* Xét $I=\int e^{2x}\sin 3x\text{d}x$
Đặt $\left\{\begin{array}{l}u=e^{2x}\\ \text{d}v=\sin 3x\text{d}x\end{array}\right.\Rightarrow \left\{\begin{array}{l}\text{d}u=2e^{2x}\text{d}x\\ v=-\frac{1}{3}\cos 3x\end{array}\right.$
Khi đó $I=-\frac{1}{3}{e}^{2x}.\cos 3x+\frac{2}{3}\int e^{2x}\cos 3x\text{d}x$ (1)
* Xét $J=\int e^{2x}\cos 3x\text{d}x$
Đặt $\left\{\begin{array}{l}{u}_1=e^{2x}\\ \text{d}{v}_1=\cos 3x\text{d}x\end{array}\right.\Rightarrow \left\{\begin{array}{l}\text{d}{u}_1=2e^{2x}\text{d}x\\ v_1=\frac{1}{3}\sin 3x\end{array}\right.$
$J=\frac{1}{3}{e}^{2x}.\sin 3x-\frac{2}{3}\int e^{2x}\sin 3x\text{d}x=\frac{1}{3}{e}^{2x}.\sin 3x-\frac{2}{3}I$(2)
Thay (2) vào (1) ta có: $I=-\frac{1}{3}{e}^{2x}.\cos 3x+\frac{2}{3}\left(\frac{1}{3}{e}^{2x}.\sin 3x-\frac{2}{3}I\right)$
Vậy $I=\frac{e^{2x}}{13}.\left(2\sin 3x-3\cos 3x\right)+C$.