Answer
$$\ln 2$$
Work Step by Step
$$\eqalign{
& \int_e^{{e^2}} {\frac{{dx}}{{x\ln x}}} \cr
& {\text{substitute }}u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = {e^2},{\text{ }}u = \ln {e^2} = 2 \cr
& {\text{if }}x = e,{\text{ }}u = \ln e = 1 \cr
& {\text{so}} \cr
& \int_e^{{e^2}} {\frac{{dx}}{{x\ln x}}} = \int_1^2 {\frac{1}{u}} du \cr
& {\text{find the antiderivative}} \cr
& = \left. {\left( {\ln \left| u \right|} \right)} \right|_1^2 \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = \ln \left( 2 \right) - \ln \left( 1 \right) \cr
& {\text{simplify}} \cr
& = \ln 2 \cr} $$
