Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = 0}^4 \mathop \smallint \limits_{y = 0}^{ - 2x + 8} {\left( {x + y + 1} \right)^{ - 2}}{\rm{d}}y{\rm{d}}x = \ln \frac{{25}}{9}$
Work Step by Step
We can consider the triangle with vertices $\left( {0,0} \right)$, $\left( {4,0} \right)$, and $\left( {0,8} \right)$ as a vertically simple region. From the figure attached, we see that the top boundary is the line $y = - 2\left( {x - 4} \right) = - 2x + 8$.
So, the domain description is given by
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 4,0 \le y \le - 2x + 8} \right\}$
We integrate $f\left( {x,y} \right) = {\left( {x + y + 1} \right)^{ - 2}}$ over the triangle as an iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = 0}^4 \mathop \smallint \limits_{y = 0}^{ - 2x + 8} {\left( {x + y + 1} \right)^{ - 2}}{\rm{d}}y{\rm{d}}x$
$ = - \mathop \smallint \limits_{x = 0}^4 \left( {{{\left( {x + y + 1} \right)}^{ - 1}}|_0^{ - 2x + 8}} \right){\rm{d}}x$
$ = - \mathop \smallint \limits_{x = 0}^4 \left( {{{\left( { - x + 9} \right)}^{ - 1}} - {{\left( {x + 1} \right)}^{ - 1}}} \right){\rm{d}}x$
$ = - \mathop \smallint \limits_{x = 0}^4 \frac{1}{{ - x + 9}}{\rm{d}}x + \mathop \smallint \limits_{x = 0}^4 \frac{1}{{x + 1}}{\rm{d}}x$
$ = \ln \left( { - x + 9} \right)|_0^4 + \ln \left( {x + 1} \right)|_0^4$
$ = \ln 5 - \ln 9 + \ln 5 - \ln 1$
$ = \ln 25 - \ln 9$
$ = \ln \frac{{25}}{9}$
