Answer
$$ - \frac{1}{{\ln x}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{x{{\left( {\ln x} \right)}^2}}}} \cr
& {\text{substitute }}u = \ln x \cr
& du = \frac{1}{x}dx \cr
& = \int {\frac{1}{{{{\left( {\ln x} \right)}^2}}}} \left( {\frac{1}{x}} \right)dx \cr
& = \int {\frac{1}{{{u^2}}}} du \cr
& {\text{write with negative exponent}} \cr
& = \int {{u^{ - 2}}} du \cr
& {\text{find the antiderivative}} \cr
& = \frac{{{u^{ - 1}}}}{{ - 1}} + C \cr
& = - \frac{1}{u} + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = \ln x \cr
& = - \frac{1}{{\ln x}} + C \cr} $$
