# 7. Integration by Parts

by M. Bourne

Sometimes we meet an integration that is the product of 2 functions. We may be able to integrate such products by using **Integration by Parts**.

If *u* and *v* are functions of *x*, the
product rule for differentiation that we met earlier gives us:

`d/(dx)(uv)=u(dv)/(dx)+v(du)/(dx)`

Rearranging, we have:

`u(dv)/(dx)=d/(dx)(uv)-v(du)/(dx)`

Integrating throughout with respect to *x*, we obtain
the formula for **integration by parts:**

`intu\ dv=uv-intv\ du`

This formula allows us to turn a complicated integral into
more simple ones. We must make sure we choose *u* and
*dv* carefully.

**NOTE: **The function *u* is chosen so
that `(du)/(dx)` is **simpler** than
*u*.

**Priorities for Choosing**
**u**

**u**

When you have a mix of functions in the expression to be integrated, use the following for your choice of `u`, in order.

1. Let `u = ln\ x`

2. Let `u = x^n`

3. Let `u = e^(nx)`

### Example 1

`intx\ sin\ 2x\ dx`

#### Solution

We need to choose `u`. In this question we don't have any of the functions suggested in the "priorities" list above.

We could let `u = x` or `u = sin\ 2x`*.* In general, we choose the one that allows `(du)/(dx)`
to be of a simpler form than *u*.

So for this example, we choose *u* = *x* and so `dv` will be the "rest" of the integral,
*dv* = sin 2*x dx*.

We have `u = x` so `du = dx`.

Also `dv = sin\ 2x\ dx` and integrating gives:

`{:(v,=intsin\ 2x\ dx),(,=(-cos\ 2x)/2):}`

Substituting these 4 expressions into the integration by parts formula, we get:

### Example 2

`intxsqrt(x+1)\ dx`

### Example 3

`intx^2\ ln\ 4x\ dx`

### Example 4

`intx\ sec^2\ x\ dx`

### Example 5

`intx^2e^(-x)dx`

### Example 6

`intln\ x\ dx`

### Example 7

`intarcsin\ x\ dx`

This time we integrated an inverse trigonometric function (as opposed to the earlier type where we obtained inverse trigonometric functions in our answer). See Integration: Inverse Trigonometric Forms.

### Alternate Method for Integration by Parts

Here's an alternative method for problems that can be done using Integration by Parts. You may find it easier to follow.

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