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