Answer
$\frac{{{x}^{n}}{{e}^{ax}}}{a}-\frac{n}{a}\int{{{x}^{n-1}}{{e}^{ax}}dx},\text{ }a\ne 0$
Work Step by Step
\[\begin{align}
& \int{{{x}^{n}}{{e}^{ax}}}dx \\
& \text{Integrate by parts} \\
& \text{Let }u={{x}^{n}}\to du=n{{x}^{n-1}}dx \\
& dv={{e}^{ax}}dx\to v=\frac{1}{a}{{e}^{ax}} \\
& \text{Using the integration by parts formula} \\
& \int{udv}=uv-\int{vdu} \\
& \int{{{x}^{n}}{{e}^{ax}}}dx=\frac{1}{a}{{x}^{n}}{{e}^{ax}}-\int{\left( \frac{1}{a}{{e}^{ax}} \right)\left( n{{x}^{n-1}} \right)dx} \\
& \int{{{x}^{n}}{{e}^{ax}}}dx=\frac{{{x}^{n}}{{e}^{ax}}}{a}-\frac{n}{a}\int{{{x}^{n-1}}{{e}^{ax}}dx},\text{ }a\ne 0 \\
\end{align}\]
