Solution
= Solution
Choose $u=x-y$ and $v=x+y$. Then $0\le u\le1$, $1\le v\le3$, while $x=(v+u)/2$, $y=(v-u)/2$ and the inverse <Jacobian determinant> has magnitude $1/2$. Hence
$$
\iint_R\left(\frac{x-y}{x+y+3}\right)^2\,dx\,dy
=\frac12\int_0^1u^2\,du\int_1^3(v+3)^{-2}\,dv
=\frac16\left(\frac14-\frac16\right)
=\boxed{\frac1{72}.}
$$
The denominator is bounded away from zero on the transformed rectangle, so no singularity or improper-integral qualification is needed.