Answer
$$9$$
Work Step by Step
$$\eqalign{
& \int_0^{\ln 10} {\int_{{e^x}}^{10} {\frac{1}{{\ln y}}} dydx} \cr
& = \int_{x = 0}^{x = \ln 10} {\int_{y = {e^x}}^{y = 0} {\frac{1}{{\ln y}}} dydx} \cr
& y = {e^x},{\text{ }}x = \ln y \cr
& \left( {{\text{Graph below}}} \right) \cr
& {\text{Switching the order of integration, the new region }}R{\text{ is}} \cr
& R = \left\{ {\left. {\left( {x,y} \right)} \right|0 \leqslant x \leqslant \ln y,{\text{ }}1 \leqslant y \leqslant 10} \right\} \cr
& {\text{Therefore,}} \cr
& \int_0^{\ln 10} {\int_{{e^x}}^{10} {\frac{1}{{\ln y}}} dydx} = \int_1^{10} {\int_0^{\ln y} {\frac{1}{{\ln y}}} dxdy} \cr
& = \int_1^{10} {\left[ {\int_0^{\ln y} {\frac{1}{{\ln y}}} dx} \right]dy} \cr
& {\text{Integrate with respect to }}x \cr
& = \int_1^{10} {\left[ {\frac{x}{{\ln y}}} \right]_0^{\ln y}dy} \cr
& = \int_1^{10} {\left[ {\frac{{\ln y}}{{\ln y}} - \frac{0}{{\ln y}}} \right]dy} \cr
& = \int_1^{10} {dy} \cr
& {\text{Integrate}} \cr
& = \left[ y \right]_1^{10} \cr
& = 10 - 1 \cr
& = 9 \cr} $$
