Partial Fractions With Linear Factors

Learning Outcomes

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

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

[latex]\displaystyle\int \frac{du}{u}=\text{ln}|u|+C\text{ and }\displaystyle\int \frac{du}{{u}^{2}+{a}^{2}}=\frac{1}{a}{\tan}^{-1}\left(\frac{u}{a}\right)+C\text{.}[/latex]

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 [latex]\displaystyle\int \frac{3x}{{x}^{2}-x - 2}dx[/latex]. However, we know from material previously developed that

[latex]\displaystyle\int \left(\frac{1}{x+1}+\frac{2}{x - 2}\right)dx=\text{ln}|x+1|+2\text{ln}|x - 2|+C[/latex].

In fact, by getting a common denominator, we see that

[latex]\frac{1}{x+1}+\frac{2}{x - 2}=\frac{3x}{{x}^{2}-x - 2}[/latex].

Consequently,

[latex]\displaystyle\int \frac{3x}{{x}^{2}-x - 2}dx=\displaystyle\int \left(\frac{1}{x+1}+\frac{2}{x - 2}\right)dx[/latex].

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 [latex]\frac{P\left(x\right)}{Q\left(x\right)}[/latex] only if [latex]\text{deg}\left(P\left(x\right)\right)<\text{deg}\left(Q\left(x\right)\right)[/latex]. In the case when [latex]\text{deg}\left(P\left(x\right)\right)\ge \text{deg}\left(Q\left(x\right)\right)[/latex], we must first perform long division to rewrite the quotient [latex]\frac{P\left(x\right)}{Q\left(x\right)}[/latex] in the form [latex]A\left(x\right)+\frac{R\left(x\right)}{Q\left(x\right)}[/latex], where [latex]\text{deg}\left(R\left(x\right)\right)<\text{deg}\left(Q\left(x\right)\right)[/latex]. We then do a partial fraction decomposition on [latex]\frac{R\left(x\right)}{Q\left(x\right)}[/latex]. The following example, although not requiring partial fraction decomposition, illustrates our approach to integrals of rational functions of the form [latex]\displaystyle\int \frac{P\left(x\right)}{Q\left(x\right)}dx[/latex], where [latex]\text{deg}\left(P\left(x\right)\right)\ge \text{deg}\left(Q\left(x\right)\right)[/latex].

Example: Integrating [latex]\displaystyle\int \frac{P\left(x\right)}{Q\left(x\right)}dx[/latex], where [latex]\text{deg}\left(P\left(x\right)\right)\ge \text{deg}\left(Q\left(x\right)\right)[/latex]

Evaluate [latex]\displaystyle\int \frac{{x}^{2}+3x+5}{x+1}dx[/latex].

Show Solution

Recall: Polynomial Long Division

Set up the division problem as the numerator divided by the denominator
Determine the first term of the quotient by dividing the leading term of the dividend by the leading term of the divisor.
Multiply the answer by the divisor and write it below the like terms of the dividend.
Subtract the bottom binomial from the top binomial.
Bring down the next term of the dividend.
Repeat steps 2–5 until reaching the last term of the dividend.
If the remainder is non-zero, express as a fraction using the divisor as the denominator.

Interactive

Visit this website for a review of long division of polynomials.

try it

Evaluate [latex]\displaystyle\int \frac{x - 3}{x+2}dx[/latex].

Hint

Show Solution

Watch the following video to see the worked solution to the above Try It

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “3.4 Partial Fractions” here (opens in new window).

To integrate [latex]\displaystyle\int \frac{P\left(x\right)}{Q\left(x\right)}dx[/latex], where [latex]\text{deg}\left(P\left(x\right)\right)<\text{deg}\left(Q\left(x\right)\right)[/latex], we must begin by factoring [latex]Q\left(x\right)[/latex].

Nonrepeated Linear Factors

If [latex]Q\left(x\right)[/latex] can be factored as [latex]\left({a}_{1}x+{b}_{1}\right)\left({a}_{2}x+{b}_{2}\right)\ldots\left({a}_{n}x+{b}_{n}\right)[/latex], where each linear factor is distinct, then it is possible to find constants [latex]{A}_{1},{A}_{2}\text{,}\ldots {A}_{n}[/latex] satisfying

[latex]\frac{P\left(x\right)}{Q\left(x\right)}=\frac{{A}_{1}}{{a}_{1}x+{b}_{1}}+\frac{{A}_{2}}{{a}_{2}x+{b}_{2}}+\cdots +\frac{{A}_{n}}{{a}_{n}x+{b}_{n}}[/latex].

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.

Example: Partial Fractions with Nonrepeated Linear Factors

Evaluate [latex]\displaystyle\int \frac{3x+2}{{x}^{3}-{x}^{2}-2x}dx[/latex].

