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\int u\,dv = uv - \int v\,du
For definite integrals: ∫abu dv=[uv]ab−∫abv du\int_a^b u\,dv = [uv]_a^b - \int_a^b v\,du.

When to use: The product rule for derivatives says (uv)′=u′v+uv′(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\int x\,e^x\,dx Let u=xu = x, dv=ex dxdv = e^x\,dx. Then du=dxdu = dx, v=exv = e^x.
=xex−∫ex dx=xex−ex+C=ex(x−1)+C= xe^x - \int e^x\,dx = xe^x - e^x + C = e^x(x - 1) + C

Notation

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

What This Formula Means

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

The product rule for derivatives says (uv)′=u′v+uv′(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 uu and vv are differentiable on [a,b][a, b], then ∫abu(x)v′(x) dx=[u(x)v(x)]ab−∫abu′(x)v(x) dx\int_a^b u(x) v'(x)\,dx = [u(x) v(x)]_a^b - \int_a^b u'(x) v(x)\,dx. Indefinite form: ∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du.

Worked Examples

Example 1

easy
Find ∫xex dx\displaystyle\int x e^x\,dx.

Answer

ex(x−1)+Ce^x(x-1) + C

First step

1
LIATE: u=xu = x, dv=ex dxdv = e^x\,dx; then du=dxdu = dx, v=exv = e^x.

Full solution

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

Example 2

hard
Find ∫exsin⁡x dx\displaystyle\int e^x \sin x\,dx.

Example 3

medium
Evaluate ∫x2ex dx\displaystyle\int x^2 e^x \, dx using integration by parts twice.

Common Mistakes

  • Picking uu and dvdv backward so the new integral is harder — use LIATE so uu differentiates toward simpler.
  • Dropping the minus sign or the uvuv term — the formula is uv−∫v duuv-\int 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 uu 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 uu 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\int u\,dv = uv - \int 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′(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 uu: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential—pick uu 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 uu 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 uu 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 uu and dvdv 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.

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

IntegralDerivative