Answer
$${\left( {\ln 3} \right)^2}$$
Work Step by Step
$$\eqalign{
& \int_1^3 {\int_1^3 {\frac{1}{{xy}}} } dydx \cr
& = \int_{16}^{25} {\left[ {\int_1^3 {\frac{1}{{xy}}} dy} \right]} dx \cr
& {\text{solve the inner integral}} \cr
& \int_1^3 {\frac{1}{{xy}}} dy = \frac{1}{x}\int_1^3 {\frac{1}{y}} dy \cr
& = \frac{1}{x}\left[ {\ln \left| y \right|} \right]_1^3 \cr
& = \frac{1}{x}\left( {\ln 3 - \ln 1} \right) \cr
& = \frac{{\ln 3}}{x} \cr
& {\text{then}} \cr
& \int_1^3 {\int_1^3 {\frac{1}{{xy}}} } dydx = \int_1^3 {\frac{{\ln 3}}{x}} dx \cr
& {\text{integrating}} \cr
& = \ln 3\left[ {\ln \left| x \right|} \right]_1^3 \cr
& = \ln 3\left( {\ln 3 - \ln 1} \right) \cr
& = {\left( {\ln 3} \right)^2} \cr} $$