Answer
$ - \frac{{\ln x}}{x} - \frac{1}{x} + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{\ln x}}{{{x^2}}}} dx \cr
& {\text{Integrating by parts}}{\text{, }} \cr
& {\text{Let }}u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr
& {\text{ }}dv = \frac{1}{{{x^2}}}dx,{\text{ }}v = \frac{{{x^{ - 2 + 1}}}}{{ - 2 + 1}} = - {x^{ - 1}} = - \frac{1}{x} \cr
& {\text{Using integration by parts formula}} \cr
& \int u dv = uv - \int {vdu} \cr
& \int {\left( {\ln x} \right)\left( {\frac{1}{{{x^2}}}} \right)} dx = \left( {\ln x} \right)\left( { - \frac{1}{x}} \right) - \int {\left( { - \frac{1}{x}} \right)\left( {\frac{1}{x}} \right)dx} \cr
& {\text{Multiply}} \cr
& \int {\frac{{\ln x}}{{{x^2}}}} dx = - \frac{{\ln x}}{x} + \int {\frac{1}{{{x^2}}}} dx \cr
& {\text{Integrate using the power rule }}\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C, \cr
& \int {\frac{{\ln x}}{{{x^2}}}} dx = - \frac{{\ln x}}{x} + \frac{{{x^{ - 2 + 1}}}}{{ - 2 + 1}} + C \cr
& \int {\frac{{\ln x}}{{{x^2}}}} dx = - \frac{{\ln x}}{x} + \frac{{{x^{ - 1}}}}{{ - 1}} + C \cr
& \int {\frac{{\ln x}}{{{x^2}}}} dx = - \frac{{\ln x}}{x} - \frac{1}{x} + C \cr} $$
