TITLE: Equating the integrals of 1/(1-xy) and 2/(1+xy) by elementary calculus?
QUESTION [11 upvotes]: The following integrals (inspired here)
are both equal to $\pi^2/6$:
$$\int_0^1\!\int_0^1 \frac{1}{1-xy}\,dx\,dy =
\int_0^1\!\int_0^1 \frac{2}{1+xy}\,dx\,dy.$$
According to conjecture 1 of Kontsevich and Zagier's article on periods, it should be possible to go from one to the other via other integrals of algebraic functions, using only the three rules of

additivity
change of variables
the Newton-Leibniz formula (a.k.a. the fundamental theorem of calculus),

What sequence of those rules, if any, yields the above identity?

REPLY [9 votes]: It is enough to set $x=u^2, y=v^2$ in the first integral, then using
$$ \frac{4uv}{1-u^2 v^2} = 2\left(\frac{1}{1-uv}-\frac{1}{1+uv}\right).$$
This proves
$$ \iint_{(0,1)^2}\frac{du\,dv}{1-uv} = \iint_{(0,1)^2}\frac{2\,du\,dv}{1+uv} $$
through elementary algebraic manipulations. That is exactly the same as proving
$$ \zeta(2) = \sum_{n\geq 1}\frac{1}{n^2} = 2\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2} = 2\,\eta(2),$$
but does not require to expand $\frac{1}{1\pm uv}$ as a geometric series, then apply termwise integration.