Answer
$$2\left( {\sqrt 2 - 1} \right)$$
Work Step by Step
$$\eqalign{
& \int_e^{{e^2}} {\frac{1}{{x\sqrt {\ln x} }}} dx \cr
& {\text{use substitution}}{\text{: }} \cr
& {\text{ }}u = \ln x,{\text{ so that }}\frac{{du}}{{dx}} = \frac{1}{x},\,\,\,dx = xdu \cr
& {\text{the new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = {e^2},{\text{ }}u = \ln {e^2} = 2 \cr
& \,\,\,\,\,\,{\text{If }}x = e,{\text{ }}u = \ln e = 1 \cr
& {\text{then}} \cr
& \int_e^{{e^2}} {\frac{1}{{x\sqrt {\ln x} }}} dx = \int_1^2 {\frac{1}{{x\sqrt u }}} \left( {xdu} \right) \cr
& = \int_1^2 {{u^{ - 1/2}}} du \cr
& {\text{integrating, we get:}} \cr
& = \left( {\frac{{{u^{1/2}}}}{{1/2}}} \right)_1^2 \cr
& = 2\left( {\sqrt u } \right)_1^2 \cr
& {\text{use the fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = 2\left( {\sqrt 2 - \sqrt 1 } \right) \cr
& {\text{simplifying, we get:}} \cr
& = 2\left( {\sqrt 2 - 1} \right) \cr} $$
