The formula
It comes straight from the product rule. Since (uv)′ = u′v + uv′, integrating both sides gives uv = ∫v du + ∫u dv, and rearranging gives the formula. So integration by parts is the product rule run backwards, in the same way that u-substitution is the chain rule run backwards.
How to use it
- Split the integrand into two parts: u, which you'll differentiate, and dv, which you'll integrate.
- Find du = u′ dx and v = ∫dv (no + C needed here).
- Substitute into uv − ∫v du.
- Integrate what's left. If it's no easier, try a different choice of u.
Choosing u: LIATE
A good rule of thumb is to choose u as whichever factor comes first in this list, because it gets simpler when differentiated:
| Letter | Type | Examples |
|---|---|---|
| L | Logarithmic | ln x, log₂ x |
| I | Inverse trig | arctan x, arcsin x |
| A | Algebraic | x, x², 3x + 1 |
| T | Trigonometric | sin x, cos x |
| E | Exponential | eˣ, 2ˣ |
It's a guideline, not a law. What matters is that ∫v du ends up simpler than the integral you started with.
Worked examples
1. ∫ x eˣ dx. Let u = x (algebraic) and dv = eˣ dx. Then du = dx and v = eˣ. So ∫x eˣ dx = x eˣ − ∫eˣ dx = x eˣ − eˣ + C = (x − 1)eˣ + C.
2. ∫ x cos x dx. Let u = x, dv = cos x dx, so du = dx and v = sin x. Then ∫x cos x dx = x sin x − ∫sin x dx = x sin x + cos x + C.
3. ∫ ln x dx. There's only one factor, so use dv = dx. With u = ln x, du = (1/x) dx and v = x. Then ∫ln x dx = x ln x − ∫x · (1/x) dx = x ln x − x + C. The same trick gives ∫arctan x dx = x arctan x − ½ ln(1 + x²) + C.
4. ∫ x² eˣ dx (twice). With u = x², dv = eˣ dx: ∫x² eˣ dx = x² eˣ − ∫2x eˣ dx. The new integral is example 1 times 2, so the answer is x² eˣ − 2(x − 1)eˣ + C = (x² − 2x + 2)eˣ + C.
5. ∫ eˣ sin x dx (the loop). Let u = sin x, dv = eˣ dx: I = eˣ sin x − ∫eˣ cos x dx. Do it again with u = cos x: ∫eˣ cos x dx = eˣ cos x + ∫eˣ sin x dx = eˣ cos x + I. So I = eˣ sin x − eˣ cos x − I. The original integral came back, so solve for it: 2I = eˣ(sin x − cos x), and I = eˣ(sin x − cos x)/2 + C.
6. A definite integral, ∫₀¹ x eˣ dx. Using example 1, [(x − 1)eˣ] from 0 to 1 = (0)e − (−1)(1) = 1. Similarly ∫₀^π x sin x dx = [−x cos x + sin x]₀^π = π.
The tabular method
When u is a polynomial that eventually differentiates to 0, a table saves time. Differentiate u down one column until you hit 0, integrate dv down the other, then multiply diagonally with alternating signs (+, −, +, …).
| Sign | Differentiate u = x³ | Integrate dv = sin x |
|---|---|---|
| + | x³ | sin x |
| − | 3x² | −cos x |
| + | 6x | −sin x |
| − | 6 | cos x |
| 0 | sin x |
Multiplying diagonally: ∫x³ sin x dx = −x³ cos x + 3x² sin x + 6x cos x − 6 sin x + C.
Common mistakes
- Choosing u = eˣ in ∫x eˣ dx. Then v = x²/2 and the new integral, ∫(x²/2)eˣ dx, is harder.
- Forgetting the minus sign in uv − ∫v du, or the alternating signs in the tabular method.
- Dropping the limits on uv in a definite integral. Evaluate [uv]ₐᵇ as well as the remaining integral.
Check any answer by differentiating it: you should get the integrand back. The integral calculator shows the steps, and the derivative calculator confirms an antiderivative. For integrals of rational functions, try partial fractions first.
Sources
- OpenStax Calculus Volume 2, 3.1: Integration by parts
- Paul's Online Notes (Lamar University): Integration by parts
Every worked example on this page was checked independently with SymPy.