Correct $\int_{0}^{1}\int_{0}^{1}{{y^n\over x}{\ln\left({x\over y}\right)}\over (1-xy)\ln(xy)}\mathrm dxdy=-{2(n+1)(n+3)\over (n+2)^3}?$ Experimental show that $(1)$
$$\int_{0}^{1}\int_{0}^{1}{{y^n\over x}{\ln\left({x\over y}\right)}\over (1-xy)\ln(xy)}\mathrm dxdy=-{2(n+1)(n+3)\over (n+2)^3}\tag1$$
$n\ne-2$
How may one verify that $(1)$ is correct?

Because the integrand contains singularities, the integral does not exist.  However, in a certain principal value sense, the following integral is zero by $x \leftrightarrow y$ symmetry arguments: 
$$
P_{xy}\int_{0}^{1}\int_{0}^{1} dx dy \, {\displaystyle{\ln\left({x\over y}\right)}\over (1-xy)\ln(xy)} = 0
$$ 
where $P_{xy}$ means we keep the integration symmetric in x and y (for example, a Riemann sum that samples the unit square symmetrically). For the same reason, your integral is zero when $n=-1$:
$$
P_{xy}\int_{0}^{1}\int_{0}^{1} dx dy \, {\displaystyle\frac{1}{xy}{\ln\left({x\over y}\right)}\over (1-xy)\ln(xy)} = 0
$$
So, your formula is correct for $n=-1$, but not for other values.
