Thomas' Calculus 13th Edition

$$\frac{3}{{10}}{\left( {x - 1} \right)^{10/3}} + \frac{6}{7}{\left( {x - 1} \right)^{7/3}} + \frac{3}{4}{\left( {x - 1} \right)^{4/3}} + C$$
\eqalign{ & \int {{x^2}{{\left( {x - 1} \right)}^{1/3}}} dx \cr & {\text{Integrate by the substitution method}} \cr & \,\,\,u = x - 1,\,\,\,\,x = u + 1,\,\,\,\,\,dx = du \cr & \int {{x^2}{{\left( {x - 1} \right)}^{1/3}}} dx = \int {{{\left( {u + 1} \right)}^2}{u^{1/3}}} du \cr & {\text{expanding}} \cr & = \int {\left( {{u^2} + 2u + 1} \right){u^{1/3}}} du \cr & = \int {\left( {{u^{7/3}} + 2{u^{4/3}} + {u^{1/3}}} \right)} du \cr & {\text{integrate by the power rule}} \cr & = \frac{{{u^{10/3}}}}{{10/3}} + 2\left( {\frac{{{u^{7/3}}}}{{7/3}}} \right) + \frac{{{u^{4/3}}}}{{4/3}} + C \cr & = \frac{3}{{10}}{u^{10/3}} + \frac{6}{7}{u^{7/3}} + \frac{3}{4}{u^{4/3}} + C \cr & \cr & {\text{Write in terms of }}x{\text{; replace }}x - 1{\text{ for }}u \cr & = \frac{3}{{10}}{\left( {x - 1} \right)^{10/3}} + \frac{6}{7}{\left( {x - 1} \right)^{7/3}} + \frac{3}{4}{\left( {x - 1} \right)^{4/3}} + C \cr}