Integration by Parts Formula: ∫u dv = uv − ∫v du, LIATE and Examples
The integration by parts formula ∫u dv = uv − ∫v du, how LIATE picks u, the definite integral version and short examples with ln x, x·eˣ and x·cos x.
Integration rules Updated 3 min read
At a glance
- Reverses
- The product rule for derivatives
- Use when
- The integrand is a product of two different kinds of function, such as x\,e^x or x\cos x
- Choosing u
- LIATE: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential
- Definite version
- \int_a^b u\,dv = \big[uv\big]_a^b - \int_a^b v\,du
- Goal
- The new integral \int v\,du should be simpler than the original
On this page
Integration by parts is the product rule for derivatives, rearranged to integrate products. It swaps one integral for another, and when u and dv are chosen well, the new integral is easier than the original.
The integration by parts formula
For differentiable functions u and v:
In function notation, with u = f(x) and dv = g'(x)\,dx:
For a definite integral, evaluate the uv term at the limits:
Where it comes from
The product rule says (uv)' = u'v + uv'. Integrate both sides and rearrange:
Choosing u with LIATE
Pick u as the factor that becomes simpler when you differentiate it, and dv as a factor you can integrate. LIATE ranks the choices for u:
| Letter | Type | Example |
|---|---|---|
| L | Logarithmic | \ln x |
| I | Inverse trig | \arctan x |
| A | Algebraic | x^2, 3x + 1 |
| T | Trigonometric | \sin x, \cos x |
| E | Exponential | e^x, 2^x |
Take u from the earliest type present. LIATE is a guideline, not a law, but it works for nearly every textbook problem.
Worked examples
Example 1: \int x\,e^x\,dx. Algebraic beats exponential, so u = x and dv = e^x\,dx. Then du = dx and v = e^x:
Example 2: \int x\cos x\,dx. Take u = x, dv = \cos x\,dx, so du = dx and v = \sin x:
Example 3: \int \ln x\,dx. There is only one factor, so take u = \ln x and dv = dx. Then du = \frac{1}{x}\,dx and v = x:
Example 4: a definite integral. Using Example 1:
Example 5: parts twice. For \int x^2 e^x\,dx, the first round with u = x^2 leaves \int 2x\,e^x\,dx, which needs a second round. The result is
Example 6: the integral comes back. For \int e^x \sin x\,dx, applying parts twice returns the original integral I with a minus sign. Solving I = e^x\sin x - e^x\cos x - I gives
Checking your answer
Differentiate with the product rule and you should get the integrand back. For Example 2:
The check is quick and catches most sign errors. A useful habit: after choosing u and dv, write u, du, v and dv in a small two-by-two box before substituting into the formula, so nothing gets lost.
Common mistakes
- Choosing u backwards. With u = e^x and dv = x\,dx, the new integral \int \frac{x^2}{2}e^x\,dx is harder than the one you started with.
- Dropping the minus sign. The formula is uv - \int v\,du, with a minus. When parts is used twice, track signs carefully.
- Forgetting to evaluate uv at the limits. In a definite integral, uv must be evaluated from a to b too.
- Adding +C to v. You may take any antiderivative for v. One +C at the very end is enough.
When a product is really a composition times an inner derivative, u-substitution is faster. For tabular integration, more LIATE practice and harder examples, read the full guide to integration by parts.
Practice it now
Integration by parts
Solve real problems on this topic with instant feedback, hints and a step-by-step walkthrough for every answer. Free, no sign-up needed to start.
Practice this lesson