Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} \frac{y}{{x + 1}}{\rm{d}}A = 8\ln 3$
Work Step by Step
We have $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} \frac{y}{{x + 1}}{\rm{d}}A$ and ${\cal R} = \left[ {0,2} \right] \times \left[ {0,4} \right]$.
Write
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} \frac{y}{{x + 1}}{\rm{d}}A = \mathop \smallint \limits_{x = 0}^2 \left( {\mathop \smallint \limits_{y = 0}^4 y{\rm{d}}y} \right)\frac{1}{{x + 1}}{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^2 \left( {\frac{1}{2}{y^2}|_0^4} \right)\frac{1}{{x + 1}}{\rm{d}}x$
$ = 8\mathop \smallint \limits_{x = 0}^2 \frac{1}{{x + 1}}{\rm{d}}x$
$ = 8\left( {\ln \left( {x + 1} \right)|_0^2} \right)$
$ = 8\left( {\ln 3 + \ln 1} \right)$
$ = 8\ln 3$
