# 3.4 Partial Fractions

### Learning Objectives

• Integrate a rational function using the method of partial fractions.
• Recognize simple linear factors in a rational function.
• Recognize repeated linear factors in a rational function.
• Recognize quadratic factors in a rational function.

We have seen some techniques that allow us to integrate specific rational functions. For example, we know that  and, as a consequence,

,

since if we let , we have and hence .

We can also evaluate using a trigonometric substitution. Indeed, if we let

(the integrand doesn’t have a square root but it still has expression , for which we used this kind of substitution), then we have that , , and .

With this, we obtain:

, that is,

.

Still, we do not yet have a technique that allows us to tackle arbitrary quotients of polynomials. For instance, it is not immediately obvious how to go about evaluating However, by bringing to a common denominator, we can verify that

and using we obtain that

In this section, we examine the method of partial fraction decomposition, which allows us to decompose rational functions into sums of simpler functions that are easier to integrate, as in the above example. In particular, formulas and will become really handy.

The key to the method of partial fraction decomposition is being able to anticipate the form that the decomposition of a rational function will take. As we shall see, this form is both predictable and highly dependent on the factorization of the denominator of the rational function. It is also extremely important to keep in mind that partial fraction decomposition can be applied to a rational function only if where stands for the degree of the polynomial . In the case when we must first perform long division to rewrite the quotient in the form where We then do a partial fraction decomposition on The following example, although not requiring partial fraction decomposition, illustrates our approach to integrals of rational functions of the form where

### Integrating where

Evaluate

#### Solution

Since we perform long division to obtain

Thus,

Visit Wikipedia to review long division of polynomials.

Evaluate

#### Hint

Use long division to obtain

To integrate where we must begin by factoring

### Nonrepeated Linear Factors

If can be factored as where each linear factor is distinct and no factor is a constant multiple of another, then it is possible to find constants satisfying

The proof that such constants exist is beyond the scope of this course.

In the next example, we see how to use partial fractions to integrate a rational function of this type.

### Partial Fractions with Nonrepeated Linear Factors

Evaluate

#### Solution

Since we begin by factoring the denominator of the integrand. We can see that Thus, there are constants and satisfying

We must now find these constants. To do so, we begin by bringing the right-hand side to a common denominator. We have:

Now, we set the numerators equal to each other, obtaining

There are two different strategies for finding the coefficients and We refer to these as the method of equating coefficients and the method of strategic substitution.

### Method of Equating Coefficients

Expand the right-hand side of (1) and then group the terms by the powers of x to rewrite it as

Equating coefficients produces the system of equations

To solve this system, we first observe that Substituting this value into the first two equations gives us the system

Multiplying the second equation by and adding the resulting equation to the first produces

which in turn implies that Substituting this value into the equation yields Thus, solving these equations yields and

It is important to note that the system produced by this method is consistent, that is, has a solution, if and only if we have set up the decomposition correctly. If the system is inconsistent, there is an error in our decomposition.

### Method of Strategic Substitution

The method of strategic substitution is based on the assumption that we have set up the decomposition correctly. If the decomposition is set up correctly, then there must be values of and that satisfy (1) for all values of That is, this equation must be true for any value of we care to substitute into it. Therefore, by choosing values of carefully and substituting them into the equation, we may find and easily. For example, if we substitute the equation reduces to Solving for yields Next, by substituting the equation reduces to or equivalently Last, we substitute into the equation and obtain Solving, we have

It is important to keep in mind that if we attempt to use this method with a decomposition that has not been set up correctly, we are still able to find values for the constants, but these constants are meaningless. If we do opt to use the method of strategic substitution, then it is a good idea to check the result by recombining the terms algebraically.

Now that we have the values of and we rewrite the original integral:

Evaluating the integral gives us

In the next example, we integrate a rational function in which the degree of the numerator is not less than the degree of the denominator.

### Dividing before Applying Partial Fractions

Evaluate

#### Solution

Since we must perform long division of polynomials. This results in

Next, we perform partial fraction decomposition on

We have

Thus,

Solving for and using either method, we obtain and

Rewriting the original integral, we have

Evaluating the integral produces

As we will see in the next example, it may be possible to apply the technique of partial fraction decomposition to a nonrational function. The trick is to convert the nonrational function to a rational function through a substitution.

### Applying Partial Fractions after a Substitution

Evaluate

#### Solution

Let’s begin by letting Consequently, After making these substitutions, we have

Applying partial fraction decomposition to gives

Therefore,

Evaluate

### Repeated Linear Factors

For some applications, we need to integrate rational expressions that have denominators with repeated linear factors—that is, there is at least one factor of the form where is a positive integer greater than or equal to If the denominator contains the repeated linear factor then the corresponding terms in the decomposition are

As we will see in our next example, the basic technique used for solving for the coefficients is the same, but it requires more algebra to determine the numerators of the partial fractions.

### Partial Fractions with Repeated Linear Factors

Evaluate

#### Solution

We have so we can proceed with the decomposition. Since is a repeated linear factor, the corresponding terms in the decomposition are going to be , and hence

Bringing to a common denominator and equating the numerators, we have

We then use the method of equating coefficients to find the values of and

Equating coefficients yields and Solving this system we obtain that and

Alternatively, we can use the method of strategic substitution. In this case, substituting and into (2) easily produces the values and At this point, it may seem that we have run out of good choices for however, since we already have values for and we can substitute in these values and choose any that we haven’t used yet. The value is a good option since it’s very easy to substitute. This way, we obtain

, and solving for we get

Now that we have the values for and we rewrite the original integral and evaluate it:

To integrate , we make a substitution , yielding , and then use the power formula to evaluate

.

