8.2.22 помогите интеграл решить

Решите задачу:

$\int\limits {(2x-1)e^{3x}} \, dx = \int\limits {(2x-1)} \, d( \frac{e^{3x}}{3} ) = \frac{1}{3} \int\limits {(2x-1)} \, d(e^{3x}) =$

$= \frac{1}{3}*[(2x-1)e^{3x}- \int\limits {e^{3x}} \, d(2x-1)] =$

$= \frac{1}{3}*[(2x-1)e^{3x}- \int\limits {e^{3x}} \, d(2x)] = \frac{1}{3}*[(2x-1)e^{3x}- 2 \int\limits {e^{3x}} \, dx] =$

$= \frac{1}{3}*[(2x-1)e^{3x}- 2 \int\limits {e^{3x}} \, d (\frac{3x}{3} )]=$

$= \frac{1}{3}*[(2x-1)e^{3x}- \frac{2}{3} \int\limits {e^{3x}} \, d(3x)] = \frac{1}{3}*[(2x-1)e^{3x}- \frac{2}{3} e^{3x}]+C =$

$= \frac{e^{3x}}{3}*[2x-1-\frac{2}{3}]+C = \frac{e^{3x}}{3}*(2x-\frac{5}{3})+C$