Calculate the double integral. \int\int_R\frac{1}{1+x+y}dA,\ R=[1,3]\time

Once again use the fact that:
${\displaystyle \int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\ln \left|f\left(x\right)\right|+C}$
It doesn't matter do you opt to solve the first integral with respect to x or with respect to y, Here, we opt to solve the integral with respect to y first:
${\displaystyle {\int }_1^3{\int }_1^2\frac{1}{1+x+y}dydx={\int }_1^3\ln (x+3)-\ln (x+2)}$
Now using the integration by parts we obtain that the integral is equal to:
${\displaystyle ((x+3)\ln (x+3)-(x+3))-((x+2)\ln (x+2)-(x+2)){\mid }_1^3=6\ln 6-5\ln 5-4\ln 4+3\ln 3}$
Result: ${\displaystyle 6\ln 6-5\ln 5-4\ln 4+3\ln 3}$