Answer
$$4\ln 2$$
Work Step by Step
$$\eqalign{
& \int_2^6 {\int_{2y}^{4y} {\frac{1}{x}} dxdy} \cr
& = \int_2^6 {\left[ {\int_{2y}^{4y} {\frac{1}{x}} dx} \right]dy} \cr
& {\text{solve the inner integral by treating }}y{\text{ as a constant}} \cr
& = \int_{2y}^{4y} {\frac{1}{x}} dx \cr
& = \left( {\ln \left| x \right|} \right)_{2y}^{4y} \cr
& {\text{evaluating the limits in the variable }}x \cr
& = \ln \left| {4y} \right| - \ln \left| {2y} \right| \cr
& {\text{use logarithmic properties}} \cr
& = \ln \left( {\frac{{4y}}{{2y}}} \right) \cr
& = \ln 2 \cr
& \cr
& \int_2^6 {\left[ {\int_{2y}^{4y} {\frac{1}{x}} dx} \right]dy} \cr
& = \ln 2\int_2^6 {dy} \cr
& {\text{integrating }} \cr
& = \ln 2\left( y \right)_2^6 \cr
& {\text{evaluating}} \cr
& = \ln 2\left( {6 - 2} \right) \cr
& {\text{simplifying}} \cr
& = 4\ln 2 \cr} $$
