Answer
$$\ln \left| {\ln x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{x\ln x}}} ,{\text{ with }}u = \ln x,{\text{ }}du = \frac{{dx}}{x} \cr
& {\text{using the indicated substitution}} \cr
& \int {\frac{{dx}}{{x\ln x}}} = \int {\frac{1}{u}du} \cr
& {\text{integrating}} \cr
& = \ln \left| u \right| + C \cr
& {\text{Back substitute }}u = \ln x \cr
& = \ln \left| {\ln x} \right| + C \cr} $$