# Integration by parts

The integral of u dv equals uv minus the integral of v du. It is the product rule run backwards, used when the integrand is a product of two different kinds of function.

Maths (university) · Part 06: Integration techniques and uses · Card 1 of 9 · Last updated 2026-10-19

## Analogy

Unpacking a parcel in two stages: you take out the big item, then deal with what is left.

## Example

**Integral of x e^x dx**

u = x, dv = e^x dx.
du = dx, v = e^x.
x e^x - integral of e^x dx.
Result: x e^x - e^x + C.

## In practice

Handles integrals such as x e^x or x sin x that substitution cannot touch.

## Practice

### Flashcards

- **State the integration by parts formula.** The integral of u dv = uv - the integral of v du.
- **What is the integral of x e^x dx?** x e^x - e^x + C.

### Interview questions

**Evaluate the integral of x e^x dx and explain how you chose u and dv.**

- Let u = x so du = dx
- Let dv = e^x dx so v = e^x
- Formula gives x e^x minus integral of e^x
- Answer x e^x - e^x + C
- Differentiate to check

## Questions people ask

### How is integration by parts related to the product rule?

Differentiating u times v gives u prime v plus u v prime. Integrate both sides and rearrange, and you get the by-parts formula. So it is the product rule in reverse, trading one integral for another that should be easier to do.

### How do I decide what to call u?

Pick u as the part that gets simpler when differentiated, such as x or a logarithm. Pick dv as the part you can integrate easily. A common order for u is logarithms, then algebraic terms like x, then trig, then exponentials last.

## Related concepts

- [Choosing u and dv](https://studyvisually.com/learn/science/maths-university/maths-university-part-6/maths-university-part-6-concept-2)
- [Partial fractions](https://studyvisually.com/learn/science/maths-university/maths-university-part-6/maths-university-part-6-concept-3)
- [Trigonometric integrals](https://studyvisually.com/learn/science/maths-university/maths-university-part-6/maths-university-part-6-concept-4)
- [Integration by substitution](https://studyvisually.com/learn/science/maths-university/maths-university-part-5/maths-university-part-5-concept-7)

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