Answer
$$\frac{2}{3}$$
Work Step by Step
$$\eqalign{
& \int_1^e {\frac{{\sqrt {\ln x} }}{x}} dx \cr
& {\text{Use substitution}}\cr
&{\text{Let }}u = \ln x,{\text{ so that }}du = \frac{1}{x}dx \cr
& {\text{The new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = e,{\text{ then }}u = \ln e = 1 \cr
& \,\,\,\,\,\,{\text{If }}x = 1,{\text{ then }}u = \ln 1 = 0 \cr
& {\text{Then}} \cr
& \int_1^e {\frac{{\sqrt {\ln x} }}{x}} dx = \int_0^1 {\sqrt u } du \cr
& = \int_0^1 {{u^{1/2}}} du \cr
& {\text{integrate}} \cr
& = \left( {\frac{{{u^{3/2}}}}{{3/2}}} \right)_0^1 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = \frac{{{1^{3/2}}}}{{3/2}} - \frac{{{0^{3/2}}}}{{3/2}} \cr
& {\text{simplifying, we get:}} \cr
& = \frac{2}{3} \cr} $$
