Integration by Parts

Core Idea

Integration by parts reverses the product rule: the integral of u dv = uv - the integral of v du. It converts one integral into another, hopefully simpler one. The LIATE rule (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) helps choose u. Common applications include integrals involving ln(x), x*e^x, x*sin(x), and arctan(x). Sometimes multiple applications or a cyclical trick are needed.

How It's Best Learned

Derive from the product rule. Practice choosing u and dv using LIATE. Work through standard types: polynomial times exponential, polynomial times trig, logarithms. Show the tabular method for repeated integration by parts. Practice the cyclical case (e.g., integral of e^x sin(x) dx).

Common Misconceptions

• Choosing u and dv poorly, making the integral harder instead of easier.
• Forgetting the minus sign in uv - integral of v du.
• Not recognizing when integration by parts needs to be applied multiple times or when a cyclical equation arises.

Explainer

Integration by parts reverses the product rule. Differentiate the product u·v and you get (uv)' = u'v + uv'. Rearranging: uv' = (uv)' − u'v. Integrate both sides and you arrive at the IBP formula: ∫u dv = uv − ∫v du. The idea is to trade one integral for another — you hope the new integral ∫v du is simpler than what you started with.

The entire game is choosing u and dv wisely. The LIATE mnemonic provides a reliable ordering: favor choosing u from whichever category comes first — Logarithmic, Inverse trigonometric, Algebraic (polynomials), Trigonometric, Exponential. The logic is that functions at the top of LIATE differentiate into simpler forms (ln(x) becomes 1/x), while exponentials and trig at the bottom integrate as easily as they differentiate. For ∫ x·eˣ dx, choose u = x (Algebraic) and dv = eˣ dx. Then du = dx, v = eˣ, and the formula yields xeˣ − ∫ eˣ dx = xeˣ − eˣ + C.

A non-obvious but important case: ∫ ln(x) dx. There is no second function in sight, but you can write it as ∫ ln(x)·1 dx and choose u = ln(x), dv = dx. This gives v = x and the new integral ∫ x·(1/x) dx = ∫ 1 dx, which trivially integrates to x. Result: x·ln(x) − x + C. The same move works for arctan(x) and arcsin(x) — when the integrand is a lone logarithm or inverse trig, use 1 as the silent second factor.

Sometimes IBP must be applied repeatedly. For ∫ x²·eˣ dx, one application reduces the power from 2 to 1; a second reduces it from 1 to 0 and the integral evaluates. The tabular method (writing successive derivatives of u in one column and successive antiderivatives of dv in another, with alternating signs) streamlines this bookkeeping.

The cyclical case is the most surprising. Applying IBP to ∫ eˣ·sin(x) dx, then applying it again to the resulting integral, produces the original integral on the right-hand side — giving I = eˣ(sin x − cos x) − I. This looks circular but is actually useful: add I to both sides, and you get 2I = eˣ(sin x − cos x), so I = eˣ(sin x − cos x)/2 + C. Recognizing the cycle and solving algebraically rather than continuing to iterate is a key IBP skill.