Answer
$$\frac{{{e^{2x}}}}{{13}}\left( {3\sin 3x + 2\cos 3x} \right) + C$$
Work Step by Step
$\begin{gathered}
\int {{e^{2x}}\cos 3x} dx \hfill \\
{\text{Use tabular integration to evaluate the integral}} \hfill \\
\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}
{{e^{2x}}{\text{ and its derivatives}}}&{}&{\cos 3x{\text{ and its integrals}}} \\
{{e^{2x}}}&{\mathop {\left( + \right)}\limits_ \searrow }&{\cos 3x} \\
{2{e^{2x}}}&{\mathop {\left( - \right)}\limits_ \searrow }&{\frac{1}{3}\sin 3x} \\
{4{e^{2x}}}&{\left( + \right)}&{ - \frac{1}{9}\cos 3x}
\end{array} \hfill \\
\end{gathered} $
$$\eqalign{
& {\text{We need to stop here because it is the same as the first row except }} \cr
& {\text{for constants }}\left( {4{\text{ on the left and }} - \frac{1}{9}{\text{ on the right}}} \right).{\text{ Then}}{\text{,}} \cr
& {\text{We interpret the table as saying }} \cr
& \int {{e^{2x}}\cos 3x} dx = \left( {{e^{2x}}} \right)\left( {\frac{1}{3}\sin 3x} \right) + \left( {2{e^{2x}}} \right)\left( { + \frac{1}{9}\cos 3x} \right) + \int {\left( {4{e^{2x}}} \right)\left( { - \frac{1}{9}\cos 3x} \right)} dx \cr
& \int {{e^{2x}}\cos 3x} dx = \frac{1}{3}{e^{2x}}\sin 3x + \frac{2}{9}{e^{2x}}\cos 3x - \frac{4}{9}\int {{e^{2x}}\cos 3x} dx \cr
& \cr
& {\text{Solve for }}\int {{e^{2x}}\cos 3x} dx \cr
& \int {{e^{2x}}\cos 3x} dx + \frac{4}{9}\int {{e^{2x}}\cos 3x} dx = \frac{1}{3}{e^{2x}}\sin 3x + \frac{2}{9}{e^{2x}}\cos 3x \cr
& \frac{{13}}{9}\int {{e^{2x}}\cos 3x} dx = \frac{1}{3}{e^{2x}}\sin 3x + \frac{2}{9}{e^{2x}}\cos 3x + C \cr
& \int {{e^{2x}}\cos 3x} dx = \frac{9}{{13}}\left( {\frac{1}{3}{e^{2x}}\sin 3x + \frac{2}{9}{e^{2x}}\cos 3x} \right) + C \cr
& \int {{e^{2x}}\cos 3x} dx = \frac{3}{{13}}{e^{2x}}\sin 3x + \frac{2}{{13}}{e^{2x}}\cos 3x + C \cr
& {\text{Factoring gives}} \cr
& \int {{e^{2x}}\cos 3x} dx = \frac{{{e^{2x}}}}{{13}}\left( {3\sin 3x + 2\cos 3x} \right) + C \cr} $$