Set up the partial fraction decomposition for (Do not solve for the coefficients or perform integration.)

### The General Method

Now that we are beginning to get the idea of how the technique of partial fraction decomposition works, let’s outline the basic method in the following problem-solving strategy.

### Problem-Solving Strategy: Partial Fraction Decomposition

To decompose the rational function use the following steps:

1. Make sure that If not, perform long division of polynomials.
2. Factor into the product of linear and irreducible quadratic factors. An irreducible quadratic is a quadratic that has no real zeros.
3. Assuming that the factors of determine the form of the decomposition of
• If can be factored as where each linear factor is distinct and no factor is a constant multiple of another, then it is possible to find constants satisfying
• If contains the repeated linear factor then the decomposition must contain
• For each irreducible quadratic factor that contains, the decomposition must include
• For each repeated irreducible quadratic factor the decomposition must include
• After the appropriate decomposition is determined, solve for the constants.
• If using the decomposition to evaluate an integral, rewrite the integrand in its decomposed form and evaluate it using previously developed techniques or integration formulas.

### Simple Quadratic Factors

Now let’s look at integrating a rational expression in which the denominator contains an irreducible quadratic factor. Recall that the quadratic is irreducible if has no real zeros—that is, if

### Rational Expressions with an Irreducible Quadratic Factor

Evaluate

#### Solution

Since factor the denominator and proceed with partial fraction decomposition. Because contains irreducible quadratic factor include as a part of the decomposition, along with for the linear term Thus, the decomposition has the form

After bringing to a common denominator and equating the numerators, we obtain the equation

Solving for and we get and

Therefore,

Substituting back into the integral, we obtain

In order to evaluate , we perform a substitution .

Note: We may rewrite if we wish to do so, since

### Partial Fractions with an Irreducible Quadratic Factor

Evaluate

#### Solution

Since the numerator is and , we can proceed with partial fraction decomposition. We start by factoring We see that the quadratic factor is irreducible since Using the decomposition described in the problem-solving strategy, we get

After bringing to a common denominator and equating the numerators, this becomes

Applying either method, we get

Rewriting we have

We can see that

but requires a bit more effort. Let’s begin by completing the square in to obtain

By letting and consequently we see that

Substituting back into the original integral and simplifying gives

Here again, we can drop the absolute value if we wish to do so, since for all

### Finding a Volume

Find the volume of the solid of revolution obtained by revolving the region enclosed by the graph of and the x-axis over the interval about the y-axis.

#### Solution

Let’s begin by sketching the region to be revolved, see the figure below. From the sketch, we see that the shell method is a good choice for solving this problem.

The volume is given by

Since we can proceed with partial fraction decomposition. Note that is a repeated irreducible quadratic. Using the decomposition described in the problem-solving strategy, we get

Finding a common denominator and equating the numerators gives

Solving, we obtain and Substituting back into the integral, we have

Set up the partial fraction decomposition for

### Key Concepts

• Partial fraction decomposition is a technique used to break down a rational function into a sum of simple rational functions that can be integrated using previously learned techniques.
• When applying partial fraction decomposition, we must make sure that the degree of the numerator is less than the degree of the denominator. If not, we need to perform long division before attempting partial fraction decomposition.
• The form the decomposition takes depends on the type of factors in the denominator. The types of factors include nonrepeated linear factors, repeated linear factors, nonrepeated irreducible quadratic factors, and repeated irreducible quadratic factors.

### Exercises

Express the given rational function as a sum or difference of simpler rational functions.

1.

2.

3.

4.

5.

#### Answer

6.

(Hint: Use long division first.)

7.

8.

9.

10.

11.

12.

13.

14.

15.

#### Answer

Evaluate the following integrals using partial fraction decomposition, when applicable.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

#### Answer

26.

(Hint: is a root of the denominator.)

#### Answer

27.

(Hint: to factor the denominator, let .)

28.

29.

30.

31.

32.

#### Answer

33.

34.

(Hint: is a root of the denominator.)

#### Answer

Use a substitution to convert the given integrals to integrals of rational functions. Then evaluate the resulting integrals using partial fractions or the methods developed earlier.

35.

36.

37.

38.

#### Answer

39.

40.

(Hint: use the simplest natural substitution.)

41.

42.

43.

44.

#### Answer

45.

46. Use the substitution to convert to an integral of a rational function, then evaluate.

#### Answer

47. Find the area of the region bounded by the curve , the x-axis, and the line

48. Find the volume of the solid generated when the region bounded by and is revolved about the x-axis.

#### Answer

49. Find the volume generated by revolving the area bounded by about the y-axis.

50. The velocity of a particle moving along a line is given by where is the time in seconds. Find the distance that the particle has traveled after sec.

#### Answer

For the following two problems, use the substitutions and

51.

52. Find the area under the curve between and (Assume the dimensions are in inches.)

#### Answer

2 in.2

53. Let Derive the formulas and

54. Evaluate

### Glossary

partial fraction decomposition
a technique used to break down a rational function into the sum of simple rational functions

## License

Calculus: Volume 2 (Second University of Manitoba Edition) Copyright © 2021 by Gilbert Strang and Edward 'Jed' Herman, modified by Varvara Shepelska is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.