Answer
$$\int_{1}^{4} \int_{1}^{e^{2}} \int_{0}^{1 / x z} \ln z \ \ d y d z d x=2 \ln 4$$
Work Step by Step
Given $$\int_{1}^{4} \int_{1}^{e^{2}} \int_{0}^{1 / x z} \ln z \ \ d y d z d x$$
So, we have
\begin{align}
I&=\int_{1}^{4} \int_{1}^{e^{2}} \int_{0}^{1 / x z} \ln z \ \ d y d z d x\\
&=\int_{1}^{4} \int_{1}^{e^{2}}[(\ln z) y]_{0}^{1 / x z} d z d x\\
&=\int_{1}^{4} \int_{1}^{e^{2}}[(\ln z) ( \frac{1}{x z}-0)] d z d x\\
&=\int_{1}^{4} \int_{1}^{e^{2}} \frac{\ln z}{x z} d z d x\\
&=\int_{1}^{4}\left[\frac{(\ln z)^{2}}{2 x}\right]_{1}^{e^{2}} d x\\
&=\int_{1}^{4}\left[\frac{(\ln e^2)^{2}-\ln 1}{2 x}\right] d x\\
&=\int_{1}^{4}\left[\frac{4}{2 x}\right] d x\\
&=\int_{1}^{4} \frac{2}{x} d x\\
&=[2 \ln |x|]_{1}^{4}\\
&=[2 \ln |4|-\ln1] \\
&=2 \ln 4
\end{align}
