Answer
$$\frac{2}{5}{y^{5/2}}{\left( {\ln y} \right)^2} - \frac{8}{{25}}{y^{5/2}}\ln y + \frac{{16}}{{125}}{y^{5/2}} + C$$
Work Step by Step
$$\eqalign{
& \int {{y^{3/2}}{{\left( {\ln y} \right)}^2}dy} \cr
& {\text{Use the method of integration by parts}} \cr
& \,\,\,\,\,\,\,\,\,{\text{Let }}\,\,\,\,\,u = {\left( {\ln y} \right)^2},\,\,\,\,\,du = \frac{{2\ln y}}{y}dy \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = {y^{3/2}},\,\,\,\,\,v = \frac{2}{5}{y^{5/2}} \cr
& \int {udv} = uv - \int {vdu} \,\,\,\, \cr
& \to \,\,\,\int {{y^{3/2}}{{\left( {\ln y} \right)}^2}dy} = \frac{2}{5}{y^{5/2}}{\left( {\ln y} \right)^2} - \int {\left( {\frac{2}{5}{y^{5/2}}} \right)\left( {\frac{{2\ln y}}{y}} \right)} dy \cr
& = \frac{2}{5}{y^{5/2}}{\left( {\ln y} \right)^2} - \int {\frac{4}{5}{y^{3/2}}\ln y} dy \cr
& \cr
& {\text{Use the method of integration by parts again}} \cr
& \,\,\,\,\,\,\,\,\,{\text{Let }}\,\,\,\,\,u = \ln y,\,\,\,\,\,du = \frac{1}{y}dy \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = \frac{4}{5}{y^{3/2}},\,\,\,\,\,v = \frac{8}{{25}}{y^{5/2}} \cr
& = \frac{2}{5}{y^{5/2}}{\left( {\ln y} \right)^2} - \left( {\frac{8}{{25}}{y^{5/2}}\ln y - \int {\left( {\frac{8}{{25}}{y^{5/2}}} \right)\left( {\frac{1}{y}} \right)dy} } \right) \cr
& = \frac{2}{5}{y^{5/2}}{\left( {\ln y} \right)^2} - \frac{8}{{25}}{y^{5/2}}\ln y + \frac{8}{{25}}\int {{y^{3/2}}dy} \cr
& = \frac{2}{5}{y^{5/2}}{\left( {\ln y} \right)^2} - \frac{8}{{25}}{y^{5/2}}\ln y + \frac{8}{{25}}\left( {\frac{{{y^{5/2}}}}{{5/2}}} \right) + C \cr
& = \frac{2}{5}{y^{5/2}}{\left( {\ln y} \right)^2} - \frac{8}{{25}}{y^{5/2}}\ln y + \frac{{16}}{{125}}{y^{5/2}} + C \cr} $$
