Answer
$\ln 2$
Work Step by Step
Let $ u=\ln t $, then $ du =\frac{1}{t}dt $ and hence $ u $ changes from $\ln e=1$ to $\ln e^2=2$; thus we have:
$$\int_{e}^{e^2} \frac{1}{t\ln t}dt=\int_{1}^{2} \frac{du }{u}=\ln u|_{1}^{2}\\
=\ln 2-\ln 1=\ln2. $$
where used the properties $\ln e=1$ and $\ln B^x=x\ln B $.
