Answer
$$\int_{3}^{4} \int_{1}^{2} \frac{1}{(x+y)^{2}}\ d y \ d x =\ln(\frac{25}{24})$$
Work Step by Step
Given $$\int_{3}^{4} \int_{1}^{2} \frac{1}{(x+y)^{2}}\ d y \ d x$$
So, we have
\begin{aligned}
I&=\int_{3}^{4} \int_{1}^{2} \frac{1}{(x+y)^{2}}\ d y \ d x\\
&=\int_{3}^{4} \int_{1}^{2} {(x+y)^{-2}}\ d y \ d x\\
&=-\int_{3}^{4} {(x+y)^{-1}}|_{1}^{2} \ \ d x\\
&=-\int_{3}^{4} {(x+2)^{-1}} \ \ d x+\int_{3}^{4} {(x+1)^{-1}} \ \ d x\\
&=-\int_{3}^{4} \frac{1} {(x+2) } \ \ d x+\int_{3}^{4} \frac{1} {(x+1) } \ \ d x\\
&= \left[ \ -\ln(x+2)+\ln(x+1)\right]_{3}^{4}\\
&= \left[ \ -\ln(4+2)+\ln(4+1)\right]- \left[ \ -\ln(3+2)+\ln(3+1)\right]\\
&= -\ln(6)+\ln(5)+\ln(5)-\ln(4)\\
&= -\ln(6)+2\ln(5) -\ln(4)\\
&=\ln(\frac{25}{24})
\end{aligned}
