Answer
$$ - \frac{{\ln x}}{x} - \frac{1}{x} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{\ln x}}{{{x^2}}}} dx \cr
& {\text{write with negative exponent}} \cr
& = \int {{x^{ - 2}}\ln x} dx \cr
& {\text{Using integration by parts method }} \cr
& \,\,\,\,\,{\text{Let }}u = \ln x,\,\,\,\,du = \frac{1}{x}dx,\,\,\,\,\,dv = {x^{ - 2}}dx,\,\,\,\,v = - {x^{ - 1}} \cr
& \cr
& {\text{Integration by parts then gives}} \cr
& \int {{x^{ - 2}}\ln x} dx = \left( {\ln x} \right)\left( { - {x^{ - 1}}} \right) - \int {\left( { - {x^{ - 1}}} \right)} \left( {\frac{1}{x}dx} \right) \cr
& \int {{x^{ - 2}}\ln x} dx = - \frac{{\ln x}}{x} + \int {{x^{ - 2}}} dx \cr
& \cr
& {\text{Integrate by the power rule}} \cr
& \int {{x^{ - 2}}\ln x} dx = - \frac{{\ln x}}{x} + \frac{{{x^{ - 1}}}}{{ - 1}} + C \cr
& \int {{x^{ - 2}}\ln x} dx = - \frac{{\ln x}}{x} - \frac{1}{x} + C \cr} $$
