## Calculus with Applications (10th Edition)

$$3\ln \left( {\frac{4}{3}} \right)$$
\eqalign{ & \iint\limits_R {\frac{3}{{{{\left( {x + y} \right)}^2}}}}dydx;\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \leqslant x \leqslant 4,\,\,\,\,\,\,\,1 \leqslant y \leqslant 6 \cr & {\text{replacing the limits for the region }}R \cr & \int_2^4 {\int_1^6 {\frac{3}{{{{\left( {x + y} \right)}^2}}}dy} dx} \cr & = \int_2^4 {\left[ {\int_1^6 {\frac{3}{{{{\left( {x + y} \right)}^2}}}dy} } \right]dx} \cr & {\text{solve the inner integral }} \cr & \int_1^6 {\frac{3}{{{{\left( {x + y} \right)}^2}}}dy} \cr & {\text{write the integrand as}} \cr & = 3\int_1^6 {{{\left( {x + y} \right)}^{ - 2}}dx} \cr & {\text{integrate using the power rule}} \cr & = 3\left[ {\frac{{{{\left( {x + y} \right)}^{ - 1}}}}{{ - 1}}} \right]_1^6 \cr & = - 3\left[ {\frac{1}{{x + y}}} \right]_1^6 \cr & {\text{evaluating the limits in the variable }}x \cr & = - 3\left[ {\frac{1}{{x + 6}} - \frac{1}{{x + 1}}} \right] \cr & = 3\left( {\frac{1}{{x + 1}} - \frac{1}{{x + 6}}} \right) \cr & {\text{then}} \cr & \cr & \int_2^4 {\left[ {\int_1^6 {\frac{3}{{{{\left( {x + y} \right)}^2}}}dy} } \right]dx} = 3\int_1^6 {\left( {\frac{1}{{x + 1}} - \frac{1}{{x + 6}}} \right)dx} \cr & {\text{integrating}} \cr & = 3\left( {\ln \left| {x + 1} \right| - \ln \left| {x + 6} \right|} \right)_2^4 \cr & = 3\left( {\ln \left| {\frac{{x + 1}}{{x + 6}}} \right|} \right)_2^4 \cr & {\text{evaluate}} \cr & = 3\left( {\ln \left| {\frac{{4 + 1}}{{4 + 6}}} \right| - \ln \left| {\frac{{2 + 1}}{{2 + 6}}} \right|} \right) \cr & {\text{simplifying}} \cr & = 3\left( {\ln \left( {\frac{1}{2}} \right) - \ln \left( {\frac{3}{8}} \right)} \right) \cr & = 3\ln \left( {\frac{{1/2}}{{3/8}}} \right) \cr & = 3\ln \left( {\frac{4}{3}} \right) \cr}