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