Answer
$\dfrac{5}{3}$
Work Step by Step
Here, we have
$ \int_{1}^2 \int_{0}^{2z} \int_{0}^{\ln x} xe^{-y} dy dx dz=- \int_{1}^2 \int_{0}^{2z} [xe^{-y}]_{0}^{\ln x} dx dz$
$\implies \int_1^2[-x+\dfrac{x^2}{2}]_0^{2z} dz= -\int^{1}_{2}[2z-2z^2] dz$
$\implies - [z^2-\dfrac{2}{3}z^3]_{1}^2=-[4-\dfrac{16}{3}-1+\dfrac{2}{3}]$
Thus, we have $\int_{1}^2 \int_{0}^{2z} \int_{0}^{\ln x} xe^{-y} dy dx dz=\dfrac{5}{3}$