Integration by parts order question + integration question In integration by parts, does it matter which of your terms is v (or, rather $f(x)$) and which term is du (or, rather $ f'(x) $?
Also, I'm having trouble finding $\int e^{2x} \cos(3x)dx $.  I keep getting $$\frac {2e^{2x}(6\sin(3x) - \cos(3x))}{37}$$ when $u=e^{2x}$ and $dv = \cos(3x) dx $.
Thanks!

Yes, it does matter. You can check Wikipedia's page on integration by parts, and in particular the LIATE rule. In general, you want $u$ to be easier to differentiate, whereas $dv$ to be easier to integrate.
In your example, however, if you follow the rule above, both functions should be $u$. You can overcome this by applying the same choice multiple times. For example, let's set $u = e^{2x}$ and $dv = \cos (3x)\ dx$. Then $du = 2e^{2x}\ dx$ and $v = \frac{\sin 3x}{3}$. So:
$$\int e^{2x} \cos (3x)\ dx = \frac{1}{3}e^{2x}\sin 3x - \frac{2}{3} \int \sin(3x) e^{2x}\ dx \tag 1$$ Integrating again, with $u = e^{2x}$ and $dv = \sin(3x)\ dx$ we get that $du = 2 e^{2x}\ dx$ and $v = - \frac{1}{3} \cos 3x$. The latter integral becomes:
$$\int \sin(3x) e^{2x}\ dx = -\frac{1}{3}e^{2x}\cos 3x + \frac{2}{3}\int e^{2x}\cos(3x)\ dx$$
Substituting back into $(1)$ we get:
$$\int e^{2x} \cos (3x)\ dx = \frac{1}{3}e^{2x}\sin 3x - \frac{2}{3} \left ( -\frac{1}{3}e^{2x}\cos 3x + \frac{2}{3}\int e^{2x}\cos(3x)\ dx\right )$$
and then
$$\int e^{2x} \cos (3x)\ dx = \frac{1}{3}e^{2x}\sin 3x + \frac{2}{9} e^{2x}\cos 3x - \frac{4}{9} \int e^{2x}\cos(3x)\ dx$$
Adding $\displaystyle \frac{4}{9} \int e^{2x}\cos(3x)\ dx$ to both sides of the equation:
$$\frac{13}{9} \int e^{2x} \cos (3x)\ dx = \frac{1}{3}e^{2x}\sin 3x + \frac{2}{9} e^{2x}\cos 3x$$
Finally:
$$\begin{align*}
\int e^{2x} \cos (3x)\ dx &= \frac{3}{13}e^{2x}\sin 3x + \frac{2}{13} e^{2x}\cos 3x + C = \\
&= \frac{1}{13} e^{2x} (3 \sin 3x + 2 \cos 3x) + C
\end{align*}$$.
