Chapter 13 - Multiple Integration - 13.1 Double Integrals over Rectangular Regions - 13.1 Exercises - Page 972: 43

$$\frac{1}{2}$$

$$\eqalign{ & \int_1^2 {\int_1^2 {\frac{x}{{x + y}}dydx} } \cr & {\text{Integrate with respect to }}y \cr & = \int_1^2 {\left[ {x\ln \left( {x + y} \right)} \right]_1^2} dx \cr & {\text{Evaluate the limits of integration}} \cr & = \int_1^2 {\left[ {x\ln \left( {x + 2} \right) - x\ln \left( {x + 1} \right)} \right]} dx \cr & = \int_1^2 {x\ln \left( {\frac{{x + 2}}{{x + 1}}} \right)} dx \cr & {\text{Integrate by parts}} \cr & u = \ln \left( {\frac{{x + 2}}{{x + 1}}} \right) \Rightarrow - \frac{1}{{\left( {x + 2} \right)\left( {x + 1} \right)}} \cr & dv = x \Rightarrow v = \frac{1}{2}{x^2} \cr & = \left[ {\frac{1}{2}{x^2}\ln \left( {\frac{{x + 2}}{{x + 1}}} \right)} \right]_1^2 + \frac{1}{2}\int_1^2 {\frac{{{x^2}}}{{\left( {x + 2} \right)\left( {x + 1} \right)}}} dx \cr & = \left[ {\frac{1}{2}{x^2}\ln \left( {\frac{{x + 2}}{{x + 1}}} \right)} \right]_1^2 + \frac{1}{2}\int_1^2 {\left( {1 - \frac{{3x + 2}}{{\left( {x + 2} \right)\left( {x + 1} \right)}}} \right)} dx \cr & = \left[ {\frac{1}{2}{x^2}\ln \left( {\frac{{x + 2}}{{x + 1}}} \right)} \right]_1^2 + \frac{1}{2}\int_1^2 {\left( {1 - \frac{4}{{x + 2}} + \frac{1}{{x + 1}}} \right)} dx \cr & = \left[ {\frac{1}{2}{x^2}\ln \left( {\frac{{x + 2}}{{x + 1}}} \right)} \right]_1^2 + \frac{1}{2}\left[ {x - 4\ln \left( {x + 2} \right) + \ln \left( {x + 1} \right)} \right]_1^2 \cr & {\text{Evaluating}} \cr & \cr & = 2\ln \frac{4}{3} - \frac{1}{2}\ln \frac{3}{2} + \frac{1}{2}\left[ {2 - 4\ln 4 + \ln 3} \right] - \frac{1}{2}\left[ {1 - 4\ln 3 + \ln 2} \right] \cr & {\text{Simplifying}} \cr & = \ln \left( {\frac{{16}}{9}} \right) - \ln \left( {\frac{{\sqrt 3 }}{{\sqrt 2 }}} \right) + \frac{1}{2}\left[ {2 - \ln 256 + \ln 3} \right] - \frac{1}{2}\left[ {1 - \ln 81 + \ln 2} \right] \cr & = \ln \left( {\frac{{16}}{9}} \right) - \ln \left( {\frac{{\sqrt 3 }}{{\sqrt 2 }}} \right) + 1 - \ln 16 + \ln \sqrt 3 - \frac{1}{2} + \ln 9 - \ln \sqrt 2 \cr & = \ln 16 - \ln 9 - \ln \sqrt 3 + \ln \sqrt 2 + \frac{1}{2} - \ln 16 + \ln \sqrt 3 + \ln 9 - \ln \sqrt 2 \cr & = \frac{1}{2} \cr} $$