Answer
$6$
Work Step by Step
We have the triple integral:
$\int^e_1 \int^{e^2}_1 \int^{e^3}_1 \dfrac{1}{xyz} dx \space dy \space dz= \int^e_1 \int^{e^2}_1 [\dfrac{\ln (x)}{yz}]^{e^3}_1 dy \space dz $
or, $=\int^e_1 \int^{e^2}_1 \dfrac{3}{yz} dy \space dz $
or, $=3 \int^e_1 [\dfrac{ln y}{z}]^{e^2}_1 dz $
Now, $\int^e_1 \int^{e^2}_1 \int^{e^3}_1 \dfrac{1}{xyz} dx \space dy \space dz=\int^e_1 \dfrac{6}{z} dz= 6 [\ln z]_1^e =6$
