Chapter 15 - Section 15.5 - Double Integrals and Applications - Exercises - Page 1135: 16

$$\frac{{115}}{{96}}$$

$$\eqalign{ & \int_1^2 {\int_{1 - 2x}^{{x^2}} {\frac{{x + 1}}{{{{\left( {2x + y} \right)}^3}}}} dydx} \cr & \int_1^2 {\left[ {\int_{1 - 2x}^{{x^2}} {\frac{{x + 1}}{{{{\left( {2x + y} \right)}^3}}}} dy} \right]dx} \cr & \int_1^2 {\left( {x + 1} \right)\left[ {\int_{1 - 2x}^{{x^2}} {\frac{1}{{{{\left( {2x + y} \right)}^3}}}} dy} \right]dx} \cr & {\text{Evaluate inner integral}} \cr & \int_{1 - 2x}^{{x^2}} {\frac{1}{{{{\left( {2x + y} \right)}^3}}}} dy = \int_{1 - 2x}^{{x^2}} {{{\left( {2x + y} \right)}^{ - 3}}} dy \cr & {\text{Recall that }}\int {{u^n}du} = \frac{{{u^{n + 1}}}}{{n + 1}} + C \cr & = \left[ {\frac{{{{\left( {2x + y} \right)}^{ - 2}}}}{{ - 2}}} \right]_{y = 1 - 2x}^{y = {x^2}} \cr & = - \frac{1}{2}\left[ {{{\left( {2x + y} \right)}^{ - 2}}} \right]_{y = 1 - 2x}^{y = {x^2}} \cr & = - \frac{1}{2}\left[ {{{\left( {2x + {x^2}} \right)}^{ - 2}} - {{\left( {2x + 1 - 2x} \right)}^{ - 2}}} \right] \cr & = - \frac{1}{2}\left[ {{{\left( {2x + {x^2}} \right)}^{ - 2}} - {{\left( 1 \right)}^{ - 2}}} \right] \cr & = - \frac{1}{2}\left[ {\frac{1}{{{{\left( {2x + {x^2}} \right)}^2}}} - 1} \right] \cr & = \frac{1}{2}\left[ {1 - \frac{1}{{{{\left( {2x + {x^2}} \right)}^2}}}} \right] \cr & {\text{Therefore,}} \cr & \int_1^2 {\left( {x + 1} \right)\left[ {\int_{1 - 2x}^{{x^2}} {\frac{1}{{{{\left( {2x + y} \right)}^3}}}} dy} \right]dx} \cr & = \frac{1}{2}\int_1^2 {\left( {x + 1} \right)\left[ {1 - \frac{1}{{{{\left( {2x + {x^2}} \right)}^2}}}} \right]dx} \cr & {\text{Distribute}} \cr & = \frac{1}{2}\int_1^2 {\left[ {\left( {x + 1} \right) - \frac{{x + 1}}{{{{\left( {2x + {x^2}} \right)}^2}}}} \right]dx} \cr & = \frac{1}{2}\int_1^2 {\left( {x + 1} \right)dx} - \frac{1}{2}\int_1^2 {\frac{{x + 1}}{{{{\left( {2x + {x^2}} \right)}^2}}}dx} \cr & = \frac{1}{2}\int_1^2 {\left( {x + 1} \right)dx} - \frac{1}{4}\int_1^2 {\frac{{2 + 2x}}{{{{\left( {2x + {x^2}} \right)}^2}}}dx} \cr & {\text{Integrating}} \cr & {\text{ = }}\left[ {\frac{{{{\left( {x + 1} \right)}^2}}}{4}} \right]_1^2 - \frac{1}{4}\left[ { - \frac{1}{{2x + {x^2}}}} \right]_1^2 \cr & {\text{ = }}\left[ {\frac{{{{\left( {x + 1} \right)}^2}}}{4} + \frac{1}{{4\left( {2x + {x^2}} \right)}}} \right]_1^2 \cr & {\text{ = }}\left[ {\frac{{{{\left( {2 + 1} \right)}^2}}}{4} + \frac{1}{{4\left( {2\left( 2 \right) + {{\left( 2 \right)}^2}} \right)}}} \right] - \left[ {\frac{{{{\left( {1 + 1} \right)}^2}}}{4} + \frac{1}{{4\left( {2\left( 1 \right) + {{\left( 1 \right)}^2}} \right)}}} \right] \cr & = \frac{{73}}{{32}} - \frac{{13}}{{12}} \cr & = \frac{{115}}{{96}} \cr} $$