Integration by parts

Nine problems, from warm-ups to the cyclic and improper integrals that show up on AP Calculus BC. Pick u, then build the answer one step at a time.

The formula

∫ u dv = uv − ∫ v du

It comes from the product rule: d(uv) = u dv + v du. Integrate both sides and rearrange. The goal is a new integral that is easier than the one you started with.

Choosing u: LIATE

LLogarithmic (ln x)
IInverse trig (arctan x)
AAlgebraic (x, x²)
TTrigonometric (sin x)
EExponential (e^x)

Let u be the function highest on the list. Whatever is left, with dx, is dv. It’s a rule of thumb, not a law: the test is whether v du is easier to integrate than u dv.

BC reminders

Definite integrals: uv is evaluated at both limits, so ∫_a^b u dv = [uv]_a^b − ∫_a^b v du.
Improper integrals: replace ∞ with t and take a limit at the end.
Check any answer by differentiating it.

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