Answer
$\ln 2(e-1)$
Work Step by Step
The iterated integral can be calculated as:
$\iint_{D} f(x,y) d A= \int_{1}^{e} \int_{0}^{y} (x+y)^{-1} \ dx \ dy \\= \int_{1}^{e} [\ln (x+y)]_0^y \ dy\\= \int_{1}^{e} (\ln 2y-\ln y) \ dy \\= \int_1^e \ln 2 \ dy \\=[y \ln 2][_1^e \\=e \ln 2 -\ln 2 \\=\ln 2(e-1)$
