Answer
$$\ln 2$$
Work Step by Step
$$\eqalign{
& \int_1^2 {\int_1^2 {\frac{1}{x}dx} dy} \cr
& \int_1^2 {\left[ {\int_1^2 {\frac{1}{x}dx} } \right]} dy \cr
& {\text{solve the inner integral}} \cr
& \int_1^2 {\frac{1}{x}dx} = \left[ {\ln \left| x \right|} \right]_1^2 \cr
& {\text{evaluate the limits}} \cr
& = \ln \left| 2 \right| - \ln \left| 1 \right| \cr
& = \ln 2 \cr
& \cr
& {\text{then}} \cr
& \int_1^2 {\left[ {\int_1^2 {\frac{1}{x}dx} } \right]} dy = \int_1^2 {\ln 2} dy \cr
& = \ln 2\int_1^2 {dy} \cr
& {\text{integrating}} \cr
& = \ln 2\left[ y \right]_1^2 \cr
& evaluating \cr
& = \ln 2\left( {2 - 1} \right) \cr
& = \ln 2 \cr} $$
