## Calculus: Early Transcendentals (2nd Edition)

$$\frac{{2{{\left( {3z + 2} \right)}^{3/2}}}}{{135}}\left( {9z + 11} \right) + C$$
\eqalign{ & \int {\left( {z + 1} \right)\sqrt {3z + 2} d} z \cr & {\text{substitute }}u = 3z + 2,{\text{ }}z = \frac{u}{3} - \frac{2}{3} \cr & {\text{ }}dz = \frac{1}{3}du \cr & \int {\left( {z + 1} \right)\sqrt {3z + 2} d} z = \int {\left( {\frac{u}{3} - \frac{2}{3} + 1} \right)\sqrt u \left( {\frac{1}{3}} \right)du} \cr & = \int {\left( {\frac{u}{3} + \frac{1}{3}} \right){u^{1/2}}\left( {\frac{1}{3}} \right)du} \cr & = \frac{1}{9}\int {\left( {u + 1} \right){u^{1/2}}du} \cr & = \frac{1}{9}\int {\left( {{u^{3/2}} + {u^{1/2}}} \right)du} \cr & {\text{find the antiderivative use the power rule}} \cr & = \frac{1}{9}\left( {\frac{2}{5}{u^{5/2}} + \frac{2}{3}{u^{3/2}}} \right) + C \cr & factoring \cr & = \frac{{2{u^{3/2}}}}{{9\left( {15} \right)}}\left( {3u + 5} \right) + C \cr & {\text{ with}}\,\,\,u = 3z + 2 \cr & = \frac{{2{{\left( {3z + 2} \right)}^{3/2}}}}{{9\left( {15} \right)}}\left( {3\left( {3z + 2} \right) + 5} \right) + C \cr & = \frac{{2{{\left( {3z + 2} \right)}^{3/2}}}}{{135}}\left( {9z + 11} \right) + C \cr}