Integral0^inf e2x cos(x) dx

Let $I = \int e^{2x} \cos(3x) \, dx$.
Choose $u = \cos(3x)$ and $dv = e^{2x} \, dx$.
Then, we find $du$ and $v$:
$du = -3\sin(3x) \, dx$
$v = \int e^{2x} \, dx = \frac{1}{2}e^{2x}$

Substitute these into the integration by parts formula:

$$I = \cos(3x) \left(\frac{1}{2}e^{2x}\right) - \int \left(\frac{1}{2}e^{2x}\right) (-3\sin(3x)) \, dx$$ $$I = \frac{1}{2}e^{2x}\cos(3x) + \frac{3}{2}\int e^{2x}\sin(3x) \, dx$$

Let $J = \int e^{2x}\sin(3x) \, dx$.
Choose $u = \sin(3x)$ and $dv = e^{2x} \, dx$.
Then, we find $du$ and $v$:
$du = 3\cos(3x) \, dx$
$v = \int e^{2x} \, dx = \frac{1}{2}e^{2x}$

Substitute these into the integration by parts formula:

$$J = \sin(3x) \left(\frac{1}{2}e^{2x}\right) - \int \left(\frac{1}{2}e^{2x}\right) (3\cos(3x)) \, dx$$ $$J = \frac{1}{2}e^{2x}\sin(3x) - \frac{3}{2}\int e^{2x}\cos(3x) \, dx$$

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

$$J = \frac{1}{2}e^{2x}\sin(3x) - \frac{3}{2}I$$

$$I = \frac{1}{2}e^{2x}\cos(3x) + \frac{3}{2}\left(\frac{1}{2}e^{2x}\sin(3x) - \frac{3}{2}I\right)$$

Distribute the $\frac{3}{2}$:

$$I = \frac{1}{2}e^{2x}\cos(3x) + \frac{3}{4}e^{2x}\sin(3x) - \frac{9}{4}I$$

Add $\frac{9}{4}I$ to both sides of the equation:

$$I + \frac{9}{4}I = \frac{1}{2}e^{2x}\cos(3x) + \frac{3}{4}e^{2x}\sin(3x)$$

Combine the terms with $I$:

$$\frac{4}{4}I + \frac{9}{4}I = \frac{13}{4}I$$

So,

$$\frac{13}{4}I = \frac{1}{2}e^{2x}\cos(3x) + \frac{3}{4}e^{2x}\sin(3x)$$

Multiply both sides by $\frac{4}{13}$ to isolate $I$:

$$I = \frac{4}{13}\left(\frac{1}{2}e^{2x}\cos(3x) + \frac{3}{4}e^{2x}\sin(3x)\right)$$ $$I = \frac{4}{13} \cdot \frac{1}{2}e^{2x}\cos(3x) + \frac{4}{13} \cdot \frac{3}{4}e^{2x}\sin(3x)$$ $$I = \frac{2}{13}e^{2x}\cos(3x) + \frac{3}{13}e^{2x}\sin(3x)$$

Finally, add the constant of integration $C$:

$$I = \frac{e^{2x}}{13}(2\cos(3x) + 3\sin(3x)) + C$$

The final answer is $\frac{e^{2x}}{13}(2\cos(3x) + 3\sin(3x)) + C$.