# Indefinite Integrals

### Topics

## Introduction to Indefinite Integrals - At A Glance:

You can think of **integration by parts** as a way to undo the product rule. While integration by substitution lets us find antiderivatives of functions that came from the chain rule, integration by parts lets us find antiderivatives of functions that came from the product rule.

Integration by parts requires learning and applying the integration-by-parts formula. Here's the formula, written in both Leibniz and Lagrange notation:

## Why the Formula is True

It's time to be skeptical. Why should

be true? Just because we wrote down an equation doesn't mean it's valid.

### Sample Problem

- What should be the derivative of with respect to
*x*?

- What should be the derivative of with respect to
*x*?

- What's the derivative of
*uv*?

Answer.

- is the family of all functions whose derivative is
*uv'*. If we take the derivative of any function in that family, we get*uv'*. So it makes sense for the derivative of to also be*uv'*. In symbols,

- is the collection of all functions whose derivative is
*vu'*. If we take the derivative of any one of those functions, we get*vu'*, so the derivative of should also be*vu'*. In symbols,

- This is a straightforward application of the product rule. The derivative of
*uv*is*uv'*+*u'v*.

Let's return to the question of why the equation

should be valid. Each side of this equation describes a family of functions. If each side of the equation describes the same family of functions, then the equation is valid.

describes the family of functions with derivative *uv*'.

The expression

describes the family of functions with derivative

Since *uv'* and *vu'* are the same, when we simplify we get

*u'v* + *uv'* – *vu'* = *u'v*.

This means the expression

also describes the family of functions with derivative *u'v*.

Since the expressions

and

describe the same family of functions, the equation

is valid.

## How to Use the Formula

Hopefully, you're now convinced that

To use this formula we have to

- pick
*u*and*v*',

- figure out what
*u '*and*v*are, and

- stick everything in the formula and simplify.

If we're using the notation

then we pick *u* and *dv*, figure out *du* and *v*, then apply the formula.