Calculate in closed form $\int_0^1 \int_0^1 \frac{dx\,dy}{1-xy(1-x)(1-y)}$ The following question has a 500 points bounty on MSE  that soon comes to an end, and no answer
as expected was given yet. How would a professional solve the problem? Wish you succcess.
https://math.stackexchange.com/questions/1261861/calculate-in-closed-form-int-01-int-01-fracdx-dy1-xy1-x1-y
 A: Not what you're really looking for, but here's a "word answer" in terms of a probabilistic interpretation.
Suppose Alice and Bob are playing a game where they are presented with $2n+1$ indistinguishable bees and have to separate them into


*

*1 queen bee,

*$n$ worker bees, and

*$n$ drones.


They both have to try to guess what groups the other has chosen.
They play this game for increasing $n=0,1,2,3,\dots$
The probability that Alice is right is
$$
\frac1{\binom{2n+1}{n}\binom{n}{1}} = \frac{(n!)^2}{(2n+1)!} = B(n+1,n+1)
$$
Thus the number you're interested in is the expected number of times Alice and Bob both guess correctly.

For the record here is the usual derivation relating the integral to the infinite series in terms of the beta function $B$.
Let $a(x)=x(1-x)$ and $b(x,y)=a(x)a(y)$, and
$$
c(x,y)=\frac1{1-b(x,y)} = \sum_{n=0}^\infty b(x,y)^n.
$$
In terms of the beta function $B$
$$
\int_0^1 a(x)^n\,dx = \frac{(n!)^2}{(2n+1)!} = B(n+1,n+1)
$$
so
$$
\int_{[0,1]^2} c(x,y)\,dx\,dy = \sum_{n=0}^\infty B(n+1,n+1)^2 = \sum_{n=1}^\infty B(n,n)^2 = \sum_{n=1}^\infty \left(\frac{\Gamma(n)^2}{\Gamma(2n)}\right)^2
$$
$$
=\sum_{n=1}^\infty \frac{\Gamma(n)^4}{\Gamma(2n)^2} = {_3}F_2\left(1,1,1; \frac32,\frac32; \frac1{16}\right).
$$
(Wolfram Alpha link for the last equation)
