Answer
$$\frac{{{z^2}}}{2}{\left( {\ln z} \right)^2} - \frac{{{z^2}}}{2}\ln z + \frac{1}{4}{z^2} + C $$
Work Step by Step
$$\eqalign{
& \int {z{{\left( {\ln z} \right)}^2}} dz \cr
& {\text{Using integration by parts method }} \cr
& \,\,\,\,\,{\text{Let }}u = {\left( {\ln z} \right)^2},\,\,\,\,du = \frac{{2\left( {\ln z} \right)}}{z}dz,\,\,\,\,dv = zdz,\,\,\,\,v = \frac{{{z^2}}}{2} \cr
& {\text{Integration by parts then gives}} \cr
& = {\left( {\ln z} \right)^2}\left( {\frac{{{z^2}}}{2}} \right) - \int {\left( {\frac{{{z^2}}}{2}} \right)\left( {\frac{{2\left( {\ln z} \right)}}{z}} \right)dz} \cr
& = \frac{{{z^2}}}{2}{\left( {\ln z} \right)^2} - \int {z\ln zdz} \cr
& \cr
& {\text{Integrate by parts again}} \cr
& \,\,\,\,\,{\text{Let }}u = \ln z,\,\,\,\,du = \frac{1}{z}dz,\,\,\,\,dv = zdz,\,\,\,\,v = \frac{{{z^2}}}{2} \cr
& = \frac{{{z^2}}}{2}{\left( {\ln z} \right)^2} - \left( {\frac{{{z^2}}}{2}\ln z - \int {\left( {\frac{{{z^2}}}{2}} \right)\left( {\frac{1}{z}dz} \right)} } \right) \cr
& = \frac{{{z^2}}}{2}{\left( {\ln z} \right)^2} - \frac{{{z^2}}}{2}\ln z + \frac{1}{2}\int z dz \cr
& = \frac{{{z^2}}}{2}{\left( {\ln z} \right)^2} - \frac{{{z^2}}}{2}\ln z + \frac{1}{2}\left( {\frac{{{z^2}}}{2}} \right) + C \cr
& = \frac{{{z^2}}}{2}{\left( {\ln z} \right)^2} - \frac{{{z^2}}}{2}\ln z + \frac{1}{4}{z^2} + C \cr} $$
