Answer
\[ = \frac{{2x}}{3}{e^{3x}} - \frac{2}{9}{e^{3x}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {2x{e^{3x}}\,\,dx} \hfill \\
\hfill \\
set\,\,\,the\,\,substitution \hfill \\
\hfill \\
u = 2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,then\,\,\,\,du = 2dx \hfill \\
dv = {e^{3x}}dx\,\,\,\,\,\,then\,\,\,\,\,v = \frac{1}{3}{e^{3x}} \hfill \\
\hfill \\
use\,\,uv - \int_{}^{} {vdu} \hfill \\
\hfill \\
we{\text{ }}replace{\text{ }}the{\text{ }}values{\text{ }}in{\text{ }}the{\text{ }}equation \hfill \\
\hfill \\
\frac{{2x}}{3}{e^{3x}} - \frac{2}{3}\int_{}^{} {{e^{3x}}dx} \hfill \\
\hfill \\
or \hfill \\
\hfill \\
\frac{{2x}}{3}{e^{3x}} - \frac{2}{{3\left( 3 \right)}}\int_{}^{} {{e^{3x}}\left( 3 \right)dx} \hfill \\
\hfill \\
integrate \hfill \\
= \frac{{2x}}{3}{e^{3x}} - \frac{2}{9}{e^{3x}} + C \hfill \\
\end{gathered} \]
