## Calculus: Early Transcendentals (2nd Edition)

$= \frac{1}{2}{x^3}{e^{2x}} - \frac{3}{4}{x^2}{e^{2x}} + \frac{{3x}}{4}{e^{2x}} - \frac{3}{8}{e^{2x}} + C$
$\begin{gathered} \int_{}^{} {{x^3}{e^{2x}}dx} \hfill \\ \hfill \\ Integrate\,\,using\,\,the\,\,formula\, \hfill \\ \hfill \\ \int_{}^{} {{x^n}{e^{ax}}dx} = \frac{1}{a}{x^n}{e^{ax}} - \frac{n}{a}\int_{}^{} {{x^{n - 1}}{e^{ax}}dx} \hfill \\ \hfill \\ Therefore, \hfill \\ \hfill \\ \int_{}^{} {{x^3}{e^{2x}}dx} = \frac{1}{2}{x^3}{e^{2x}} - \frac{3}{2}\int_{}^{} {{x^2}{e^{2x}}dx} \hfill \\ \hfill \\ = \frac{1}{2}{x^3}{e^{2x}} - \frac{3}{2}\,\,\left[ {\frac{1}{2}{x^2}{e^{2x}} - \int_{}^{} {x{e^{2x}}dx} } \right] \hfill \\ \hfill \\ = \frac{1}{2}{x^3}{e^{2x}} - \frac{3}{2}\,\,\left[ {\frac{1}{2}{x^2}{e^{2x}} - \,\,\left[ {\frac{x}{2}{e^{2x}} - \frac{{{e^{2x}}}}{4}} \right]} \right] \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \frac{1}{2}{x^3}{e^{2x}} - \frac{3}{4}{x^2}{e^{2x}} + \frac{{3x}}{4}{e^{2x}} - \frac{3}{8}{e^{2x}} + C \hfill \\ \end{gathered}$