## Calculus: Early Transcendentals (2nd Edition)

$= \frac{1}{{{a^2}}}\,\left( {\frac{{\,{{\left( {ax + b} \right)}^{n + 2}}}}{{n + 2}} - \frac{{b\,{{\left( {ax + b} \right)}^{n + 1}}}}{{n + 1}}} \right) + C$
$\begin{gathered} \int_{}^{} {x\,{{\left( {ax + b} \right)}^n}dx} \hfill \\ \hfill \\ set \hfill \\ u = ax + b\,\,\,\,\,then\,\,\,\,x = \frac{{u - b}}{a} \hfill \\ du = adx\,\,\,\,\,then\,\,\,\,\,\,\,\,dx = \frac{{du}}{a} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \int_{}^{} {x\,{{\left( {ax + b} \right)}^n}dx} = \frac{1}{a}\int_{}^{} {\,\left( {\frac{{u - b}}{a}} \right){u^n}\frac{{du}}{a}} \hfill \\ \hfill \\ integrating \hfill \\ \hfill \\ \int_{}^{} {x\,{{\left( {ax + b} \right)}^n}dx} = \frac{1}{{{a^2}}}\int_{}^{} {\,\left( {{u^{n + 1}} - b{u^n}} \right)du} \hfill \\ \hfill \\ \frac{1}{{{a^2}}}\,\left( {\frac{{{u^{n + 2}}}}{{n + 2}} - \frac{{b{u^{n + 1}}}}{{n + 1}}} \right) + C \hfill \\ \hfill \\ \,substitute\,\,back\,\,u = ax + b \hfill \\ \hfill \\ = \frac{1}{{{a^2}}}\,\left( {\frac{{\,{{\left( {ax + b} \right)}^{n + 2}}}}{{n + 2}} - \frac{{b\,{{\left( {ax + b} \right)}^{n + 1}}}}{{n + 1}}} \right) + C \hfill \\ \end{gathered}$