Answer
$$ - \frac{1}{2}{\left( {\ln x} \right)^{ - 2}} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\left( {\ln x} \right)}^{ - 3}}}}{x}} dx \cr
& {\text{integrate by the substitution method}} \cr
& {\text{set }}u = \ln x{\text{ then }}\frac{{du}}{{dx}} = \frac{1}{x},\,\,\,\,dx = xdu \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{{{{\left( {\ln x} \right)}^{ - 3}}}}{x}} dx = \int {\frac{{{u^{ - 3}}}}{x}} \left( {xdu} \right) \cr
& {\text{cancel common terms}} \cr
& = \int {{u^{ - 3}}} du \cr
& {\text{integrating}}{\text{,}} \cr
& = \frac{{{u^{ - 2}}}}{{ - 2}} + C \cr
& = - \frac{1}{2}{u^{ - 2}} + C \cr
& {\text{replace }}\ln x{\text{ for }}u \cr
& = - \frac{1}{2}{\left( {\ln x} \right)^{ - 2}} + C \cr} $$
