Integration by parts has one formula and one decision: ∫u dv = uv − ∫v du. The formula is mechanical. The decision — which factor to call u and which to call dv — is where students get stuck. LIATE is a heuristic for making that decision quickly. It works because of a specific property of the function types involved, not by coincidence.

The Formula

∫ u dv = uv − ∫ v du

Choose u (to differentiate) and dv (to integrate). The goal: after one application, the remaining integral ∫v du should be simpler than what you started with.

The LIATE Rule — How to Choose u

LIATE gives the order of preference for u. Choose the function that appears first:

L — Logarithms (ln x, log x) I — Inverse trig (arcsin, arctan) A — Algebraic (xⁿ, polynomials) T — Trigonometric (sin, cos, tan) E — Exponential (eˣ, aˣ)

If you have a log × polynomial: u = ln x (L before A). Polynomial × exponential: u = polynomial (A before E). LIATE fails occasionally — it is a guideline, not a law.

Example 1 — ∫ x·eˣ dx (A before E)

📋 Solution
Chooseu = x, dv = eˣ dx → du = dx, v = eˣ
Applyx·eˣ − ∫eˣ dx = x·eˣ − eˣ + C
Answereˣ(x − 1) + C

Example 2 — ∫ x·sin(x) dx (A before T)

📋 Solution
Chooseu = x, dv = sin(x) dx → du = dx, v = −cos(x)
Apply−x·cos(x) − ∫(−cos x) dx = −x·cos(x) + sin(x) + C
Answer−x·cos(x) + sin(x) + C

Example 3 — ∫ ln(x) dx (L alone — use u = ln x, dv = dx)

📋 Solution
Chooseu = ln x, dv = dx → du = dx/x, v = x
Applyx·ln(x) − ∫x·(1/x)dx = x·ln(x) − ∫1 dx
Answerx·ln(x) − x + C

Example 4 — ∫ x²·eˣ dx (Apply IBP Twice)

📋 Solution
First IBPu = x², dv = eˣ dx → x²eˣ − ∫2x·eˣ dx
Second IBP∫2x·eˣ dx = 2(x·eˣ − eˣ) = 2eˣ(x−1)
Answereˣ(x² − 2x + 2) + C

The Tabular Method (For Repeated IBP)

When you need to apply IBP multiple times and one factor is a polynomial, the tabular method saves time. Create two columns: differentiate u repeatedly until zero, integrate dv repeatedly.

📋 ∫ x³·sin(x) dx using Tabular Method
Derivatives of x³x³ → 3x² → 6x → 6 → 0
Integrals of sin(x)sin x → −cos x → −sin x → cos x → sin x
Multiply diagonally (+−+−)+x³(−cos x) − 3x²(−sin x) + 6x(cos x) − 6(sin x)
Answer−x³cos x + 3x²sin x + 6x cos x − 6 sin x + C

When to Use Integration by Parts

Use IBP when you have a product of two different function types that u-substitution cannot handle: polynomial × trig, polynomial × exponential, polynomial × log, inverse trig alone, or log alone.

⚠ IBP Loop Warning

Sometimes IBP returns you to the original integral. This is not a mistake — it is a trick. Set the repeated integral equal to I and solve algebraically: e.g. ∫eˣsin x dx = eˣsin x − eˣcos x − ∫eˣsin x dx → 2I = eˣ(sin x − cos x) → I = eˣ(sin x − cos x)/2 + C.

Frequently Asked Questions

Does LIATE always work?▾
Usually, but not always. The real question is: after applying IBP, is the new integral simpler? If LIATE gives you a harder integral, try the opposite assignment. Understanding why LIATE works (logarithms simplify when differentiated; exponentials don't change form) helps you know when to break the rule.
How many times do I apply IBP?▾
As many times as needed until the integral is solvable directly. For ∫xⁿ·eˣ dx, you need n applications. The tabular method handles this efficiently without repeating the setup each time.
AM
Dr. Aisha Malik, PhD Mathematics
Senior Lecturer in Applied Mathematics · 12 years teaching calculus
Dr. Malik holds a PhD in Applied Mathematics from the University of Edinburgh and has taught calculus to over 4,000 students. She has reviewed this article for mathematical accuracy and pedagogical clarity.
Technically reviewed by: Prof. James Chen, Stanford Mathematics Department · April 2026