Answer
$$\frac{{{{\left( {x - 1} \right)}^9}}}{9} + \frac{{{{\left( {x - 1} \right)}^8}}}{4} + \frac{{{{\left( {x - 1} \right)}^7}}}{7} + C$$
Work Step by Step
$$\eqalign{
& \int {{x^2}{{\left( {x - 1} \right)}^6}} dx \cr
& {\text{Integrate by substitution method }}\left( {{\text{See note 1 in the book}}} \right) \cr
& {\text{ }}u = x - 1{\text{ }} \Rightarrow {\text{ }}x = u + 1{\text{ }} \Rightarrow {\text{ }}dx = du \cr
& {\text{Substituting}} \cr
& \int {{x^2}{{\left( {x - 1} \right)}^6}} dx = \int {{{\left( {u + 1} \right)}^2}{u^6}} du \cr
& {\text{Expanding}} \cr
& \int {\left( {{u^2} + 2u + 1} \right)} {u^6}du \cr
& \int {\left( {{u^8} + 2{u^7} + {u^6}} \right)} du \cr
& {\text{Integrate use the power rule }}\int {{u^n}du = \frac{{{u^{n + 1}}}}{{n + 1}} + C} \cr
& \int {\left( {{u^8} + 2{u^7} + {u^6}} \right)} du = \frac{{{u^9}}}{9} + \frac{{{u^8}}}{4} + \frac{{{u^7}}}{7} + C \cr
& {\text{Write in terms of }}x,{\text{ }}u = x - 1 \cr
& \frac{{{{\left( {x - 1} \right)}^9}}}{9} + \frac{{{{\left( {x - 1} \right)}^8}}}{4} + \frac{{{{\left( {x - 1} \right)}^7}}}{7} + C \cr} $$
