We will integrate rational functions using partial fraction decompositions.
Partial Fractions
In this section we will learn an important method for integrating rational functions. Recall that a rational function is a ratio of polynomials. The key background skill is the ability to factor polynomials. We will look at three examples that cover the scenarios that we will encounter in our integrals.
First, consider the cubic polynomial . We can factor out an and then factor again using the difference of two perfect squares:
This is a complete factorization consisting of three different linear factors, and . We can say that the original cubic polynomial factored into distinct linear factors.
Second, consider the cubic polynomial . This polynomial factors as but this is not a complete factorization. The quadratic, , can be factored since it is a perfect square. The complete factorization is In this case, our original cubic polynomial factored into a linear factor and a repeated linear factor.
Last, consider the cubic polynomial . This polynomial factors as . The quadratic does not factor (unless you are willing to use imaginary numbers) and is called an irreducible quadratic polynomial. So we say that the original cubic polynomial factored into a linear factor and an irreducible quadratic factor.
Now we will investigate the three decomposition forms we will encounter in our integrals.
- Distinct Linear Factors
A rational function whose denominator factors into a product of distinct linear factors can be decomposed into a sum of rational functions whose numerators are constants and whose denominators are the factors of the denominator. For example, where and are constants. - Repeated Linear Factor
The repeated linear factor will contribute a number of terms to the partial fraction decomposition equal to the multiplicity of the factor. Each of these terms will have a constant in the numerator and a distinct power of the repeated factor in the denominator. For example, where and are constants. Note how the repeated linear factor contributes two terms whose denominators are and and whose numerators were constants. - Irreducible Quadratic Factor
An irreducible quadratic factor in the denominator will contribute one term to the decomposition whose numerator is a linear polynomial and whose denominator is the irreducible quadratic factor. For example, where and are constants.
Before we move on to integrating the special rational functions seen in the above decompositions, we will look at one more example consisting of all three of the ideas mentioned above. Find the partial fraction decomposition of The denominator is already in a completely factored form. This form consists of a linear factor , a repeated linear factor and an irreducible quadratic . The partial fraction decomposition will consist of one term for the factor and three terms for the factor . Each of these terms will have a constant in the numerator. The decomposition will also contain one term for the irreducible quadratic, and this term will have a linear polynomial in the numerator and the irreducible quadratic in the denominator. The form of the decomposition is: where and are constants.
Special Integrals
The point of a partial fraction decomposition is to replace a complicated rational function by a sum of simpler ones which we can integrate. We will now look at some of these simpler forms and learn how to integrate them.
example 1 Compute the indefinite integral Using a -substitution with and , we get
In future examples, we will not perform the substitution. Instead, we will just write
example 2 Compute the following indefinite integral, assuming that : Using a -substitution with
and , we get
In short, we use the power rule with exponent ‘’ and we multiply by .
example 3 Compute the indefinite integral Using a substitution with and , so that
and , we get
In summary,
We are now ready to tackle more general scenarios.
Distinct Linear Factors
example 5
Compute the indefinite integral We begin by factoring the denominator and writing the partial fraction decomposition: We can now integrate using the results of the special integrals section above: We have used for the constant of integration since the letter was already in use. We will convert back to in our final answer, below. Now, we must determine the values of the parameters and using algebraic methods. Going back to the original decomposition, we have Multiplying both sides of this equation by we have Now, we will plug in the values and for to determine and respectively. Letting in the equation above eliminates the and terms, yielding so that . Next, letting eliminates the and terms, yielding so that . Finally, letting eliminates the and terms, yielding so that . We can now write our final answer:
Here is a detailed, lecture style video on distinct linear factors:
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(problem 5a)
Compute
The partial fraction decomposition has the form
The answer has the form
The coefficients are
(problem 5b)
Compute
The partial fraction decomposition has the form
The answer has the form
The coefficients are
Repeated Linear Factors
example 6 Compute the indefinite integral We begin by factoring the denominator
and writing the partial fraction decomposition: We can now integrate using the
results of the special integrals section above: Now, we must determine the values of
the parameters and using algebraic methods. Going back to the original
decomposition, we have Multiplying both sides of this equation by gives Letting
will eliminate the and terms, allowing us to find : which implies that . If we let the
and terms get eliminated and we are left with: Choosing another -value will leave
us with another equation involving . Since and , choosing yields and so . The final
answer is thus:
Here is a detailed, lecture style video on repeated linear factors:
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(problem 6a)
Compute
The partial fraction decomposition has the form
The answer has the form
The coefficients are
(problem 6b)
Compute
The partial fraction decomposition has the form
The answer has the form
The coefficients are
(problem 6c)
Compute
The partial fraction decomposition has the form
The answer has the form
The coefficients are
Irreducible Quadratic Factors
example 7
Compute the indefinite integral:
We begin by factoring the denominator and writing the partial fraction decomposition: We can now integrate using the results of the special integrals section above: Now, we must determine the values of the parameters and using algebraic methods. Going back to the original decomposition, we have Multiplying both sides of this equation by gives Rather than plugging in certain values for , we can find and by another method, called “equating coefficients.” We will rewrite the right hand side in standard polynomial form: Since these polynomials are equal, their coefficients must match. Hence The third of these equations gives , and using this in the first equation gives also. The final answer is thus:
Here is a detailed, lecture style video on irreducible quadratic factors:
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