Answer
$$\frac{3}{2}$$
Work Step by Step
$$\eqalign{
& \int_e^{{e^2}} {\frac{{\ln x}}{x}} dx \cr
& {\text{Set }}u = \ln x \to du = \frac{1}{x}dx \cr
& u = \ln x,{\text{ for }}x = e,\,\,u = 1 \cr
& u = \ln x,{\text{ for }}x = {e^2},\,\,u = 2 \cr
& {\text{using the substitution}} \cr
& \int_e^{{e^2}} {\frac{{\ln x}}{x}} dx = \int_1^2 u du \cr
& {\text{integrate}} \cr
& = \frac{1}{2}\left[ {{u^2}} \right]_1^2 \cr
& {\text{evaluate}} \cr
& = \frac{1}{2}\left[ {{{\left( 2 \right)}^2} - {{\left( 1 \right)}^2}} \right] \cr
& = \frac{1}{2}\left( {4 - 1} \right) \cr
& = \frac{3}{2} \cr} $$