Integration by Parts (IBP)

For indefinite integrals, $\int udv=uv-\int vdu$

For example, calculate $\int x{e}^{x}dx$ using IBP.

$u$	$v$
$u=x$	$v=e^x$
$du=dx$	$dv=e^{x}dx$

If we’re working with a definite integral, we can be even more efficient…?

Derivation

The MATH213 teacher quickly showed how we got this, didn’t finish reasoning about it.

It comes from the Product Rule

$(fg{)}^{\prime }=f^{\prime }g+f{g}^{\prime }$ $\int (fg{)}^{\prime }dt=\int f^{\prime }g+f{g}^{\prime }dt$ $\int f^{\prime }gdt=\int (fg{)}^{\prime }dt-\int f{g}^{\prime }dt$ $\int f^{\prime }gdt=fg-\int f{g}^{\prime }dt$ Let $dv=f^{\prime }dt$ and $du=g^{\prime }dt$, then

$\int udv=uv-\int vdu$ (this is confusing because we are using f and g), rewrite it using u and v

$\text{Product Rule:}(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$ $\text{Integrate Both Sides:}\int (uv{)}^{\prime }dx=\int (u^{\prime }v+u{v}^{\prime })dx$ $\text{Simplify:}uv=\int u^{\prime }vdx+\int u{v}^{\prime }dx$ $\text{Rearrange:}\int u{v}^{\prime }dx=uv-\int u^{\prime }vdx$

$\text{Let }dv=v^{\prime }dx,\text{and }du=u^{\prime }dx,\text{then}$ $\text{Integration by Parts Formula:}\int udv=uv-\int vdu$