Past exam of the mathematical methods II course of the University of Oxford / 2016 resit / 9 / c / Solution

Choose $u=x-y$ and $v=x+y$. Then $0\le u\le 1$, $1\le v\le 3$, while $x=(v+u)/2$, $y=(v-u)/2$ and the inverse Jacobian determinant has magnitude $1/2$. HenceThe denominator is bounded away from zero on the transformed rectangle, so no singularity or improper-integral qualification is needed.