Question

How do you express `(2x+ 3)/(x^4-9x^2)` in partial fractions?

Answer 1

`(2x+3)/(x^4-9x^2)=-2/(9x)+1/(3x^2)-1/(18(x+3))-1/(6(x-3))`

Explanation:

Looking to the denominator `x^4-9x^2=x^2(x+3)(x-3)` then the fraction admits an expansion such that
`(2x+3)/(x^4-9x^2)=A/x+B/x^2+C/(x+3)+D/(x-3)` so
doing `f = (2x+3)/(x^4-9x^2)-A/x-B/x^2-C/(x+3)-D/(x-3)=0` and calculanting `f(x^4-9x^2) = (D+C-A)x^3+(3D+B-3C)x^2+(9A+2)x+3-9B =0` for all values of x, we get:
`A=-2/9,B=1/3,C=-1/18,D=-1/6`