Evaluate the integral integral e3x 5x \,dx using integration by parts twice.

Let $u = \cos 5x$ and $dv = e^{3x} \,dx$.
Then, we find $du$ and $v$:
$du = -5 \sin 5x \,dx$
$v = \int e^{3x} \,dx = \frac{1}{3} e^{3x}$

Substitute these into the integration by parts formula:

$$I = (\cos 5x) \left(\frac{1}{3} e^{3x}\right) - \int \left(\frac{1}{3} e^{3x}\right) (-5 \sin 5x) \,dx$$ $$I = \frac{1}{3} e^{3x} \cos 5x + \frac{5}{3} \int e^{3x} \sin 5x \,dx$$

Let $I_1 = \int e^{3x} \sin 5x \,dx$.
Let $u = \sin 5x$ and $dv = e^{3x} \,dx$.
Then, we find $du$ and $v$:
$du = 5 \cos 5x \,dx$
$v = \int e^{3x} \,dx = \frac{1}{3} e^{3x}$

Substitute these into the integration by parts formula:

$$I_1 = (\sin 5x) \left(\frac{1}{3} e^{3x}\right) - \int \left(\frac{1}{3} e^{3x}\right) (5 \cos 5x) \,dx$$ $$I_1 = \frac{1}{3} e^{3x} \sin 5x - \frac{5}{3} \int e^{3x} \cos 5x \,dx$$

Notice that the integral on the right side is our original integral $I$.
So, we have:

$$I_1 = \frac{1}{3} e^{3x} \sin 5x - \frac{5}{3} I$$

$$I = \frac{1}{3} e^{3x} \cos 5x + \frac{5}{3} \left( \frac{1}{3} e^{3x} \sin 5x - \frac{5}{3} I \right)$$ $$I = \frac{1}{3} e^{3x} \cos 5x + \frac{5}{9} e^{3x} \sin 5x - \frac{25}{9} I$$

Add $\frac{25}{9} I$ to both sides:

$$I + \frac{25}{9} I = \frac{1}{3} e^{3x} \cos 5x + \frac{5}{9} e^{3x} \sin 5x$$ $$\left(1 + \frac{25}{9}\right) I = \frac{1}{3} e^{3x} \cos 5x + \frac{5}{9} e^{3x} \sin 5x$$ $$\left(\frac{9+25}{9}\right) I = \frac{1}{3} e^{3x} \cos 5x + \frac{5}{9} e^{3x} \sin 5x$$ $$\frac{34}{9} I = \frac{1}{3} e^{3x} \cos 5x + \frac{5}{9} e^{3x} \sin 5x$$

Multiply both sides by $\frac{9}{34}$:

$$I = \frac{9}{34} \left( \frac{1}{3} e^{3x} \cos 5x + \frac{5}{9} e^{3x} \sin 5x \right)$$ $$I = \frac{9}{34} \cdot \frac{1}{3} e^{3x} \cos 5x + \frac{9}{34} \cdot \frac{5}{9} e^{3x} \sin 5x$$ $$I = \frac{3}{34} e^{3x} \cos 5x + \frac{5}{34} e^{3x} \sin 5x$$

$$I = \frac{e^{3x}}{34} (3 \cos 5x + 5 \sin 5x) + C$$

The final answer is $\frac{e^{3x}}{34} (3 \cos 5x + 5 \sin 5x) + C$.