Answer
$$\frac{{{{\ln }^3}2}}{6}$$
Work Step by Step
$$\eqalign{
& \int_0^{\ln 2} {\int_{{e^y}}^2 {\frac{y}{x}dx} dy} \cr
& {\text{Integrate with respect to }}x \cr
& = \int_0^{\ln 2} {\left[ {y\ln \left| x \right|} \right]_{{e^y}}^2dy} \cr
& {\text{Evaluating the limits}} \cr
& = \int_0^{\ln 2} {\left[ {y\ln \left| 2 \right| - y\ln \left| {{e^y}} \right|} \right]dy} \cr
& = \int_0^{\ln 2} {\left( {y\ln 2 - {y^2}} \right)dy} \cr
& {\text{Integrate}} \cr
& = \left[ {\frac{{\ln 2}}{2}{y^2} - \frac{1}{3}{y^3}} \right]_0^{\ln 2} \cr
& {\text{Evaluating}} \cr
& = \left[ {\frac{{\ln 2}}{2}{{\left( {\ln 2} \right)}^2} - \frac{1}{3}{{\left( {\ln 2} \right)}^3}} \right] - \left[ {\frac{{\ln 2}}{2}{{\left( 0 \right)}^2} - \frac{1}{3}{{\left( 0 \right)}^3}} \right] \cr
& = \frac{{{{\ln }^3}2}}{2} - \frac{{{{\ln }^3}2}}{3} \cr
& = \frac{{{{\ln }^3}2}}{6} \cr} $$
