Question

An integral a day:
Day 5.
\[-4 \int_0^1\int_0^1 \frac{dxdy}{ln(xy)(1+(xy)^2)}\]

Answer 1

do a change of variables...w=xy

then use Fubini's theorem to change the order of integration...then it is trivial

Answer 2

answer is \(\pi\)

Answer 3

Mathematica lets me down again...it is unable to do this integral

Answer 4

Good job zarkon, Here's the math behind his words:
\[w=xy~~~~dw=ydx~~~w(0)=0 ~~~~w(1)=y\]\[-4\int_0^1\int_0^y \frac{dw~dy}{y\ln(w)(1+w^2)}\]\[-4\int_0^1\int_w^1\frac{dy~dw}{y\ln(w)(1+w^2)}\]\[-4\int_0^1 \frac{dw}{\ln(w)(1+w^2)}(\ln(1)-\ln(w))\]\[4\int_0^1 \frac{dw}{1+w^2}\]\[\arctan(w)|_0^1\]\[ \pi\]