The integration by parts formula is:
$$ \int udv = uv - \int vdu $$

## Deriving the Integration by Parts Formula
$$ \frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x) $$
1. Integrating both sides, we get:
$$\int \frac{d}{dx} (f(x)g(x))dx = \int [f'(x)g(x) + f(x)]$$
2. Through the distributive property of integrals,
$$ = \int f'(x)g(x)dx + \int f(x)g'(x)dx $$
3. An integral cancels out an antiderivative, therefore:
$$f(x)g(x) = \int f'(x)g(x)dx + \int f(x)g'(x)dx $$
4. Moving terms around:
$$ \int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx$$

Now, let $u = f(x)$ and $v = g(x)$, then $dv = g'(x)dx$ and $du = f'(x)dx$. 

# Examples
>  Evaluate the below antiderivative using integration by parts.
$$\int xe^{2x}dx$$
1. Define $u$ to be a value you can take the derivative of easily, in this case $u = x$. The rest of the integral will be set to $dv$, in this case, $dv = e^{2x}dx$. 
	- $u = x$
	- $du = \frac{d}{dx}(x)= 1dx$ 
	- $dv = e^{2x}dx$
	- $v = \frac{1}{2}e^{2x}$ - The antiderivative of $dv$.
2. Looking back at the integration by parts formula, we know that:
	$$ \int udv = uv - \int v du $$