Answer
$$\ln \left| {\ln v} \right| + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{dv}}{{v\ln v}}} \cr
& {\text{Integrate by the substitution method}} \cr
& {\text{set }}u = \ln v{\text{ then }}\frac{{du}}{{dv}} = \frac{1}{v},\,\,\,\,dv = vdu \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{{dv}}{{v\ln v}}} = \int {\frac{1}{{vu}}} \left( {vdu} \right) \cr
& {\text{cancel common terms}} \cr
& = \int {\frac{1}{u}} du \cr
& {\text{integrating}}{\text{,}} \cr
& = \ln \left| u \right| + C \cr
& {\text{replace }}\ln v{\text{ for }}u \cr
& = \ln \left| {\ln v} \right| + C \cr} $$
