Answer
$$6 \ln 6-2 \ln 2-5 \ln 5 $$
Work Step by Step
\begin{align*}
\int_{1}^{2} \int_{0}^{4} \frac{d y d x}{x+y}&=\int_{1}^{2}\left(\int_{0}^{4} \frac{d y}{x+y}\right) d x\\
&=\left.\int_{1}^{2} \ln (x+y)\right|_{y=0} ^{4} d x\\
&=\int_{1}^{2}(\ln (x+4)-\ln x) d x
\end{align*}
Use
\begin{aligned}
\int_{1}^{2} \int_{0}^{4} \frac{d y d x}{x+y} &=(x+4)(\ln (x+4)-1)-\left.x(\ln x-1)\right|_{1} ^{2} \\
&=6(\ln 6-1)-2(\ln 2-1)-(5(\ln 5-1)-(\ln 1-1)) \\
&=6 \ln 6-2 \ln 2-5 \ln 5
\end{aligned}
