Related to Partial Fractions: Heaviside Cover-up Method

Relevant Links:
Heaviside cover-up method (Wikipedia.org)
Partial Fraction (Wikipedia.org)

This post acts more as a supplement to the Wikipedia.com post. I would suggest at first you do both the Heaviside method and the original method (expanding and equating coefficients) to make sure you're getting the right answers. Ultimately, the Heaviside cover-up method will help you reduce the amount of math you have to do, but you will sometimes still need the original method.

After setting up and cross-multiplying, you can plug in values to eliminate most factors. This allows you to solve for some of the coefficients. Unfortunately without going into complex numbers, the method is a little restricted, but every little bit helps.

Example 1:
As Related to Example 2 on page 486:
$\frac{x^{2}+2x-1}{x(2x-1)(x+2)}=\frac{A}{\vphantom{2}x}+\frac{B}{2x-1}+\frac{C}{x+2}$
$x^{2}+2x-1=A(2x-1)(x+2)+Bx(x+2)+Cx(2x-1)$
For $x=0$ we get $-1=A\cdot(-1)\cdot2$. So $A=\frac{1}{2}$. For $x=\frac{1}{2}$ we get $\frac{1}{4}=B\cdot\frac{1}{2}\cdot\frac{5}{2}$. So $B=\frac{1}{5}$. For $x=-2$ we get $-1=C\cdot(-2)\cdot(-5)$. So $C=-\frac{1}{10}$.

Example 2: (similar or same one done in class on 20120221)
The expression $$\frac{x^2-1}{(x-1)(x-2)(x-3)}$$ simplifies to $$\frac{x+1}{(x-2)(x-3)}$$. After simplification, we set up the partial fraction decomposition as $$\frac{x+1}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3}$$. Cross multiplying we get $$x+1=A(x-3)+B(x-2)$$. If $x=3$ then $4=B\cdot 1$, so $B=4$. If $x=2$ then $3=A\cdot (-1)$, so $A=-3$.

Example 3: (similar or same one done in section 2 on 20120221)
Consider the expression $$\frac{1}{(x-1)(x+2)^2}$$. Then the set-up is $$\frac{1}{(x-1)(x+2)^2}=\frac{A}{(x-1)^\vphantom{2}}+\frac{B}{(x+2)^\vphantom{2}}+\frac{C}{(x+2)^2}$$.
Cross multiplying we obtain $$1=A(x+2)^2 +B(x-1)(x+2)+C(x-1)$$. If $x=-2$ then $1=C(-3)$ so $C=-1/3$. If $x=1$ then $1=A\cdot 9$, so $A=1/9$. From here we can expand and equate coefficients. But we could actually plug in any value of $x$ we want and proceed to solve for $B$. If $x=0$, then $1=4/9-2B+1/3$, so $-2B=2/9$, so $B=-1/9$.

More examples to come as time allows. Perhaps you can request examples. Though I'm not sure anybody is even reading this.