Answer
$$-\frac{1}{{\ln x}} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{x{{\left( {\ln x} \right)}^2}}}} dx \cr
& {\text{Integrate by substitution method}} \cr
& {\text{Let }}u = \ln x,\,\,\,\,du = \frac{1}{x}dx,\,\,\,\,dx = xdu \cr
& {\text{Then}}{\text{,}} \cr
& \int {\frac{1}{{x{{\left( {\ln x} \right)}^2}}}} dx = \int {\frac{1}{{x{u^2}}}} \left( {xdu} \right) \cr
& {\text{Cancel common factor }}x \cr
& = \int {\frac{1}{{{u^2}}}} du \cr
& {\text{integrating}} \cr
& = - \frac{1}{u} + C \cr
& {\text{replacing }}u = \ln x \cr
& = - \frac{1}{{\ln x}} + C \cr} $$
