Integration by Parts

Derivation of Integration by Parts

Recall the product rule:

(uv)' = u'v + uv'

or

uv' = (uv)' - u'v

Integrating both sides, we have that

 Theorem:  Integration by Parts Let u and v be differentiable functions, then

Examples

A)

 u = x du = dx dv = ex  dx v = ex

Hence we have

=   xex - ex + C

Exercise

Evaluate

Integration By Parts Twice

Example

Solve

 u = x2 du = 2xdx dv = ex  dx v = ex

We have

We just did this integral in the last example, so our solution is

x2ex - 2[xex - ex] + C

The By Parts Trick

Example

Evaluate

 u =  ex du = ex dx dv = sinx dx v = -cosx

We have

Now try integration by parts again:

 u =  ex du = ex dx dv = cosx dx v = sinx

It seems as if we are back to where we started, however with a clever move, the answer appears.

Let

Then we have

I = -excosx + ex sinx -  I

Adding I to both sides we get

2I = -excosx + ex sinx

So that

-excosx + ex sinx
I =                                      +  C
2

We can conclude that

Other By Parts

Example:

 u = arctanx 1 du =                   dx            1 + x2 dv = dx v =  x

We get

letting

u = 1 + x2,    du = 2x dx,     x dx = du/2

Exercise:

Evaluate

When to Use Integration By Parts

1. When there is a mix of two types of functions such as trig and exponentials, poly and trig, etc.

2. Strange functions such as arctrig and ln.

3. When all else fails.

e-mail Questions and Suggestions