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