Today: 7.4: Integration of rational functions and
Supp. 4: Partial fraction expansion
Next: 7.7: Approximate integration 
Our goal today is to compute integrals of
the form
by decomposing
.
This is called partial fraction expansion.
Note that
, where
,
are complex conjugates.
Types of rational functions
. To do a
partial fraction expansion, first make sure
using long division. Then there are four possible situation,
each of increasing generality (and difficulty):
 is a product of distinct linear factors;
 is a product of linear factors, some of which are repeated;
 is a product of distinct irreducible quadratic factors,
along with linear factors some of which may be repeated; and,
 is has repeated irreducible quadratic factors, along with
possibly some linear factors which may be repeated.
The general partial fraction expansion theorem is beyond the
scope of this course. However, you might find the following
special case and its proof interesting.
Proof.
Since
and
are relatively prime, using the Euclidean
algorithm (long division), we can find polynomials
and
such that
Dividing both sides by
and multiplying by
yields
which completes the proof.
Example 5.5.3
Compute
First do long division. Get quotient of
and remainder of
.
This means that
Since we have distinct linear factors, we know
that we can write
for real numbers
.
A clever way to find
is to substitute
appropriate values in, as follows.
We have
Setting
on both sides we have (taking a limit):
Likewise, we have
Thus
Example 5.5.4
Compute the partial fraction expansion of
.
By the partial fraction theorem, there are constants
such that
Note that there's no possible way this could work
without the
term, since otherwise
the common denominator would be
.
We have
This method will not get us
!
For example,
While true this is useless.
Instead, we use that we know and ,
and evaluate at another value of , say 0.
so
.
Thus finally,
Example 5.5.5
Let's compute
.
Notice that
is a factor, since
is a root.
We have
There exist constants
such that
Then
You could find
by factoring the quadratic over
the complex numbers and getting complex number
answers. Instead, we evaluate
at a couple of values.
For example, at
we get
so
.
Next, use
to get
.
so
Finally,
It remains to compute
First, complete the square to get
Let
, so
and
.
Then
Finally, we put it all together and get
Discuss second quiz problem.
Problem: Compute
using
complex exponentials.
The answer is
Here's how to get it.
Simplify the inside part requires some imagination:

William Stein
20060315