Answer
$$2\left( {5 - e} \right)$$
Work Step by Step
$$\eqalign{
& \int_1^{\ln 5} {\int_0^{\ln 3} {{e^{x + y}}} } dxdy \cr
& = \int_1^{\ln 5} {\left[ {\int_0^{\ln 3} {{e^{x + y}}dx} } \right]} dy \cr
& {\text{use }}{e^{m + n}} = {e^m}{e^n} \cr
& = \int_1^{\ln 5} {\left[ {\int_0^{\ln 3} {{e^x}{e^y}dx} } \right]} dy \cr
& {\text{solve the inner integral}}{\text{, treat }}y{\text{ as a constant}} \cr
& = \int_0^{\ln 3} {{e^x}{e^y}dx} \cr
& = {e^y}\int_0^{\ln 3} {{e^x}dx} \cr
& = {e^y}\left[ {{e^x}} \right]_0^{\ln 3} \cr
& {\text{evaluating the limits for }}x \cr
& = {e^y}\left[ {{e^{\ln 3}} - {e^0}} \right] \cr
& = {e^y}\left( {3 - 1} \right) \cr
& = 2{e^y} \cr
& \cr
& \int_1^{\ln 5} {\left[ {\int_0^{\ln 3} {{e^x}{e^y}dx} } \right]} dy = \int_1^{\ln 5} {2{e^y}} dy \cr
& = 2\int_1^{\ln 5} {{e^y}} dy \cr
& {\text{integrating}} \cr
& = 2\left( {{e^y}} \right)_1^{\ln 5} \cr
& = 2\left( {{e^{\ln 5}} - {e^1}} \right) \cr
& {\text{simplifying}} \cr
& = 2\left( {5 - e} \right) \cr} $$
