9. Partial Fractions

c1. Computing Integrals

We have seen that a partial fraction expansion expresses a rational function \(\dfrac{p(x) }{q(x) }\) as the sum of a polynomial and rational functions of four basic types:

\(\dfrac{\text{Constant}}{\text{Linear}}\qquad \dfrac{A}{x-a}\) \(\dfrac{\text{Constant}}{\text{Linear}^n}\qquad \dfrac{A}{(x-a)^n}\qquad n>1\)
\(\dfrac{\text{Linear}}{\text{Quadratic}}\quad \dfrac{B(x-a)+C}{(x-a)^2+b^2}\)   \(\dfrac{\text{Linear}}{\text{Quadratic}^n}\quad \dfrac{B(x-a)+C}{\left((x-a)^2+b^2\right)^n}\quad n>1\)

For the purpose of doing the integrals, the terms with quadratic denominators need to be broken up into two pieces. So there are really six basic terms:

\(\dfrac{\text{Constant}}{\text{Linear}}\qquad \dfrac{A}{x-a}\) \(\dfrac{\text{Constant}}{\text{Linear}^n}\qquad \dfrac{A}{(x-a)^n}\qquad n>1\)
\(\dfrac{\text{Linear}}{\text{Quadratic}}\quad \dfrac{B(x-a)}{(x-a)^2+b^2}\)   \(\dfrac{\text{Linear}}{\text{Quadratic}^n}\quad \dfrac{B(x-a) }{\left((x-a)^2+b^2\right)^n}\quad n>1\)
\(\dfrac{\text{Constant}}{\text{Quadratic}}\quad \dfrac{C}{(x-a)^2+b^2}\)   \(\dfrac{\text{Constant}}{\text{Quadratic}^n}\quad \dfrac{C}{\left((x-a)^2+b^2\right)^n}\quad n>1\)

To compute the integrals of these terms you use the following techniques: (The comment buttons reveal examples.)

Compute:   \(\displaystyle \int \dfrac{5}{x-2}\,dx\).

Let \(u=x-2\). Then \(du=dx\) and: \[\begin{aligned} \int \dfrac{5}{x-2}\,dx &=\int \dfrac{5}{u}\,du =5\ln |u|+C \\ &=5\ln | x-2|+C \end{aligned}\] With experience, you can probably do this without the \(u\)-substitution.

Compute:   \(\displaystyle \int \dfrac{5}{(x-2)^3}\,dx\).

Let \(u=x-2\). Then \(du=dx\) and: \[\begin{aligned} \int \dfrac{5}{(x-2)^3}\,dx &=\int \dfrac{5}{u^3}\,du =\int 5u^{-3}\,du \\ &=5\dfrac{u^{-2}}{-2}+C=-\dfrac{5}{2}\dfrac{1}{u^2}+C \\ &=-\dfrac{5}{2}\dfrac{1}{(x-2)^2}+C \end{aligned}\] With experience, you can probably do this without the \(u\)-substitution: \[\begin{aligned} \int \dfrac{5}{(x-2)^3}\,dx &=\int 5(x-2)^{-3}\,dx =5\dfrac{(x-2)^{-2}}{-2}+C \\ &=-\dfrac{5}{2}\dfrac{1}{(x-2)^2}+C \end{aligned}\]

Compute:   \(\displaystyle \int \dfrac{7(x-5)}{(x-5)^2+9}\,dx\).

Let \(u=(x-5)^2+9\). Then \(du=2(x-5)\,dx\) and: \[\begin{aligned} \int \dfrac{7(x-5) }{(x-5)^2+9}\,dx &=\dfrac{7}{2}\int \dfrac{1}{u}\,du=\dfrac{7}{2}\ln | u|+C \\ &=\dfrac{7}{2}\ln \left((x-5)^2+9\right)+C \end{aligned}\]

Compute:   \(\displaystyle \int \dfrac{7(x-5) }{\left((x-5)^2+9\right)^4}\,dx\).

