9. Partial Fractions

c1. Computing Integrals

a. General Formulas

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 six basic types. Here are their integrals and how to compute them:

Caution: Do not try to memorize these integral formulas. Rather, remember the three techniques.

• \(\displaystyle \dfrac{\text{Constant}}{\text{Linear}} \qquad \int \dfrac{1}{x-a}\,dx=\ln | x-a|+C \)
  
  Substitute   \(u=x-a\)  

  Substitute \(u=x-a\). Then \(du=dx\) and: \[\begin{aligned} \int \dfrac{1}{x-a}\,dx &=\int \dfrac{1}{u}\,du=\ln | u|+C \\ &=\ln | x-a|+C \end{aligned}\]



• \(\displaystyle \dfrac{\text{Constant}}{\text{Linear}^n} \qquad \int\dfrac{1}{(x-a)^n}\,dx =\dfrac{-1}{n-1}\dfrac{1}{(x-a)^{n-1}}+C\,\)   if  \(\,n>1\)
  
  Substitute   \(u=x-a\)  

  Substitute \(u=x-a\). Then \(du=dx\) and: \[\begin{aligned} \int &\dfrac{1}{(x-a)^n}\,dx =\int u^{-n}\,du =\dfrac{u^{-n+1}}{-n+1}+C \\ &=\dfrac{-1}{n-1}\dfrac{1}{u^{n-1}}+C =\dfrac{-1}{n-1}\dfrac{1}{(x-a)^{n-1}}+C \end{aligned}\]



• \(\displaystyle \dfrac{\text{Linear}}{\text{Quadratic}} \qquad \int \dfrac{(x-a) }{(x-a)^2+b^2}\,dx =\dfrac{1}{2}\ln\left((x-a)^2+b^2\right)+C\)
  
  Substitute   \(u=(x-a)^2+b^2\)  

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



• \(\displaystyle \dfrac{\text{Linear}}{\text{Quadratic}^n} \qquad \int \dfrac{(x-a) }{\left((x-a)^2+b^2\right)^n}\,dx =\dfrac{-1}{2(n-1)}\dfrac{1}{\left((x-a)^2+b^2\right)^{n-1}}+C\)   if   \(n>1\)
  
  Substitute   \(u=(x-a)^2+b^2\)  

  Substitute \(u=(x-a)^2+b^2\). Then \(du=2(x-a)\,dx\) and: \[\begin{aligned} \int &\dfrac{x-a}{\left((x-a)^2+b^2\right)^n}\,dx =\dfrac{1}{2}\int u^{-n}\,du \\ &=\dfrac{1}{2}\dfrac{u^{-n+1}}{-n+1}+C =\dfrac{-1}{2(n-1)}\dfrac{1}{u^{n-1}}+C \\ &=\dfrac{-1}{2(n-1)}\dfrac{1}{\left((x-a)^2+b^2\right)^{n-1}}+C \end{aligned}\]



• \(\displaystyle \dfrac{\text{Constant}}{\text{Quadratic}} \qquad \int \dfrac{1}{(x-a)^2+b^2}\,dx =\dfrac{1}{b}\arctan \left(\dfrac{x-a}{b}\right)+C\)
  
  Substitute   \(x-a=b\tan\theta\)  

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



• \(\displaystyle \dfrac{\text{Constant}}{\text{Quadratic}^n} \qquad \int \dfrac{1}{\left((x-a)^2+b^2\right)^n}\,dx =\dfrac{1}{b^{2n-1}}\int \cos^{2n-2}\theta\,d\theta\)   if   \(n>1\)
  
  Substitute\(^\dagger\)   \(x-a=b\tan\theta\)  
  \(^\dagger\)Complete the integral and then substitute back using \(\tan\theta=\dfrac{x-a}{b}\) and hence \[ \sin\theta=\dfrac{x-a}{\sqrt{(x-a)^2+b^2}} \qquad \text{and} \qquad \cos\theta=\dfrac{b}{\sqrt{(x-a)^2+b^2}} \]

  Substitute \(x-a=b\tan\theta\). Then \(dx=b\sec^2\theta\,d\theta\) and: \[\begin{aligned} \int &\dfrac{1}{\left((x-a)^2+b^2\right)^n}\,dx =\int \dfrac{1}{\left(b^2\tan^2\theta+b^2\right)^n}\,b\sec^2\theta\,d\theta \\ &=\dfrac{1}{b^{2n-1}}\int \dfrac{1}{\left(\tan^2\theta+1\right)^n}\sec^2\theta\,d\theta \\ &=\dfrac{1}{b^{2n-1}}\int \dfrac{1}{\sec^{2n-2}\theta}\,d\theta =\dfrac{1}{b^{2n-1}}\int \cos^{2n-2}\theta\,d\theta \end{aligned}\] Complete the integral using the reduction formula: \[ \int \cos^n x\,dx =\dfrac{1}{n}\cos^{n-1}x\sin x+\dfrac{n-1}{n}\int \cos^{n-2}x\,dx \] and then substitute back.