Answer
$$\ln 3$$
Work Step by Step
$$\eqalign{
& \int_1^2 {\int_2^3 {\left( {\frac{1}{x} + \frac{1}{y}} \right)} dxdy} \cr
& \int_1^2 {\left[ {\int_2^3 {\left( {\frac{1}{x} + \frac{1}{y}} \right)} dx} \right]dy} \cr
& {\text{Evaluate inner integral}} \cr
& \int_2^3 {\left( {\frac{1}{x} + \frac{1}{y}} \right)} dx = \left[ {\ln \left| x \right| + \frac{x}{y}} \right]_{x = 2}^{x = 3} \cr
& = \left[ {\ln \left| 3 \right| + \frac{3}{y}} \right] - \left[ {\ln \left| 2 \right| + \frac{2}{y}} \right] \cr
& = \ln \left| 3 \right| + \frac{3}{y} - \ln \left| 2 \right| - \frac{2}{y} \cr
& = \ln \left( {\frac{3}{2}} \right) + \frac{1}{y} \cr
& {\text{Therefore,}} \cr
& \int_1^2 {\int_2^3 {\left( {\frac{1}{x} + \frac{1}{y}} \right)} dxdy} = \int_1^2 {\left[ {\ln \left( {\frac{3}{2}} \right) + \frac{1}{y}} \right]} dy \cr
& {\text{Integrating}} \cr
& {\text{ = }}\left[ {\ln \left( {\frac{3}{2}} \right)y + \ln \left| y \right|} \right]_1^2 \cr
& = \left[ {\ln \left( {\frac{3}{2}} \right)\left( 2 \right) + \ln \left| 2 \right|} \right] - \left[ {\ln \left( {\frac{3}{2}} \right)\left( 1 \right) + \ln \left| 1 \right|} \right] \cr
& = 2\ln \left( {\frac{3}{2}} \right) + \ln 2 - \ln \left( {\frac{3}{2}} \right) \cr
& = \ln \left( {\frac{3}{2}} \right) + \ln 2 \cr
& = \ln 3 \cr} $$
