Integration by Parts Formula

The integration by parts formula ∫ u dv = uv - ∫ v du is the reverse of the product rule.

The Formula

∫u dv=uv−∫v du
For definite integrals: ∫abu dv=[uv]ab−∫abv du.

When to use: The product rule for derivatives says (uv)′=u′v+uv′. Rearranging and integrating gives integration by parts. The idea is to trade your original integral for a (hopefully easier) one. You're transferring the derivative from one factor to the other.

Quick Example

∫x ex dx Let u=x, dv=ex dx. Then du=dx, v=ex.
=xex−∫ex dx=xex−ex+C=ex(x−1)+C

Notation

The LIATE rule helps choose u: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential—pick u from earlier in this list.

What This Formula Means

An integration technique based on the product rule: ∫u dv=uv−∫v du. Used when the integrand is a product of two functions.

The product rule for derivatives says (uv)′=u′v+uv′. Rearranging and integrating gives integration by parts. The idea is to trade your original integral for a (hopefully easier) one. You're transferring the derivative from one factor to the other.

Formal View

If u and v are differentiable on [a,b], then ∫abu(x)v′(x) dx=[u(x)v(x)]ab−∫abu′(x)v(x) dx. Indefinite form: ∫u dv=uv−∫v du.

Worked Examples

Example 1

easy
Find ∫xex dx.

Answer

ex(x−1)+C

First step

1
LIATE: u=x, dv=ex dx; then du=dx, v=ex.

Full solution

  1. 2
    ∫xex dx=xex−∫ex dx=xex−ex+C.
  2. 3
    Factor: ex(x−1)+C.
LIATE places algebraic before exponential, so u=x. One IBP step reduces the remaining integral to something immediate.

Example 2

hard
Find ∫exsin⁡x dx.

Example 3

medium
Evaluate ∫x2ex dx using integration by parts twice.

Common Mistakes

  • Picking u and dv backward so the new integral is harder — use LIATE so u differentiates toward simpler.
  • Dropping the minus sign or the uv term — the formula is uv−∫v du, both pieces required.
  • Trying parts when substitution fits — if an inner derivative is present, u-substitution is the right tool.

Why This Formula Matters

Integration by parts handles products that substitution can't — polynomial times exponential, anything times a logarithm or inverse trig. It encodes a strategic trade: you swap your integral for a new one, and choosing u wisely (LIATE) makes the new integral easier, while a poor choice makes it harder. Recognizing it by "Is the integrand a product of unlike functions where differentiating one factor simplifies it, with no inner-derivative match for substitution?" — rather than by familiar numbers — is what lets a student tell it apart from u-substitution and product rule and liate choice of u in a mixed problem set.

Frequently Asked Questions

What is the Integration by Parts formula?

An integration technique based on the product rule: ∫u dv=uv−∫v du. Used when the integrand is a product of two functions.

How do you use the Integration by Parts formula?

The product rule for derivatives says (uv)′=u′v+uv′. Rearranging and integrating gives integration by parts. The idea is to trade your original integral for a (hopefully easier) one. You're transferring the derivative from one factor to the other.

What do the symbols mean in the Integration by Parts formula?

The LIATE rule helps choose u: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential—pick u from earlier in this list.

Why is the Integration by Parts formula important in Math?

Integration by parts handles products that substitution can't — polynomial times exponential, anything times a logarithm or inverse trig. It encodes a strategic trade: you swap your integral for a new one, and choosing u wisely (LIATE) makes the new integral easier, while a poor choice makes it harder. Recognizing it by "Is the integrand a product of unlike functions where differentiating one factor simplifies it, with no inner-derivative match for substitution?" — rather than by familiar numbers — is what lets a student tell it apart from u-substitution and product rule and liate choice of u in a mixed problem set.

What do students get wrong about Integration by Parts?

The procedure for integration by parts is the easy part; the trap is picking u and dv backward so the new integral is harder. Asking "Is the integrand a product of unlike functions where differentiating one factor simplifies it, with no inner-derivative match for substitution?" first is what keeps a correct-looking calculation from being attached to the wrong concept.

What should I learn before the Integration by Parts formula?

Before studying the Integration by Parts formula, you should understand: integral, derivative.

Want the Full Guide?

This formula is covered in depth in our complete guide:

How to Integrate Rational Functions: Long Division and Partial Fractions →