# How do I find intxe^(2x)dx using integration by parts?

Jan 26, 2015

Since $\left(f g\right) ' = f ' g + f g '$, by integrating one has that $f g = \setminus \int f ' g + \setminus \int f g '$.

In your case, it is useful to see $f \left(x\right) = x$, and $g ' \left(x\right) = {e}^{2 x}$.
This means:
$f \left(x\right) = x \setminus \implies f ' \left(x\right) = 1$
$g ' \left(x\right) = {e}^{2 x} \setminus \implies g \left(x\right) = {e}^{2 x} / 2$

So we have that $\setminus \int f g ' = f g - \setminus \int f ' g$, that in our cases is

$\setminus \int x {e}^{2 x} = x {e}^{2 x} / 2 - \setminus \int {e}^{2 x} / 2$, where the last integral is easily ${e}^{2 x} / 4$. So, we have that

$\setminus \int x {e}^{2 x} = x {e}^{2 x} / 2 - {e}^{2 x} / 4$