Answer
$$\frac{3}{{2\sqrt e }} - \frac{2}{e}$$
Work Step by Step
$$\eqalign{
& \int_{\sqrt e }^e {\frac{{\ln x}}{{{x^2}}}} dx \cr
& \int_{\sqrt e }^e {{x^{ - 2}}\ln x} dx \cr
& {\text{substitute }}u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr
& dv = {x^{ - 2}}dx,{\text{ }}v = \frac{{{x^{ - 1}}}}{{ - 1}} \cr
& {\text{applying integration by parts}}{\text{, we have}} \cr
& \int_{\sqrt e }^e {{x^{ - 2}}\ln x} dx = \left. {\left( {\frac{{{x^{ - 1}}}}{{ - 1}}\ln x} \right)} \right|_{\sqrt e }^e - \int_{\sqrt e }^e {\left( {\frac{{{x^{ - 1}}}}{{ - 1}}} \right)\left( {\frac{1}{x}dx} \right)} \cr
& \int_{\sqrt e }^e {{x^{ - 2}}\ln x} dx = \left. {\left( { - \frac{{\ln x}}{x}} \right)} \right|_{\sqrt e }^e + \int_{\sqrt e }^e {\frac{1}{{{x^2}}}} dx \cr
& {\text{integrating}} \cr
& \int_{\sqrt e }^e {{x^{ - 2}}\ln x} dx = \left. {\left( { - \frac{{\ln x}}{x}} \right)} \right|_{\sqrt e }^e - \left. {\left( {\frac{1}{x}} \right)} \right|_{\sqrt e }^e \cr
& \int_{\sqrt e }^e {{x^{ - 2}}\ln x} dx = \left. {\left( { - \frac{{\ln x + 1}}{x}} \right)} \right|_{\sqrt e }^e \cr
& {\text{evaluate limits}} \cr
& = \left( { - \frac{{\ln e + 1}}{e}} \right) + \left( {\frac{{\ln \sqrt e + 1}}{{\sqrt e }}} \right) \cr
& {\text{simplify}} \cr
& = - \frac{2}{e} + \frac{{3/2}}{{\sqrt e }} \cr
& = \frac{3}{{2\sqrt e }} - \frac{2}{e} \cr} $$