Let \(u=(x-5)^2+9\). Then \(du=2(x-5)\,dx\) and: \[\begin{aligned} \int \dfrac{7(x-5) }{\left((x-5)^2+9\right)^4}\,dx =\dfrac{7}{2}\int \dfrac{1}{u^4}\,du=-\dfrac{7}{6}\dfrac{1}{u^3}+C \\ =-\dfrac{7}{6}\dfrac{1}{\left((x-5)^2+9\right)^3}+C \end{aligned}\]

Compute:   \(\displaystyle \int \dfrac{4}{(x-5)^2+9}\,dx\).

Let \(x-5=3\tan\theta\). Then \(dx=3\sec^2\theta\,d\theta\) and: \[\begin{aligned} \int \dfrac{4}{(x-5)^2+9}\,dx &=\int \dfrac{4}{9\tan^2\theta+9}\,3\sec^2\theta\,d\theta \\ &=\dfrac{4}{3}\int \dfrac{\sec^2\theta}{\tan^2\theta+1}\,d\theta =\dfrac{4}{3}\int 1\,d\theta \\ &=\dfrac{4}{3}\theta+C=\dfrac{4}{3}\arctan \left(\dfrac{x-5}{3}\right)+C \end{aligned}\]

This example is probably harder than anything you will be asked to do in this course. It is included here for completeness.

Compute:   \(\displaystyle \int \dfrac{4}{\left((x-5)^2+9\right)^4}\,dx\).

Let \(x-5=3\tan\theta\). Then \(dx=3\sec^2\theta\,d\theta\) and: \[\begin{aligned} \int &\dfrac{4}{\left((x-5)^2+9\right)^4}\,dx =\int \dfrac{4}{(9\tan^2\theta+9)^4}\,3\sec^2\theta\,d\theta \\ &=\dfrac{4}{3^7}\int \dfrac{\sec^2\theta}{(\tan^2\theta+1)^4}\,d\theta =\dfrac{4}{3^7}\int \dfrac{1}{\sec^6\theta}\,d\theta \\ &=\dfrac{4}{3^7}\int \cos^6\theta\,d\theta \qquad \text{(A bunch of work here.)} \\ &=\dfrac{4}{3^7}\left[\dfrac{1}{6}\cos^5\theta\sin\theta +\dfrac{5}{24}\cos^3\theta\sin\theta+\dfrac{5}{16}\cos\theta\sin\theta +\dfrac{5}{16}\theta\right]+C \end{aligned}\] Since \(\tan\theta=\dfrac{x-5}{3}\), we have: (Draw a triangle with opposite side \(x-5\) and adjacent side \(3\).) \[ \sin\theta=\dfrac{x-5}{\sqrt{(x-5)^2+9}} \quad \text{and} \quad \cos\theta=\dfrac{3}{\sqrt{(x-5)^2+9}} \] Thus \[\begin{aligned} \int \dfrac{4}{\left((x-5)^2+9\right)^4}\,dx =&\dfrac{4}{3^7}\left[ \dfrac{1}{6}\dfrac{3^5(x-5)}{\left((x-5)^2+9\right)^3} +\dfrac{5}{24}\dfrac{3^3(x-5)}{\left((x-5)^2+9\right)^2}\right. \\ &+\left. \dfrac{5}{16}\dfrac{3(x-5)}{(x-5)^2+9} +\dfrac{5}{16}\arctan \left(\dfrac{x-5}{3}\right) \right]+C \end{aligned}\]

The ability to use the substitution \(u=(x-a)^2+b^2\) in the \(\dfrac{\text{Linear}}{\text{Quadratic}}\) and \(\dfrac{\text{Linear}}{\text{Quadratic}^n}\) terms is precisely why we use the numerator \(B(x-a)+C\) for the quadratic denominators rather than \(Bx+C\).

If you want to see the general integrals, they are on the attached page. However, DO NOT try to memorize the integral formulas. Rather, remember the three techniques.

You are now ready to compute the integral of any rational function. The next page has several complete examples and exercises.

You can also practice computing integrals using Partial Fraction Expansions by using the following Maplet (requires Maple on the computer where this is executed):

Partial Fractions: Evaluating the Integral   Rate It

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Supported in part by NSF Grant #1123255