Integration by Parts: Formula, LIATE & Examples

The formula

\int u d v = uv - \int v d u

Choose $u$ as the factor to differentiate and $d v$ as the part to integrate. The goal is to make the new integral $\int v d u$ easier than the original.

Why the formula works

Start with the product rule:

\frac{d}{d x} (uv) = u \frac{d v}{d x} + v \frac{d u}{d x}

Integrate both sides and rearrange:

Product rule

d (uv) = u d v + v d u

Integrate

uv = \int u d v + \int v d u

Rearrange

\int u d v = uv - \int v d u

Before you start: You should know the product rule, standard derivatives and antiderivatives, definite-integral evaluation, and basic algebra. This guide covers choosing a split, definite integrals, repeated use, and verification.

Choose u so differentiation makes progress

LIATE is a useful preference order: logarithmic, inverse trigonometric, algebraic, trigonometric, exponential. It is a heuristic, not a theorem. The real test is whether the remaining integral becomes easier.

Current integral Integral of x times e to the x $\int x e^{x} d x$

Which factor should be u?

Choose u

x

e^{x}

Choose one factor, then check whether differentiation makes the new integral simpler.

How to integrate by parts in four steps

1
Split the integrand
Choose $u$ to differentiate and include every remaining factor, including $d x$ , in $d v$ .
2
Compute du and v
Differentiate $u$ and integrate $d v$ . If finding $v$ is harder than the original problem, reconsider the split.
3
Substitute into the formula
Write $uv - \int v d u$ carefully. The most common sign error happens here.
4
Finish and verify
Evaluate the new integral, apply bounds if present, and differentiate an indefinite answer to recover the original integrand.

When to stop

If the new integral is essentially unchanged but has a different sign, algebra may let you solve for the original integral. If it becomes more complex, back up and choose a different split.

Integration by parts examples

Example 1

Polynomial times exponential

Evaluate $\int x e^{x} d x$ . Let $u = x$ and $d v = e^{x} d x$ . Then $d u = d x$ and $v = e^{x}$ .

\int x e^{x} d x = x e^{x} - \int e^{x} d x = x e^{x} - e^{x} + C

Check: $\frac{d}{d x} (x e^{x} - e^{x}) = e^{x} + x e^{x} - e^{x} = x e^{x}$ .

Example 2

A single logarithm still has two factors

Write $\int ln x d x$ as $\int (ln x) (1) d x$ . For $x > 0$ , choose $u = ln x$ and $d v = d x$ .

\int ln x d x = x ln x - \int 1 d x = x ln x - x + C

The restriction $x > 0$ matches the real-valued logarithm used in the problem.

Example 3

Apply the bounds after finding an antiderivative

Evaluate $\int_{0}^{1} x e^{x} d x$ using the antiderivative from Example 1.

[x e^{x} - e^{x}]_{0}^{1} = (e - e) - (0 - 1) = 1

Check: the integrand is nonnegative on $[0, 1]$ , so the positive value is consistent with its signed area.

Example 4

Repeat the method for a polynomial

For $\int x^{2} cos x d x$ , choose $u = x^{2}$ and $d v = cos x d x$ . The remaining integral $\int 2 x sin x d x$ needs integration by parts once more.

\int x^{2} cos x d x = x^{2} sin x + 2 x cos x - 2 sin x + C

Differentiation cancels the extra terms and leaves $x^{2} cos x$ .

Verify an integral in the calculator

Enter the integrand, keep the bounds empty for an antiderivative, and inspect the derivative check.

Open the integral calculator

Common integration by parts mistakes

Choosing dv that is hard to integrate

You must be able to find v. A split that sends a difficult factor into dv often makes no progress.

Forgetting the minus sign

The formula is $uv - \int v d u$ . Write the formula before substituting to protect the sign.

Treating LIATE as an absolute rule

LIATE suggests a first choice for u. Always judge the new integral, especially for cyclic or unusual products.

Applying definite bounds inconsistently

Either use the definite formula throughout or find an antiderivative first and apply the bounds at the end.

Choose a split, then solve

1. Evaluate

\int x cos x d x

Choose $u = x$ and $d v = cos x d x$ . The result is $x sin x + cos x + C$ .

2. Evaluate

\int x e^{2 x} d x

With $u = x$ and $d v = e^{2 x} d x$ , the result is $\frac{1}{2} x e^{2 x} - \frac{1}{4} e^{2 x} + C$ .

3. Evaluate

\int_{0}^{π} x sin x d x

An antiderivative is $- x cos x + sin x$ . Evaluating from 0 to $π$ gives $π$ .

4. Evaluate

\int arctan x d x

Choose $u = arctan x$ and $d v = d x$ . The result is $x arctan x - \frac{1}{2} ln (1 + x^{2}) + C$ .

Where integration by parts fits

The College Board places integration by parts in AP Calculus BC Unit 6, Topic 6.11. It is not listed as an AP Calculus AB topic. University Calculus II courses usually teach it after substitution and before more advanced integration strategies.

Review the chain rule if nested derivatives are slowing down your choice of du, or return to the main calculus solver to check another problem.