Chapter 15 - Multiple Integrals - Review - Exercises - Page 1103: 55

$-\ln 2$

Let us consider that $x-y =u $ and $y=x+y$ Therefore, $x=\dfrac{u+v}{2}; y=\dfrac{v-u}{2}$ Now, we have: $Jacobin =|\dfrac{1}{2}|$ and $\iint_{R}\dfrac{x-y}{x+y} dA=\dfrac{1}{2} \iint_{D} uv^{-1} dA=\dfrac{1}{2} \int_{-2}^{0} \int_2^4 uv^{-1} dv \space du$ $\implies =\dfrac{1}{2} \times [\dfrac{u^2}{2}]_{-2}^0 \times [\ln v]_{2}^4$ $\implies =-\ln 2$