Answer
$$\frac{1}{8}\ln \left( {1 + 4{{\ln }^2}y} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\ln y}}{{y + 4y{{\ln }^2}y}}} dy \cr
& {\text{Factoring the denomiantor}}{\text{, the GCF is }}y \cr
& = \int {\frac{{\ln y}}{{y\left( {1 + 4{{\ln }^2}y} \right)}}} dy \cr
& {\text{Integrate by the substitution method}}{\text{,}} \cr
& \,\,{\text{Let }}u = 1 + 4{\ln ^2}y,\,\,\,du = 0 + 8\ln y\left( {\frac{1}{y}} \right)dy,\,\,\,\,dy = \frac{{ydu}}{{8\ln y}} \cr
& \,\,{\text{Write the integrand in terms of }}u \cr
& \int {\frac{{\ln y}}{{y\left( {1 + 4{{\ln }^2}y} \right)}}} dy = \int {\frac{{\ln y}}{{y\left( u \right)}}} \left( {\frac{{ydu}}{{8\ln y}}} \right) \cr
& = \int {\frac{1}{{\left( u \right)}}} \left( {\frac{{du}}{8}} \right) \cr
& = \frac{1}{8}\int {\frac{1}{u}} du \cr
& {\text{Integrating}} \cr
& = \frac{1}{8}\ln \left| u \right| + C \cr
& {\text{Write in terms of }}y,{\text{ substitute }}1 + 4{\ln ^2}y{\text{ for }}y \cr
& = \frac{1}{8}\ln \left| {1 + 4{{\ln }^2}y} \right| + C \cr
& = \frac{1}{8}\ln \left( {1 + 4{{\ln }^2}y} \right) + C \cr} $$
