Answer
$-\ln 2$
Work Step by Step
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 $\iint_{R}\dfrac{x-y}{x+y} dA=\dfrac{1}{2} \iint_{D} uv^{-1} dA=\dfrac{1}{2} \int_{-2}^{0} \int_2^4 uv^{-1} dv \space du$
$\implies =\dfrac{1}{2} \times [\dfrac{u^2}{2}]_{-2}^0 \times [\ln v]_{2}^4$
$\implies =-\ln 2$
