Methods of partial fractions

Partial fractions are useful in the study of discrete maths, control theory, differential equations, and calculus.

Methods:
1. Observation
For simple rational functions, we can guess the partial fraction decomposition.
$$\frac{1}{(x+1)(x+2)} = \frac{1}{x+1} -\frac{1}{x+2} \\ \frac{x}{(x+2)(x+3)} = \frac{3}{x+3} - \frac{2}{x+2}$$
2. Method of undetermined coefficients: solving a system of linear equations
2a. Comparing coefficients

Example:
$$\frac{x^2}{(x-1)(x^2+4)^2}=\frac{A}{x-1}+\frac{Bx+C}{x^2+4}+\frac{Dx+E}{(x^2+4)^2} \\\\ x^2=A(x^2+4)^2+(Bx+C)(x-1)(x^2+4)+(Dx+E)(x-1)$$ Put $x=1$, we have $A=\dfrac{1}{25}$. Now compare the coefficients of $x^4$, $$A+B=0\Rightarrow B=-\frac{1}{25}.$$ Compare the coefficients of $x^3$, $$-B+C=0 \Rightarrow C=-\frac{1}{25}.$$ Compare those of $x^2$, $$8A-C+4B+D=1 \Rightarrow D=1+\frac{4}{25}-\frac{1}{25}-\frac{8}{25}=\frac{4}{5}.$$ Compare the constant term, $$16A-4C-E=0 \Rightarrow E=\frac{16}{25}+\frac{4}{25}=\frac{4}{5}.$$
2b. Substitution
$$\dfrac{x^2+4}{x(x+2)(3x-2)}=\dfrac{A}{x}+\dfrac{B}{x+2}+\dfrac{C}{3x-2}\\x^2+4=A(x+2)(3x-2)+Bx(3x-2)+Cx(x+2)$$ Substitute $x=0,-2,\dfrac{2}{3}$, we have $A=-1,B=\dfrac{1}{2},C=\dfrac{5}{2}$ respectively.

2c. Differentiation

3. Heaviside’s cover-up technique (only a shorthand of substitution)

Note: This method does not work all the time. Sometimes, it gives only partial results.

Example:

To find A, cover up the factor $x^2$ and put $x=0$. Similarly, cover up $x+1$ and put $x=-1$ to find B. Finally, subtract $\large \frac{A}{x^2}$ and $\large \frac{B}{x+1}$ from the original rational function to get C. Alternatively, one can differentiate the equation $x-5 = A(x+1) + Bx² + Cx(x+1)$ twice, yielding $0 = 2(B+C) \Rightarrow C = -B = 6$.

This method is essentially substitution. If you multiply both sides by $x^2(x+1)$, you see how it works. This cover-up technique is just an easy way of finding pfd without the fuss of writing many steps.

Example:
$$\frac{5x+6}{(x^2+4)(x-2)} = \frac{Ax+B}{x^2+4} + \frac{C}{x-2}$$ Using the cover-up technique, we know $C = 2$.
Now what we can do is to simplify $\dfrac{5x+6}{(x^2+4)(x-2)}-\dfrac{2}{x-2}$ or solve two simultaneous equations to find A and B. Alternatively, we can express $x^2+4$ as $(x-2i)(x+2i)$ and use the cover-up technique. However, the calculations will be longer and can be more prone to mistakes.

The cover-up method gives partial results too if a linear factor is repeated. It works only for the highest power of the linear factor.

Reference:
http://www.cimt.plymouth.ac.uk/journal/man.pdf
http://meikleriggs.org.uk/CUR/CUR.pdf

4. Limit

Example:
$$\begin{align}F(x)&=\frac{6x-1}{(1+x)(1-3x)^2}=\frac{A}{1+x}+\frac{B}{(1-3x)^2}+\frac{C}{1-3x}\\ A&=\frac{6x-1}{(1-3x)^2}\bigg|_{x=-1}=-\frac{7}{16}\\ B&=\frac{6x-1}{1+x}\bigg|_{x=\frac{1}{3}}=\frac{3}{4}\\ \large 0 &=\lim\limits_{x \to +\infty} \frac{x(6x-1)}{(1+x)(1-3x)^2}\\ & =\lim\limits_{x \to +\infty} (\frac{Ax}{1+x}+\frac{Bx}{(1-3x)^2}+\frac{Cx}{1-3x})\\ &=A-\frac{1}{3}C \\ \therefore C&=3A=-\frac{21}{16}\end{align}$$
5. Keily’s Method (Useful for repeated linear factors)

Example:
$$\begin{align} \frac{3}{(x+1)^2(x+2)}&=\frac{1}{x+1}\frac{3}{(x+1)(x+2)}\\&=\frac{1}{x+1}(\frac{3}{x+1}+\frac{-3}{x+2})\\&=\frac{3}{(x+1)^2}-\frac{3}{(x+1)(x+2)}\\&=\frac{3}{(x+1)^2}-3(\frac{1}{x+1}-\frac{1}{x+2})\\&=\frac{3}{(x+1)^2}-\frac{3}{x+1}+\frac{3}{x+2}\end{align}$$
Reference:
Pages 5, 6

6. Algebraic Identity

Let p(x) be a function and a, b distinct scalars.
$$\frac{1}{(p(x)+a)(p(x)+b)}=\frac{1}{b-a}(\frac{1}{p(x)+a}-\frac{1}{p(x)+b})$$
Example 1:
$$\frac{x}{(x^2+1)(x^2+4)}=\frac{x}{4-1}(\frac{1}{x^2+1}-\frac{1}{x^2+4})=\frac{1}{3}(\frac{x}{x^2+1}-\frac{x}{x^2+4})$$
Example 2:
$$\begin{align}\frac{1}{s^2(s+a)}&=\frac{s+a}{s^2(s^2-a^2)}\\\\ &=\frac{s+a}{a^2}(\frac{1}{s^2-a^2}-\frac{1}{s^2})\\\\ &=\frac{1}{a^2}\frac{1}{s-a}-\frac{1}{a^2}\frac{s+a}{s^2}\\\\ &=\frac{1}{a^2}\frac{1}{s-a}-\frac{1}{a^2}\frac{1}{s}(1+\frac{a}{s})\\\\ &=\frac{1}{a^2}\frac{1}{s-a}-\frac{1}{a^2}\frac{1}{s}-\frac{1}{a}\frac{1}{s^2}\end{align}$$
More methods

7. Algebraic manipulations

Example: $$\begin{align}\bigg(\dfrac{z+1}{z-1}\bigg)^3&=\dfrac{1}{(z-1)^3}[(z-1)+2]^3\\&=1+\dfrac{6}{z-1}+\dfrac{12}{(z-1)^2}+\dfrac{8}{(z-1)^3}\end{align}$$
Application of partial fractions:
1. Integration

2. Differential equations
Inverse Laplace Transform by Partial Fraction Decomposition

Example:

More examples on pages 4-6