<s>[INST] Evaluation of $\int_{0}^{1}{\int_{0}^{1}{\frac{\ln \left( 1-x \right)-\ln \left( 1-y \right)}{x-y-1}dx}dy}$ I trying to verify the following:
$$
\int_{0}^{1}{\int_{0}^{1}{\frac{\ln \left( 1-x \right)-\ln \left( 1-y \right)}{x-y-1}dx}dy}=\frac{{{\pi }^{2}}}{12}+{{\ln }^{2}}\left( 2 \right)
$$
I thought the solution may be similar to this problem but i can't handle the integral in a similar way?!?!?!
 [/INST] With 

$$\int\frac{\ln(x+z)}{x-a}dx = \text{Li}_2\left(\frac{x+z}{a+z}\right) - \ln(x+z)\,\text{Li}_1\left(\frac{x+z}{a+z}\right) + C $$ 

we get 
$$\int\limits_0^1\frac{\ln(1-x)}{x-y-1}dx = \int\limits_0^{-1}\frac{\ln(x+1)}{x+y+1}dx = -\text{Li}_2\left(-\frac{1}{y}\right)$$ 
and the other part is: 
$$\int\limits_0^1\frac{-\ln(1-y)}{x-y-1}dx = -\ln(1-y)\ln\left(\frac{y}{y+1}\right)$$
Now we need
$$\int -\text{Li}_2\left(-\frac{1}{y}\right)dy = -y\,\text{Li}_2\left(-\frac{1}{y}\right) + y\ln\left(1+\frac{1}{y}\right) + \ln(y+1) + C$$ and 
$$\int -\ln(1-y)\ln\left(\frac{y}{y+1}\right)dy = \\ 2\,\text{Li}_2\left(\frac{1-y}{2}\right) + \text{Li}_2(y) + (y + (1-y)\ln(1-y))\ln\left(\frac{y}{y+1}\right) \\ + 2\ln(1-y)\ln\left(\frac{y+1}{2}\right) - \ln(y+1) + C$$
which leads to: 
$$\int\limits_0^1\int\limits_0^1 \frac{\ln(1-x) - \ln(1-y)}{x-y-1} dxdy = \int\limits_0^1 -\text{Li}_2\left(-\frac{1}{y}\right)dy + \int\limits_0^1 -\ln(1-y)\ln\left(\frac{y}{y+1}\right)dy \\ = \left(\frac{\pi^2}{12} + \ln 4\right) + \left((\ln 2)^2 - \ln 4\right) = \frac{\pi^2}{12} + (\ln 2)^2$$
</s>