Answer
$$\frac{{3\ln 2}}{2} + \frac{1}{{16}}e - 1$$
Work Step by Step
$$\eqalign{
& \int_1^{\ln 8} {\int_1^{\sqrt z } {\int_{\ln y}^{\ln 2y} {{e^{x + {y^2} - z}}} dxdydz} } \cr
& = \int_1^{\ln 8} {\int_1^{\sqrt z } {\int_{\ln y}^{\ln 2y} {{e^{{y^2} - z}}{e^x}} dxdydz} } \cr
& {\text{Integrate with respect to }}x \cr
& = \int_1^{\ln 8} {\int_1^{\sqrt z } {\left[ {{e^{{y^2} - z}}{e^x}} \right]_{\ln y}^{\ln 2y}dydz} } \cr
& = \int_1^{\ln 8} {\int_1^{\sqrt z } {\left[ {{e^{{y^2} - z}}\left( {{e^{\ln 2y}} - {e^{\ln y}}} \right)} \right]dydz} } \cr
& = \int_1^{\ln 8} {\int_1^{\sqrt z } {\left[ {{e^{{y^2} - z}}\left( {2y - y} \right)} \right]dydz} } \cr
& = \int_1^{\ln 8} {\int_1^{\sqrt z } {y{e^{{y^2} - z}}dydz} } \cr
& {\text{Integrate with respect to }}y \cr
& = \frac{1}{2}\int_1^{\ln 8} {\left[ {{e^{{y^2} - z}}} \right]} _1^{\sqrt z }dz \cr
& = \frac{1}{2}\int_1^{\ln 8} {\left[ {{e^{{{\left( {\sqrt z } \right)}^2} - z}} - {e^{{{\left( 1 \right)}^2} - z}}} \right]} dz \cr
& = \frac{1}{2}\int_1^{\ln 8} {\left( {{e^0} - {e^{1 - z}}} \right)} dz \cr
& = \frac{1}{2}\int_1^{\ln 8} {\left( {1 - {e^{1 - z}}} \right)} dz \cr
& {\text{Integrate}} \cr
& = \frac{1}{2}\left[ {z + {e^{1 - z}}} \right]_1^{\ln 8} \cr
& = \frac{1}{2}\left[ {\ln 8 + e{e^{ - \ln 8}}} \right] - \frac{1}{2}\left[ {1 + {e^{1 - 1}}} \right] \cr
& = \frac{1}{2}\left[ {\ln 8 + \frac{1}{8}e} \right] - 1 \cr
& = \frac{{3\ln 2}}{2} + \frac{1}{{16}}e - 1 \cr} $$
