Short on time? This page is the recap. For the derivation, the graph and worked examples, use the full study guide. BC-only material. Prerequisites (product rule 2.8, FTC 6.7, basic antiderivatives 6.8, substitution 6.9) are on the Calculus BC roadmap.
Recap
- Integration by parts undoes the product rule: d/dx[uv] = u′v + uv′, so ∫ uv′ dx = uv − ∫ vu′ dx.
- It trades ∫ u dv for ∫ v du. Worth it only if the new integral is easier.
- Try substitution first. Use parts for products of unrelated functions (x cos x, x eˣ, ln x).
- No + C when you find v; one + C at the end of an indefinite integral.
Key relationships
| Form | Formula | Note |
|---|---|---|
| Indefinite | ∫ u dv = uv − ∫ v du | Choose u so du is simpler; dv must be integrable |
| Definite | ∫ from a to b of u dv = [uv] from a to b − ∫ from a to b of v du | Limits go on both pieces |
| Single hard function | ∫ ln x dx = x ln x − x + C | Take u = ln x, dv = dx |
| Repeated | ∫ x² eˣ dx = eˣ(x² − 2x + 2) + C | Each step lowers the power of x |
| Self-returning | ∫ eˣ sin x dx = ½ eˣ(sin x − cos x) + C | Apply twice, keep the same type of u, solve for the integral |
Choosing u: the honest version
- LIATE (Log, Inverse trig, Algebraic, Trig, Exponential) is a starting guess.
- Then test it: can you integrate dv? Is du simpler than u?
- Counter-example to LIATE: ∫ x³ e^(x²) dx needs u = x², dv = x e^(x²) dx.
Mistakes to avoid
- Product of integrals. ∫ fg dx ≠ (∫ f dx)(∫ g dx).
- Lost minus signs, in the formula or inside v (∫ sin x dx = −cos x).
- Missing chain-rule factor in v. dv = e^(2x) dx gives v = ½ e^(2x).
- Limits on one piece only in a definite integral.
- Swapping u halfway through a self-returning integral.
- Forgetting to divide after solving 2I = …
Quick self-check
- Find ∫ x sin x dx. (−x cos x + sin x + C)
- Evaluate ∫ from 0 to π/2 of x cos x dx. (π/2 − 1, about 0.571)
- Find ∫ x ln x dx. (½x² ln x − ¼x² + C, with u = ln x, dv = x dx)
Next: practice questions.