Answer
$$\ \iint \limits_{R} \frac{1}{1+x+y} d A = 6\ln 6-5 \ln 5 - 4 \ln 4 +3\ln 3 $$
Work Step by Step
Given$$\ \iint \limits_{R} \frac{1}{1+x+y} d A, \quad R=[1,3] \times[1,2]$$
So, we get \begin{aligned} I&= \int_{1}^{2} \int_{1}^{3} \frac{1}{1+x+y} dx \ dy \\
&=\int_{1}^{2} \left[ \ln(1+x+y) \right]_{1}^{3} dy \\
&=\int_{1}^{2} \left[ \ln(1+3+y) \right] dy - \int_{1}^{2} \left[ \ln(1+1+y) \right] dy \\
&=\int_{1}^{2} \left[ \ln(4+y) - \ln(2+y) \right] dy \\
\end{aligned}
So by partition technique Let $$u= \ln(4+y) - \ln(2+y)\\
\Rightarrow du=\left( \frac{1}{4+y}- \frac{1}{2+y}\right)dy$$ $$dv= dy \Rightarrow v=y $$ So, we get \begin{aligned} I&=uv|_{1}^{2}- \int_{1}^{2} vdu\\
&= y\ln(4+y)|_{1}^{2} - y \ln(2+y)|_{1}^{2}- \int_{1}^{2} \left( \frac{y}{4+y}- \frac{y}{2+y}\right)dy\\
&= 2\ln 6- \ln 5 - 2 \ln 4 +\ln 3- \int_{1}^{2} \left( \frac{4+y-4}{4+y}- \frac{2+y-2}{2+y}\right)dy\\
&= 2\ln 6- \ln 5 - 2 \ln 4 +\ln 3- \int_{1}^{2} \left( 1-\frac{ 4}{4+y}- 1+\frac{ 2}{2+y}\right)dy\\
&= 2\ln 6- \ln 5 - 2 \ln 4 +\ln 3- \int_{1}^{2} \left( -\frac{ 4}{4+y}+\frac{ 2}{2+y}\right)dy\\
&= 2\ln 6- \ln 5 - 2 \ln 4 +\ln 3- \left( -4 \ln (4+y) +2\ln (2+y)\right)_{1}^{2}\\
&= 2\ln 6- \ln 5 - 2 \ln 4 +\ln 3- \left( -4 \ln 6 +2 \ln 4+4\ln 5- 2 \ln3\right)\\
&= 2\ln 6- \ln 5 - 2 \ln 4 +\ln 3+4 \ln 6 -2 \ln 4-4\ln 5+ 2 \ln3 \\
&= 6\ln 6-5 \ln 5 - 4 \ln 4 +3\ln 3 \\
\end{aligned}
