Question

How do you express `1/(x^3-4x)` in partial fractions?

Answer 1

You can analyze it as follows

`1/[x^3-4x]=1/(x(x-2)(x+2))=A/x+B/(x-2)+C/(x+2)`

Now try various values of `x` except `0,-2,2` to find the values of
constants `A,B,C` finally you get

`1/[x^3-4x]=-1/(4x)+1/[8*(x-2)]+1/[8*(x+2)]`