Answer
$\frac{{2{{\left( {z - 1} \right)}^{5/2}}}}{5} + \frac{{2{{\left( {z - 1} \right)}^{3/2}}}}{3} + C$
Work Step by Step
$$\eqalign{
& \int {z\sqrt {z - 1} } dz,{\text{ }}u = z - 1 \cr
& {\text{Let }}u = z - 1 \to z = u + 1,{\text{ }}dz = du \cr
& {\text{Applying the substitution}}{\text{, we obtain}} \cr
& \int {z\sqrt {z - 1} } dz = \int {\left( {u + 1} \right)\sqrt u } du{\text{ }} \cr
& = \int {\left( {u + 1} \right){u^{1/2}}} du{\text{ }} \cr
& = \int {\left( {{u^{3/2}} + {u^{1/2}}} \right)} du{\text{ }} \cr
& {\text{Integrate apply the power rule }}\int {{u^n}du} = \frac{{{u^{n + 1}}}}{{n + 1}}{\text{ + C }} \cr
& = \frac{{{u^{5/2}}}}{{5/2}} + \frac{{{u^{3/2}}}}{{3/2}} + C \cr
& = \frac{{2{u^{5/2}}}}{5} + \frac{{2{u^{3/2}}}}{3} + C \cr
& {\text{Write in terms of }}t,{\text{ substitute }}u = z - 1 \cr
& = \frac{{2{{\left( {z - 1} \right)}^{5/2}}}}{5} + \frac{{2{{\left( {z - 1} \right)}^{3/2}}}}{3} + C \cr} $$
