- 3.4.1. Integrate a rational function using the method of partial fractions.
- 3.4.2. Recognize simple linear factors in a rational function.
- 3.4.3. Recognize repeated linear factors in a rational function.
- 3.4.4. 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
However, we do not yet have a technique that allows us to tackle arbitrary quotients of this type. Thus, it is not immediately obvious how to go about evaluating However, we know from material previously developed that
In fact, by getting a common denominator, we see that
In this section, we examine the method of partial fraction decomposition, which allows us to decompose rational functions into sums of simpler, more easily integrated rational functions. Using this method, we can rewrite an expression such as: as an expression such as
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 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
Since we perform long division to obtain
Visit this website for a review of long division of polynomials.
To integrate where we must begin by factoring
Nonrepeated Linear Factors
If can be factored as where each linear factor is distinct, then it is possible to find constants satisfying
The proof that such constants exist is beyond the scope of this course.
In this next example, we see how to use partial fractions to integrate a rational function of this type.
Partial Fractions with Nonrepeated Linear Factors
Since we begin by factoring the denominator of We can see that Thus, there are constants and satisfying
We must now find these constants. To do so, we begin by getting a common denominator on the right. Thus,
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.
Rewrite Equation 3.8 in the form
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 if and only if we have set up the decomposition correctly. If the system is inconsistent, there is an error in our decomposition.
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 Equation 3.8 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
Since we must perform long division of polynomials. This results in
Next, we perform partial fraction decomposition on We have
Solving for and using either method, we obtain and
Rewriting the original integral, we have
Evaluating the integral produces
As we 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
Let’s begin by letting Consequently, After making these substitutions, we have
Applying partial fraction decomposition to gives
Repeated Linear Factors
For some applications, we need to integrate rational expressions that have denominators with repeated linear factors—that is, rational functions with 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 decomposition must contain
As we 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
We have so we can proceed with the decomposition. Since is a repeated linear factor, include in the decomposition. Thus,
After getting 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 yields and
Alternatively, we can use the method of strategic substitution. In this case, substituting and into Equation 3.9 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 value for not previously used. The value is a good option. In this case, we obtain the equation or, equivalently,
Now that we have the values for and we rewrite the original integral and evaluate it:
Set up the partial fraction decomposition for (Do not solve for the coefficients or complete the 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.
To decompose the rational function use the following steps:
- Make sure that If not, perform long division of polynomials.
- Factor into the product of linear and irreducible quadratic factors. An irreducible quadratic is a quadratic that has no real zeros.
- Assuming that the factors of determine the form of the decomposition of
- If can be factored as where each linear factor is distinct, 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.
- Last, rewrite the integral in its decomposed form and evaluate it using previously developed techniques or integration formulas.
- If can be factored as where each linear factor is distinct, then it is possible to find constants satisfying
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
Since factor the denominator and proceed with partial fraction decomposition. Since contains the irreducible quadratic factor include as part of the decomposition, along with for the linear term Thus, the decomposition has the form
After getting a common denominator and equating the numerators, we obtain the equation
Solving for and we get and
Substituting back into the integral, we obtain
Note: We may rewrite if we wish to do so, since
Partial Fractions with an Irreducible Quadratic Factor
We can start by factoring We see that the quadratic factor is irreducible since Using the decomposition described in the problem-solving strategy, we get
After obtaining 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 on 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.
Let’s begin by sketching the region to be revolved (see Figure 3.11). 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
Section 3.4 Exercises
Express the rational function as a sum or difference of two simpler rational expressions.
(Hint: Use long division first.)
Use the method of partial fractions to evaluate each of the following integrals.
Evaluate the following integrals, which have irreducible quadratic factors.
Use the method of partial fractions to evaluate the following integrals.
(Hint: Use the rational root theorem.)
Use substitution to convert the integrals to integrals of rational functions. Then use partial fractions to evaluate the integrals.
(Give the exact answer and the decimal equivalent. Round to five decimal places.)
Use the given substitution to convert the integral to an integral of a rational function, then evaluate.
Graph the curve over the interval Then, find the area of the region bounded by the curve, the x-axis, and the line
Find the volume of the solid generated when the region bounded by and is revolved about the x-axis.
The velocity of a particle moving along a line is a function of time given by Find the distance that the particle has traveled after sec.
Solve the initial-value problem for x as a function of t.
Find the x-coordinate of the centroid of the area bounded by
(Round the answer to two decimal places.)
Find the volume generated by revolving the area bounded by about the y-axis.
Find the area bounded by (Round the answer to the nearest hundredth.)
Evaluate the integral
For the following problems, use the substitutions and
Find the area under the curve between and (Assume the dimensions are in inches.)
Given derive the formulas and