Trade a hard product for an easier one: Integration by parts is the product rule run backwards. It swaps the integral of a product for a new, hopefully simpler, integral.
Pick one factor to be u (you differentiate it) and the rest to be dv (you integrate it). The trick is choosing u so that differentiating it makes things SIMPLER. The order to try is LIATE: Log, Inverse-trig, Algebra (powers of x), Trig, Exponential — whatever comes first is your u.
IB-style question — x times a trig function
A small object's displacement integral takes the form below, with a power of x times a cosine.
Find ∫ x cos x dx.
Step by step
- By LIATE, Algebra (x) beats Trig (cos x), so let u = x and dv = cos x dx.
- Differentiate u, integrate dv.
- Apply ∫ u dv = uv − ∫ v du.
- The new integral is easy.
Final answer
∫ x cos x dx = x sin x + cos x + C.
Log first; repeat parts for x²-type products: ln x is top of LIATE (L) — make it u even when there seems to be 'nothing' to integrate: take dv = dx so v = x. This is how ∫ ln x dx is done.
For a product like x²eˣ, one round of parts leaves an x·eˣ integral — so do parts again. A power of xⁿ needs parts n times, each round dropping the power by one.
IB-style question — x times a logarithm
Find ∫ x ln x dx.
Step by step
- By LIATE, Log beats Algebra, so let u = ln x and dv = x dx.
- Differentiate the log, integrate the power.
- Apply the formula; the new integral's x cancels nicely.
- Finish the easy integral.
Final answer
∫ x ln x dx = (x²/2) ln x − x²/4 + C.
IB-style question — repeated parts (x²eˣ)
Find ∫ x² eˣ dx.
Step by step
- Let u = x² (Algebra), dv = eˣ dx. Then du = 2x dx, v = eˣ.
- The leftover ∫ 2x eˣ dx still needs parts — use ∫ x eˣ dx = x eˣ − eˣ.
- Substitute that back in.
- Tidy up.
Final answer
∫ x² eˣ dx = eˣ(x² − 2x + 2) + C.