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

Calculate the double integral.
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broliY
Once again use the fact that:
$\int \frac{{f}^{\prime }\left(x\right)}{f\left(x\right)}dx=\mathrm{ln}|f\left(x\right)|+C$
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:
${\int }_{1}^{3}{\int }_{1}^{2}\frac{1}{1+x+y}dydx={\int }_{1}^{3}\mathrm{ln}\left(x+3\right)-\mathrm{ln}\left(x+2\right)$
Now using the integration by parts we obtain that the integral is equal to:
$\left(\left(x+3\right)\mathrm{ln}\left(x+3\right)-\left(x+3\right)\right)-\left(\left(x+2\right)\mathrm{ln}\left(x+2\right)-\left(x+2\right)\right){\mid }_{1}^{3}=6\mathrm{ln}6-5\mathrm{ln}5-4\mathrm{ln}4+3\mathrm{ln}3$
Result: $6\mathrm{ln}6-5\mathrm{ln}5-4\mathrm{ln}4+3\mathrm{ln}3$