Show me how to evaluate $\int_0^1\int_0^1\frac{[-\ln(x)]^s}{1-xy}dxdy=\frac{\zeta(s+2)}{\Gamma(s+2)}$ Double integrals
$$\int_0^1\int_0^1\frac{[-\ln(x)]^s}{1-xy}dxdy=\frac{\zeta(s+2)}{\Gamma(s+2)} \tag1$$
$$\int_0^1\int_0^1\frac{[-\ln(xy)]^s}{1-xy}dxdy=\zeta(s+2)\Gamma(s+2) \tag2$$
Where $\sum_{n=0}^{\infty}\frac{1}{(n+1)^s}=\zeta(s)$, valid for $\Re(s)>1$ and $\Gamma(n+1)=n!$ valid for all non-negative integers and rational arguments.
I came across these two double integrals during the time I was on Wolfram integrator, was trying to search for something. It didn't gave me the closed form, just the numerical values and the rest I had to find the closed form base on these values.
I don't know how to prove these integrals, can somebody show me how to prove it, so I can learn from it, so next time I can independently do it myself.
I hope these closed form are correct.
P.s
Please, try not to miss too many steps. Thank you!
 A: For the first integral, we can write 
$$\begin{align}
I(s)&=\int_0^1\int_0^1 \frac{\left(-\log(x)\right)^s}{1-xy}\,dx\,dy\\\\
&=\sum_{n=0}^\infty\left(\int_0^1 y^n \,dy \int_0^1 x^n \log^s(1/x)\,dx \right)\\\\
&=\sum_{n=0}^\infty \frac{1}{n+1}\int_0^1 x^n \log^s(1/x)\,dx\\\\
&=\sum_{n=0}^\infty \frac{1}{n+1} \int_0^\infty t^s\,e^{-(n+1)t}\,dt\\\\
&=\sum_{n=0}^\infty \frac{1}{(n+1)^{s+2}}\int_0^\infty t^s\,e^{-t}\,dt\\\\
&=\zeta(s+2)\Gamma(s+1) 
\end{align}$$

For the second integral, we can write 
$$\begin{align}
J(s)&=\int_0^1\int_0^1 \frac{\left(-\log(xy)\right)^s}{1-xy}\,dx\,dy\\\\
&=\sum_{n=1}^\infty \int_0^1  \frac1y \int_0^y t^n\log^s(1/t)\,dt \,dy\\\\
&=\sum_{n=1}^\infty \int_0^1  t^n\log^s(1/t) \int_t^1 \frac1y \,dy  \,dt\\\\
&=\sum_{n=1}^\infty \int_0^1  t^n\log^{s+1}(1/t)\,dt\\\\
&=\sum_{n=1}^\infty \int_0^1  e^{-(n+1)u}u^{s+1}\,du\\\\
&=\sum_{n=1}^\infty \frac{1}{(n+1)^{s+2}}\int_0^1  e^{-u}u^{s+1}\,du\\\\
&=\zeta(s+2)\Gamma(s+2)
\end{align}$$
as was to be shown!
A: I can't comment.
See this article:
http://arxiv.org/pdf/math/0506319.pdf
Correlaries 3.1 and 3.2 have your answers
