Answer
$$\frac{{118}}{3}$$
Work Step by Step
$$\eqalign{
& \int_1^3 {\int_x^{{x^2}} {\int_0^{\ln z} {x{e^y}dydzdx} } } \cr
& = \int_1^3 {\int_x^{{x^2}} {\left[ {\int_0^{\ln z} {x{e^y}dy} } \right]dzdx} } \cr
& {\text{Solve the inner integral}}{\text{, integrate with respect to }}y,{\text{ treat }}z{\text{ and }}x{\text{ as constants}} \cr
& \int_0^{\ln z} {x{e^y}dy} = \left[ {x{e^y}} \right]_0^{\ln z} \cr
& {\text{Evaluate the limits in the variable }}y \cr
& = x{e^{\ln z}} - x{e^0} \cr
& = xz - z \cr
& {\text{Then}}{\text{,}} \cr
& \int_1^3 {\int_x^{{x^2}} {\left[ {\int_0^{\ln z} {x{e^y}dy} } \right]dzdx} } = \int_1^3 {\int_x^{{x^2}} {\left( {xz - x} \right)dzdx} } \cr
& = \int_1^3 {\left( {\int_x^{{x^2}} {\left( {xz - x} \right)dz} } \right)dx} \cr
& {\text{Integrate with respect to }}z,{\text{ treat }}x{\text{ as a constant}} \cr
& \int_x^{{x^2}} {\left( {xz - x} \right)dz} = \left( {\frac{{x{z^2}}}{2} - xz} \right)_x^{{x^2}} \cr
& = \left( {\frac{{x{{\left( {{x^2}} \right)}^2}}}{2} - x\left( {{x^2}} \right)} \right) - \left( {\frac{{x{{\left( x \right)}^2}}}{2} - x\left( x \right)} \right) \cr
& = \frac{{{x^5}}}{2} - {x^3} - \frac{{{x^3}}}{2} + {x^2} \cr
& = \frac{{{x^5}}}{2} - \frac{{3{x^3}}}{2} + {x^2} \cr
& \int_1^3 {\left( {\int_x^{{x^2}} {\left( {xz - x} \right)dz} } \right)dx} = \int_1^3 {\left( {\frac{{{x^5}}}{2} - \frac{{3{x^3}}}{2} + {x^2}} \right)dx} \cr
& {\text{Integrate}} \cr
& = \left( {\frac{{{x^6}}}{{12}} - \frac{{3{x^4}}}{8} + \frac{{{x^3}}}{3}} \right)_1^3 \cr
& {\text{Evaluate}} \cr
& = \left( {\frac{{{{\left( 3 \right)}^6}}}{{12}} - \frac{{3{{\left( 3 \right)}^4}}}{8} + \frac{{{{\left( 3 \right)}^3}}}{3}} \right) - \left( {\frac{{{{\left( 1 \right)}^6}}}{{12}} - \frac{{3{{\left( 1 \right)}^4}}}{8} + \frac{{{{\left( 1 \right)}^3}}}{3}} \right) \cr
& = \frac{{315}}{8} - \frac{1}{{24}} \cr
& = \frac{{118}}{3} \cr} $$
