Answer
$$\frac{1}{2}$$
Work Step by Step
$$\eqalign{
& \int_1^e {\frac{{\ln x}}{x}} dx \cr
& {\text{Let }}u = \ln x,{\text{ }}du = \frac{1}{2}dx \cr
& {\text{The new limits of integration are}} \cr
& x = e \to u = 1 \cr
& x = 1 \to u = 0 \cr
& {\text{Substitute and integrate}} \cr
& \int_1^e {\frac{{\ln x}}{x}} dx = \int_0^1 u du \cr
& = \left[ {\frac{{{u^2}}}{2}} \right]_0^1 \cr
& = \frac{{{{\left( 1 \right)}^2}}}{2} - \frac{{{{\left( 0 \right)}^2}}}{2} \cr
& = \frac{1}{2} \cr} $$