Show Solution

Since [latex]\text{deg}\left(3x+2\right)<\text{deg}\left({x}^{3}-{x}^{2}-2x\right)[/latex], we begin by factoring the denominator of [latex]\frac{3x+2}{{x}^{3}-{x}^{2}-2x}[/latex]. We can see that [latex]{x}^{3}-{x}^{2}-2x=x\left(x - 2\right)\left(x+1\right)[/latex]. Thus, there are constants [latex]A[/latex], [latex]B[/latex], and [latex]C[/latex] satisfying

[latex]\frac{3x+2}{x\left(x - 2\right)\left(x+1\right)}=\frac{A}{x}+\frac{B}{x - 2}+\frac{C}{x+1}[/latex].

We must now find these constants. To do so, we begin by getting a common denominator on the right. Thus,

[latex]\frac{3x+2}{x\left(x - 2\right)\left(x+1\right)}=\frac{A\left(x - 2\right)\left(x+1\right)+Bx\left(x+1\right)+Cx\left(x - 2\right)}{x\left(x - 2\right)\left(x+1\right)}[/latex].

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

[latex]3x+2=A\left(x - 2\right)\left(x+1\right)+Bx\left(x+1\right)+Cx\left(x - 2\right)[/latex].

There are two different strategies for finding the coefficients [latex]A[/latex], [latex]B[/latex], and [latex]C[/latex]. We refer to these as the method of equating coefficients and the method of strategic substitution.

Rule: Method of Equating Coefficients

Rewrite the previous equation in the form

[latex]3x+2=\left(A+B+C\right){x}^{2}+\left(\text{-}A+B - 2C\right)x+\left(-2A\right)[/latex].

Equating coefficients produces the system of equations

[latex]\begin{array}{ccc}\hfill A+B+C& =\hfill & 0\hfill \\ \hfill -A+B - 2C& =\hfill & 3\hfill \\ \hfill -2A& =\hfill & 2.\hfill \end{array}[/latex]

To solve this system, we first observe that [latex]-2A=2\Rightarrow A=-1[/latex]. Substituting this value into the first two equations gives us the system

[latex]\begin{array}{ccc}\hfill B+C& =\hfill & 1\hfill \\ \hfill B - 2C& =\hfill & 2.\hfill \end{array}[/latex]

Multiplying the second equation by [latex]-1[/latex] and adding the resulting equation to the first produces

[latex]-3C=1[/latex],

which in turn implies that [latex]C=-\frac{1}{3}[/latex]. Substituting this value into the equation [latex]B+C=1[/latex] yields [latex]B=\frac{4}{3}[/latex]. Thus, solving these equations yields [latex]A=-1[/latex], [latex]B=\frac{4}{3}[/latex], and [latex]C=-\frac{1}{3}[/latex].

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.

Rule: 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 [latex]A[/latex], [latex]B[/latex], and [latex]C[/latex] that satisfy the equation for all values of [latex]x[/latex]. That is, this equation must be true for any value of [latex]x[/latex] we care to substitute into it. Therefore, by choosing values of [latex]x[/latex] carefully and substituting them into the equation, we may find [latex]A[/latex], [latex]B[/latex], and [latex]C[/latex] easily. For example, if we substitute [latex]x=0[/latex], the equation reduces to [latex]2=A\left(-2\right)\left(1\right)[/latex]. Solving for [latex]A[/latex] yields [latex]A=-1[/latex]. Next, by substituting [latex]x=2[/latex], the equation reduces to [latex]8=B\left(2\right)\left(3\right)[/latex], or equivalently [latex]B=\frac{4}{3}[/latex]. Last, we substitute [latex]x=-1[/latex] into the equation and obtain [latex]-1=C\left(-1\right)\left(-3\right)[/latex]. Solving, we have [latex]C=-\frac{1}{3}[/latex].

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 [latex]A[/latex], [latex]B[/latex], and [latex]C[/latex], we rewrite the original integral:

[latex]\displaystyle\int \frac{3x+2}{{x}^{3}-{x}^{2}-2x}dx=\displaystyle\int \left(\text{-}\frac{1}{x}+\frac{4}{3}\cdot \frac{1}{\left(x - 2\right)}-\frac{1}{3}\cdot \frac{1}{\left(x+1\right)}\right)dx[/latex].

Evaluating the integral gives us

[latex]\displaystyle\int \frac{3x+2}{{x}^{3}-{x}^{2}-2x}dx=\text{-}\text{ln}|x|+\frac{4}{3}\text{ln}|x - 2|-\frac{1}{3}\text{ln}|x+1|+C[/latex].

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.

