Integration by Parts Formula

The Formula

\int u\,dv = uv - \int v\,du
For definite integrals: \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'. 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

\int x\,e^x\,dx Let u = x, dv = e^x\,dx. Then du = dx, v = e^x.
= xe^x - \int e^x\,dx = xe^x - e^x + C = e^x(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: \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'. 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 \int_a^b u(x) v'(x)\,dx = [u(x) v(x)]_a^b - \int_a^b u'(x) v(x)\,dx. Indefinite form: \int u\,dv = uv - \int v\,du.

Worked Examples

Example 1

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

Solution

  1. 1
    LIATE: u = x, dv = e^x\,dx; then du = dx, v = e^x.
  2. 2
    \int xe^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C.
  3. 3
    Factor: e^x(x-1) + C.

Answer

e^x(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 \displaystyle\int e^x \sin x\,dx.

Example 3

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

Common Mistakes

  • Choosing u and dv poorly: if you let u = e^x and dv = x\,dx in \int xe^x\,dx, the new integral \int \frac{x^2}{2}e^x\,dx is HARDER. Use LIATE: pick u = x (algebraic) and dv = e^x\,dx.
  • Forgetting the minus sign: the formula is uv \mathbf{-} \int v\,du. Dropping the negative is a common algebraic error that changes the entire answer.
  • Not recognizing when to stop: for \int e^x\cos x\,dx, after applying IBP twice, the original integral reappears—set up an equation and solve for it instead of continuing to integrate by parts forever.

Why This Formula Matters

Essential for integrating products like x\sin x, x^2 e^x, \ln x, and e^x\cos x. Also used to derive reduction formulas and in proving theoretical results. Together with u-substitution, it handles the majority of integration problems.

Frequently Asked Questions

What is the Integration by Parts formula?

An integration technique based on the product rule: \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'. 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?

Essential for integrating products like x\sin x, x^2 e^x, \ln x, and e^x\cos x. Also used to derive reduction formulas and in proving theoretical results. Together with u-substitution, it handles the majority of integration problems.

What do students get wrong about Integration by Parts?

Sometimes you need to apply integration by parts TWICE (as with \int e^x\sin x\,dx) and then solve for the original integral algebraically when it reappears.

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 →