1} \dfrac{(xy)^{u-1}}{1-xyz}(-\ln(xy))^s dxdy =\Gamma(s+2)\Phi(z,s+2,u)
$$ $\displaystyle \Phi$ denoting the Lerch transcendent function: $$ \Phi(z,s,u)= \sum_{k=0}^{\infty}\frac{z^k}{(k+u)^{s}}.$$
Update: a proof of the last step may be found here.
A: 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:
$$\begin{align}
I
&=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{\ln{(1+xy)}}{1-xy}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{1}{x}\int_{0}^{x}\mathrm{d}u\,\frac{\ln{(1+u)}}{1-u}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{1}{x}\left[-\operatorname{Li}_2{\left(\frac{x+1}{2}\right)}-\ln{\left(\frac{1-x}{2}\right)}\ln{(1+x)}-\frac12\ln^2{2}+\frac12\zeta{(2)}\right]\\
&=-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(\frac{1-x}{2}\right)}\ln{(1+x)}}{x}-\int_{0}^{1}\mathrm{d}x\,\frac{\zeta{(2)}-\ln^2{2}-2\operatorname{Li}_2{\left(\frac{x+1}{2}\right)}}{2x}\\
&=\frac58 \zeta{(3)}+\frac12\zeta{(2)}\ln{2}-\frac{13}{8}\zeta{(3)}+\zeta{(2)}\ln{2}\\
&=\frac32\zeta{(2)}\ln{2}-\zeta{(3)}.
\end{align}$$
