# 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.

Source: https://practicecalculus.com/rules/integration-by-parts-rule
Updated: 2026-11-03

$$ \int u\,dv = uv - \int v\,du $$

## 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

## Key takeaways

- Integration by parts trades $\int u\,dv$ for $uv - \int v\,du$.
- Pick $u$ as the factor that gets simpler when differentiated; LIATE is a good default order.
- Some integrals need parts twice, and some come back to the original integral, which you then solve for.

**Integration by parts** is the [product rule](/rules/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 $v$:

$$\int u\,dv = uv - \int v\,du$$

In function notation, with $u = f(x)$ and $dv = g'(x)\,dx$:

$$\int f(x)\,g'(x)\,dx = f(x)\,g(x) - \int f'(x)\,g(x)\,dx$$

For a definite integral, evaluate the $uv$ term at the limits:

$$\int_a^b u\,dv = \big[uv\big]_a^b - \int_a^b v\,du$$

## Where it comes from

The product rule says $(uv)' = u'v + uv'$. Integrate both sides and rearrange:

$$uv = \int v\,du + \int u\,dv \quad\Longrightarrow\quad \int u\,dv = uv - \int v\,du$$

## 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 $v = e^x$:

$$\int x\,e^x\,dx = x\,e^x - \int e^x\,dx = x\,e^x - e^x + C$$

**Example 2: $\int x\cos x\,dx$.** Take $u = x$, $dv = \cos x\,dx$, so $du = dx$ and $v = \sin x$:

$$\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x + \cos x + C$$

**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 $v = x$:

$$\int \ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}\,dx = x\ln x - x + C$$

**Example 4: a definite integral.** Using Example 1:

$$\int_0^1 x\,e^x\,dx = \big[x\,e^x - e^x\big]_0^1 = (e - e) - (0 - 1) = 1$$

**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

$$\int x^2 e^x\,dx = e^x\left(x^2 - 2x + 2\right) + C$$

**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

$$\int e^x\sin x\,dx = \frac{e^x\left(\sin x - \cos x\right)}{2} + C$$

## Checking your answer

Differentiate with the product rule and you should get the integrand back. For Example 2:

$$\frac{d}{dx}\left[x\sin x + \cos x\right] = \sin x + x\cos x - \sin x = x\cos x$$

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

1. **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.
2. **Dropping the minus sign.** The formula is $uv - \int v\,du$, with a minus. When parts is used twice, track signs carefully.
3. **Forgetting to evaluate $uv$ at the limits.** In a definite integral, $uv$ must be evaluated from $a$ to $b$ too.
4. **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](/rules/u-substitution-rule) is faster. For tabular integration, more LIATE practice and harder examples, read the full guide to [integration by parts](/blog/integration-by-parts).

## Frequently asked questions

### What is the integration by parts formula?

It is $\int u\,dv = uv - \int v\,du$. You split the integrand into a factor $u$ to differentiate and a factor $dv$ to integrate.

### What does LIATE stand for?

Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. Choose $u$ from the type that appears earliest in the list; the rest is $dv$.

### How do you integrate ln x?

Take $u = \ln x$ and $dv = dx$, so $du = \frac{1}{x}dx$ and $v = x$. Then $\int \ln x\,dx = x\ln x - \int 1\,dx = x\ln x - x + C$.

### When should I use integration by parts instead of u-substitution?

Use substitution when one factor is the derivative of an inner function. Use parts when you have a product of unrelated functions, such as a polynomial times an exponential, log or trig function.