We shall describe a method for computing the integral of any rational function, and we shall find that the result can always be expressed in terms of polynomials, rational functions, inverse tangents, and logarithms.

The basic idea of the method is to decompose a given rational function into a sum of simpler fractions (called partial fractions) that can be integrated by the techniques discussed earlier. We shall describe the general procedure by means of a number of simple examples that illustrate all the essential features of the method.

EXAMPLE 1. In this example we begin with two simple fractions,
and
which we know how to integrate, and see what happens when we form a linear
combination of these fractions. For example, if we take twice the first
fraction plus three times the second, we obtain

If now, we read this formula from right to left, it tells us that the rational
function
given by
has been expressed as a linear combination of
and
.
Therefore, we may evaluate the integral of
by writing

EXAMPLE 2. The foregoing example suggests a procedure for dealing with
integrals of the form
For example, to evaluate
,
we try to express the integral as a linear combination of
and
by writing

with constants
and
to be determined. If we can choose
and
so that the above equation is an identity, then the integral of the fraction
on the left is equal to the sum of the integrals of the simpler fractions on
the right. To find
and
,
we multiply both sides by
to remove the fractions. This gives us

At this stage there are two methods commonly used to find
and
.
One method is to equate coefficients of like powers of
.
This leads to the equations
and
.
Solving this pair of simultaneous equations, we obtain
and
The other method involves the substitution of two values of
in the equation and leads to another pair of equations for
and
.
In this particular case, the presence of the factors
and
suggests that we use the values
I
and
.
When we put
in the equation, the coefficient of
vanishes, and we find
,
or
Similarly, we can make the coefficient of
vanish by putting
This gives us
,
or
.
In any event, we have found values of
and
to satisfy the equation, so we have

It is clear that the method described in Example 2 also applies to integrals of the form in which is a linear polynomial and is a quadratic polynomial that can be factored into distinct linear factors with real coefficients, say In this case the quotient can be expressed as a linear combination of and , and integration of leads to a corresponding combination of the logarithmic terms and .

The foregoing examples involve rational functions
in which the degree of the numerator is less than that of the denominator. A
rational function with this property is said to be a *proper* rational
function. If
is improper, that is, if the degree of
is not less than that of
,
then we can express
as the sum of a polynomial and a proper rational function. In fact, we simply
divided by
to obtain

where
and
are polynomials (called the *quotient* and *remainder*,
respectively) such that the remainder has degree less than that of
.
For example,

Therefore, in the study of integration technique, there is no loss in
generality if we restrict ourselves to *proper* rational functions, and
from now on we consider
,
where
has degree less than that of
.

A general theorem in algebra states that every proper rational function can be
expressed as a finite sum of fractions of the forms

where
and
are positive integers and
are constants with
The condition
means that the quadratic polynomial
cannot be factored into linear factors with real coefficients or, what amounts
to the same thing, the quadratic equation
has no real roots. Such a quadratic factor is said to be *irreducible*.
When a rational function has been so expressed, we say that it has been
decomposed into **partial fractions**. Therefore the problem of
integrating this rational function reduces to that of integrating its partial
fractions. These may be easily dealt with by the techniques described in the
examples which follow.

We shall not bother to prove that partial-fraction decompositions always exist. Instead, we shall show (by means of examples) how to obtain the partial fractions in specific problems. In each case that arises the partial-fraction decomposition can be verified directly.

It is convenient to separate the discussion into cases depending on the way in which the denominator of the quotient can be factored.

CASE 1. *The denominator is a product of distinct linear factors.*
Suppose that
splits into
distinct linear factors, say

Now notice that a linear combination of the form

may be expressed as a single fraction with the common denominator
and the numerator of this fraction will be a polynomial of degree
involving the
's.
Therefore, if we can find
's
to make this numerator equal to
,
we shall have the decomposition

and the integral of
will be equal to
In the next example, we work out a case with

EXAMPLE 3. Integrate

Solution. Since
,
the denominator is a product of distinct linear factors, and we try to find
and
such that

Clearing the fractions, we obtain

When
we find
so
When
we obtain
,
and when
we find
or
Therefore
we have

CASE 2. The denominator is a product of linear factors, some of which are repeated. We illustrate this case with an example.

EXAMPLE 4. Integrate .

Solution. Here we try to find
so that

We need both
and
as well as
in order to get a polynomial of degree two in the numerator and to have as
many constants as equations when we try to determine the
's.
Clearing the fractions, we obtain

Substituting
,
we find
,
so
.
When
we obtain
and
We need one more equation to determine
Since there are no other choices of
that will make any factor vanish, we choose a convenient
that will help to simplify the calculations. For example, the choice
leads to the equation
from which we find
An alternative method is to differentiate both sides of the equation and then
substitute a convenient x. Differentiation leads to the equation

and, if we put
we
find
so
as before. Therefore we have found
's
to satisfy the equation, so we have

If, on the left of the equation the factor
had appeared instead of
,
we would have added an extra term
on the right. More generally, if a linear factor
appears
times in the denominator, then for this factor we must allow for a sum of
terms, namely

where the
's
are constants. A sum of this type is to be used for each repeated linear
factor.

CASE 3. The denominator contains irreducible quadratic factors, none of which are repeated.

EXAMPLE 5. Integrate .

Solution. The denominator can be split as the product
,
where
is irreducible, and we try a decomposition of the form

In the fraction with denominator
,
we have used a linear polynomial
in the numerator in order to have as many constants as equations when we solve
for
Clearing the fractions and solving for
and
we find
and
Therefore we have

The first integral on the right is
.
To evaluate the second integral, we write

If we let
and
,
the last integral is

Therefore, we have

CASE 4. *The denominator contains irreducible quadratic factors, some of
which are repeated.* Here the situation is analogous to Case 2. In the
partial fraction decomposition of
we allow first of all, a sum of the form

for each linear factor, as already described. In addition. if an irreducible
quadratic factor
is repeated
times, we allow a sum of
terms, namely

where each numerator is linear.

EXAMPLE 6. Integrate

Solution. We write

Clearing the fractions and solving for
and
we find that

Therefore, we have

The foregoing examples are typical of what happens in general. The problem of
integrating a proper rational function reduces to that of calculating
integrals of the forms

The first integral is
if
and
if
To
treat the other two, we express the quadratic as a sum of two squares by
writing

where
and
.
(This is possible because
.)
The substitution
reduces the problem to that of computing

The first of these is
if
,
and
if
When
,
the second integral is evaluated by the formula

The case
may be reduced to the case
by repeated application of the recursion formula

which is obtained by integration by parts. This discussion shows that every
rational function may he integrated in terms of polynomials, rational
functions, inverse tangents, and logarithms.

A function of two variables defined by an equation of the form

is called a *polynomial in two variables.* The quotient of two such
polynomials is called a *rational function of two variables.* Integrals
of the form
where
is a rational function of two variables, may be reduced by the substitution
to integrals of the form
where
is a rational function of one variable. The latter integral may be evaluated
by the techniques just described. We illustrate the method with a particular
example.

EXAMPLE 1. Integrate .

Solution. The substitution
gives us

and

Therefore,
we have

where
and
.
The method of partial fractions leads to

and,
since
,
we obtain

The
final answer may be simplified somewhat by using suitable trigonometric
identities. First we note that
so the numerator of the last fraction is
In the denominator we write

We
may combine the term
with the arbitrary constant and rewrite as follows: