Answer
$\left( {\ln y} \right)\ln \left( {\ln y} \right) - \left( {\ln y} \right) + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{\ln \left( {\ln y} \right)}}{y}} dy \cr
& = \int {\ln \left( {\ln y} \right)\left( {\frac{1}{y}} \right)} dy \cr
& {\text{Let }}t = \ln y,{\text{ }}dt = \frac{1}{y}dy \cr
& {\text{Substituting}} \cr
& \int {\ln \left( {\ln y} \right)\left( {\frac{1}{y}} \right)} dy = \int {\ln t} dt \cr
& {\text{Integrating by parts}} \cr
& {\text{Let }}u = \ln t,{\text{ }}du = \frac{1}{t}dt \cr
& dv = dt,{\text{ }}v = t \cr
& {\text{Using integration by parts formula}} \cr
& \int {\ln t} dt = t\ln t - \int {t\left( {\frac{1}{t}} \right)} dt \cr
& = t\ln t - \int {dt} \cr
& = t\ln t - t + C \cr
& {\text{Write in terms of }}y,{\text{ substitute }}\ln y{\text{ for }}t \cr
& = \left( {\ln y} \right)\ln \left( {\ln y} \right) - \left( {\ln y} \right) + C \cr} $$