where and are polynomials. It's possible to express
as a sum of simpler fractions provided that the degree of
is less than the degree of Such a rational function
is calledproper. If is improper, then we must take
the preliminary step of dividing into (by long division)
until a remainder is obtained such that
The division statement is

where and are also polynomials.

The next step is to factor the denominator as far as
possible. It can be shown that any polynomial can be factored
as a product of linear factors (of the form ) and irreducible
quadratic factors (of the form where ).
For instance,

The third step is to express the proper rational function
as a sum of partial fractions of the form

A theorem in algebra guarantees that it is always possible to do this.

There are four possible cases:

Case I: The denominator is a product of distinct linear
factors.

This means that we can write

where no factor is repeated (and no factor is a constant multiple of
another). For example,

In this case the partial fraction theorem states that there
exist constants such that

(1)

For instance,

Here is a more complicated example:

where

Case II: The denominator is a product of linear
factors, some of which are repeated.

This means that we can write

where For example,

In this case instead of the single term
in (1), we would use

(2)

For instance,

Here is a more complicated example:

where

Case III: The denominator
contains irreducible quadratic factors, none of which is repeated.

For example,

In this case if has the factor where
then, in addition to the partial fractions in (1) and (2),
the expression for will have a term of the form

(3)

For instance,

Here is a more complicated example:

where

Case IV: The denominator
contains a repeated irreducible quadratic factor.

For example,

In this case if has the factor where
then, instead of a single partial fraction (3),
the sum

(4)

occurs in the partial fraction decomposition of

For instance,

Here is a more complicated example:

where

EXAMPLES:
1. Evaluate

Solution: This is a Type I integral. Therefore first we find constants
and such that

We have

therefore

(5)

We now can proceed in two different ways:

Method 1: If we expand the parentheses on the right-hand side of (5)
and collect like terms, we get

hence

Method 2: If we put in (5), we get

Similarly, if we put in (5), we get

Finally, if we put in (5), we get

So, we have

therefore

2. Evaluate

Solution: Since

this is a Type I integral. Therefore first we find constants
and such that

We have

therefore

(6)

We now can proceed in two different ways:

Method 1: If we expand the parentheses on the right-hand side of (6)
and collect like terms, we get

hence

Method 2: If we put in (6), we get

Similarly, if we put in (6), we get

So, we have

therefore

3. Evaluate

Solution: Since the degree of the numerator is greater than the degree of
the denominator, we first perform the long division. This
enables us to write

4. Evaluate

Solution: Since we have a repeated linear factor and no
irreducible quadratic factors, this is a Type II integral.
Therefore first we find constants and such that

We have

therefore

(7)

We now can proceed in two different ways:

Method 1: If we expand the parentheses on the right-hand side of (7)
and collect like terms, we get

hence

Method 2: If we put in (7), we get

Similarly, if we put in (7), we get

So, we have

therefore

5. Evaluate

Solution: We have

Since we have a repeated linear factor and no
irreducible quadratic factors, this is a Type II integral.
Therefore first we find constants and such that

We have

therefore

(8)

We now can proceed in two different ways:

Method 1: If we expand the parentheses on the right-hand side of (8)
and collect like terms, we get

hence

Method 2: If we put in (8), we get

Similarly, if we put in (8), we get

Finally, if we put in (8), we get

So, we have

therefore

6. Evaluate

Solution: We have

This is a Type III integral.
Therefore first we find constants and such that

We have

therefore

(9)

We now can proceed in two different ways:

Method 1: If we expand the parentheses on the right-hand side of (9)
and collect like terms, we get

hence

Method 2: If we put in (9), we get

so
Similarly, if we put in (9), we get

so
Finally, plugging in , we get

So, we have

hence

Note that

therefore

7. Evaluate

Solution: We have

This is a Type III integral.
Therefore first we find constants and such that

We have

therefore

(10)

We now can proceed in two different ways:

Method 1: If we expand the parentheses on the right-hand side of (10)
and collect like terms, we get

hence

Method 2: If we put in (10), we get

Similarly, if we put in (10), we get

Finally, plugging in , we get

From and it follows that

So, we have

therefore (see Appendix B)

8. Evaluate

Solution: This is a Type IV integral.
Therefore first we find constants and such that

We have

therefore

(11)

We now can proceed in two different ways:

Method 1: If we expand the parentheses on the right-hand side of (11)
and collect like terms, we get

hence

Method 2: If we put in (11), we get
Plugging in and three other different values of into (11),
we get and

So, we have

hence

Note that

therefore

9. Evaluate

Solution: Since

this is a Type IV integral.
Therefore first we find constants and such that

We have

therefore

(12)

We now can proceed in two different ways:

Method 1: If we expand the parentheses on the right-hand side of (12)
and collect like terms, we get

hence

Method 2: If we put in (12), we get
Plugging in four other different values of into (12),
we get and