Why does the hard-looking double integral $\int_{0}^{1}\int_{0}^{1}\frac{x(1-x)y(1+y)}{(1-xy)\ln(xy)}dxdy=-\frac{1}{2}$? (1) originate from here problem 11322
(1)
$$\int_{0}^{1}\int_{0}^{1}\frac{x(1-x)y(1+y)}{(1-xy)\ln(xy)}dxdy=-\frac{1}{2}$$
On my recent post see here Marco Cantarini and hints from other proved the hard-looking integral. 
We manage to find another hard-looking double integral type that yield simple answer via wolfram integrator.
Can anyone help us to prove (1)?
 A: If we perform a change of variables $x=e^{-u},y=e^{-v}$ we may easily check that
$$J(a,b) = \iint_{(0,1)^2}\frac{x^a y^b}{\log(xy)}\,dx\,dy = -\frac{1}{a-b}\,\log\left(\frac{1+a}{1+b}\right) \tag{1}$$
from which it follows that
$$\begin{eqnarray*}K(a,b) &=& \iint_{(0,1)^2}\frac{x^a y^b(1-x)(1+y)}{\log(xy)}\,dx\,dy\\ &=& \frac{\log\left(\frac{2+a}{1+b}\right)}{1+a-b}+\frac{\log\left(\frac{1+a}{2+b}\right)}{1+b-a}+\frac{\log\left(\frac{2+a}{2+b}\right)-\log\left(\frac{1+a}{1+b}\right)}{a-b} \tag{2}\end{eqnarray*}$$
and by letting $b\to a$
$$ H(a) = \iint_{(0,1)^2}\frac{x^a y^a(1-x)(1+y)}{\log(xy)}\,dx\,dy = \frac{1}{2+a}-\frac{1}{1+a}\tag{3}$$
so, just like in the other problem, the original integral is related with a telescopic series:

$$\begin{eqnarray*}I=\iint_{(0,1)^2}\frac{xy}{1-xy}\cdot\frac{(1-x)(1+y)}{\log(xy)}\,dx\,dy &=& \sum_{n\geq 1} H(n) \\ &=& -\sum_{n\geq 1}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)=\color{red}{-\frac{1}{2}}\tag{4}\end{eqnarray*} $$

as wanted. The only non-trivial step is $(1)$, that follows from Frullani's theorem.
There is also a symmetry trick. By exchanging $x$ and $y$ in the original integral and considering the arithmetic mean of the two integrals we get:
$$I=\iint_{(0,1)^2}\frac{xy}{\log(xy)}\,dx\,dy = J(1,1) = -\left.\frac{d}{dz}\log(z+1)\right|_{z=1} = \color{red}{-\frac{1}{2}}.\tag{5}$$