Example: Dividing before Applying Partial Fractions

Evaluate [latex]\displaystyle\int \frac{{x}^{2}+3x+1}{{x}^{2}-4}dx[/latex].

Show Solution

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.

Example: Applying Partial Fractions after a Substitution

Evaluate [latex]\displaystyle\int \frac{\cos{x}}{{\sin}^{2}x-\sin{x}}dx[/latex].

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try it

Evaluate [latex]\displaystyle\int \frac{x+1}{\left(x+3\right)\left(x - 2\right)}dx[/latex].

Hint

Show Solution

Watch the following video to see the worked solution to the above Try It

You can view the transcript for this segmented clip of “3.4 Partial Fractions” here (opens in new window).

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 [latex]{\left(ax+b\right)}^{n}[/latex], where [latex]n[/latex] is a positive integer greater than or equal to [latex]2[/latex]. If the denominator contains the repeated linear factor [latex]{\left(ax+b\right)}^{n}[/latex], then the decomposition must contain

[latex]\frac{{A}_{1}}{ax+b}+\frac{{A}_{2}}{{\left(ax+b\right)}^{2}}+\cdots +\frac{{A}_{n}}{{\left(ax+b\right)}^{n}}[/latex].

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.

Example: Partial Fractions with Repeated Linear Factors

Evaluate [latex]\displaystyle\int \frac{x - 2}{{\left(2x - 1\right)}^{2}\left(x - 1\right)}dx[/latex].

Show Solution

We have [latex]\text{degree}\left(x - 2\right)<\text{degree}\left({\left(2x - 1\right)}^{2}\left(x - 1\right)\right)[/latex], so we can proceed with the decomposition. Since [latex]{\left(2x - 1\right)}^{2}[/latex] is a repeated linear factor, include [latex]\frac{A}{2x - 1}+\frac{B}{{\left(2x - 1\right)}^{2}}[/latex] in the decomposition. Thus,

[latex]\frac{x - 2}{{\left(2x - 1\right)}^{2}\left(x - 1\right)}=\frac{A}{2x - 1}+\frac{B}{{\left(2x - 1\right)}^{2}}+\frac{C}{x - 1}[/latex].

After getting a common denominator and equating the numerators, we have

[latex]x - 2=A\left(2x - 1\right)\left(x - 1\right)+B\left(x - 1\right)+C{\left(2x - 1\right)}^{2}[/latex].

We then use the method of equating coefficients to find the values of [latex]A[/latex], [latex]B[/latex], and [latex]C[/latex].

[latex]x - 2=\left(2A+4C\right){x}^{2}+\left(-3A+B - 4C\right)x+\left(A-B+C\right)[/latex].

Equating coefficients yields [latex]2A+4C=0[/latex], [latex]-3A+B - 4C=1[/latex], and [latex]A-B+C=-2[/latex]. Solving this system yields [latex]A=2[/latex], [latex]B=3[/latex], and [latex]C=-1[/latex].

Alternatively, we can use the method of strategic substitution. In this case, substituting [latex]x=1[/latex] and [latex]x=\frac{1}{2}[/latex] into the previous equation easily produces the values [latex]B=3[/latex] and [latex]C=-1[/latex]. At this point, it may seem that we have run out of good choices for [latex]x[/latex], however, since we already have values for [latex]B[/latex] and [latex]C[/latex], we can substitute in these values and choose any value for [latex]x[/latex] not previously used. The value [latex]x=0[/latex] is a good option. In this case, we obtain the equation [latex]-2=A\left(-1\right)\left(-1\right)+3\left(-1\right)+\left(-1\right){\left(-1\right)}^{2}[/latex] or, equivalently, [latex]A=2[/latex].

Now that we have the values for [latex]A[/latex], [latex]B[/latex], and [latex]C[/latex], we rewrite the original integral and evaluate it:

[latex]\begin{array}{cc}\hfill {\displaystyle\int \frac{x - 2}{{\left(2x - 1\right)}^{2}\left(x - 1\right)}dx}& ={\displaystyle\int \left(\frac{2}{2x - 1}+\frac{3}{{\left(2x - 1\right)}^{2}}-\frac{1}{x - 1}\right)dx}\hfill \\ & =\text{ln}|2x - 1|-\frac{3}{2\left(2x - 1\right)}-\text{ln}|x - 1|+C.\end{array}[/latex]

Try It

Set up the partial fraction decomposition for [latex]\displaystyle\int \frac{x+2}{{\left(x+3\right)}^{3}{\left(x - 4\right)}^{2}}dx[/latex]. (Do not solve for the coefficients or complete the integration.)

Hint

Show Solution

Watch the following video to see the worked solution to the above Try It

You can view the transcript for this segmented clip of “3.4 Partial Fractions” here (opens in new window).

Module 3: Techniques of Integration