Question

Intro. problem for Heaviside cover up

Answer 1

http://i.imgur.com/q69NYyv.png

Answer 2

Problem: Decompose 1/(x^3-x) into partial fractions

Solution: I factored the bottom into x(x-1)(x+1)

For the first term in the denominator I have A/x. Covering up x does x=0? If so then A=0 but the answer says A is equal to -1

Answer 3

how do you get A = 0 by your coverup?

Answer 4

\[\frac{1}{x(x-1)(x+1)}\]
we 'coverup' the bad zero ...
\[\frac{1}{\cancel{0}(0-1)(0+1)}=\frac{1}{-1(1)}\]

Answer 5

let x = -1, and x=1 for the other coverups ...