Otdfbt

Faster method to calculate the exact solution of the following integral (based on the ideas of Fedor Petrov: https://mathoverflow.net/users/4312/fedor-petrov, Double integral with logarithms, URL (version: 2019-04-15): https://mathoverflow.net/q/328126):

$$Jequiv int_0^1int_0^1fracln x-ln yx-ydxdy .$$

Since
$$fleft( x,y right)=fracln x-ln yx-y=fleft( y,x right),$$
the surface $fleft( x,y right) $ is symmetric with respect to the bisector plane $x = y$; so,

$$fracJ2=int_0^1dxint_0^xfracln x-ln yx-ydy.$$

With the change of variable$$yequiv tx, tin left( 0, 1 right),$$
the integral
$$int_0^xfracln x-ln yx-ydy,$$
is transformed into the following one that does not depend on $x$,
$$ Iequiv -int_0^1fracln t1-t,dt.$$

The integration on the unit square $(0, 0), (1, 0), (1, 1), (0,1)$ is reduced to the integration on the triangle $(0, 0), (1, 0), (1, 1).$

To solve the integral $I$ we will carry out the new change of variable,
$$sequiv 1-t,$$
by which $I$ is transformed into the integral that defines the dilogarithm, whose value for $s = 1$ coincides with the Riemann zeta $zeta left( 2 right)$,whose value is well known:

$$I=int_1^0fracln left( 1-s right)s,ds=textLtexti_2left( 1 right)=zeta left( 2 right)=fracpi ^26.$$
Therefore, the solution to the proposed integral is
$$J=fracpi ^23.$$

Note. I've tried it by polylogarithmic transformations, but I couldn't get the result $fracpi ^23.$

By symmetry we have $J/2=int_0^1 dx int_0^x f(x, y) dy$ where $f(x, y) $ is your integrand. Integrating against $y$ for fixed $x$ we denote $y=tx$, $t$ varies from 0 to 1 and the integral against $y$ reads as $-int_0^1 fraclog t 1-tdt$. It does not depend on $x$ and is well known to be equal to $pi^2/6$ (you may use the geometric progression expansion $frac11 - t =sum_n>0 t^n-1$ and integrate term-wise to get $sum 1/n^2$).