Answer
$-\ln 2$
Work Step by Step
Suppose that $u=x-y ; y=x+y$
This implies that $x=\dfrac{u+v}{2}; y=\dfrac{v-u}{2}$
Now, $Jacobin =|\dfrac{1}{2}|$
Therefore, $\iint_{R}\dfrac{x-y}{x+y} dA=\dfrac{1}{2} \iint_{D} uv^{-1} dA$
or, $=\dfrac{1}{2} \int_{-2}^{0} \int_2^4 uv^{-1} dv \space du$
or, $ =\dfrac{1}{2} \times [\dfrac{u^2}{2}]_{-2}^0 \times [\ln v]_{2}^4$
or, $=-\ln 2$
