How do I solve this via partial fraction decomposition?

Question

(x^2+9)/(x^4-2x^2-8)

Answer 1

The bottom factors to (x^2-4)(x^2+2)
= (x+2)x-2)x^2+4)

so let
(x^2+9)/(x^4-2x^2-8) = A/(x+2) + B/(x-2) + C/(x^2+2)

( A(x-2)(x^2+2) + B(x+2)(x^2+2) + C(x+2)(x-2) ) = x^2 + 9
since the denominators on the left and right are the same.

This must be true for all values of x

let x=2 ---> 24B = 13 or B = 13/24
let x=-2 --> - 24A = 13 or A = -13/24
let x = 1 --> -3A + 9B - 3C = 10
sub in the values of A and B to get
C = -7/6

so (x^2+9)/(x^4-2x^2-8)
= -13/(24(x+2)) + 13/(24(x-2)) - 7/(6(x^2+2))

Answer 2

Thank you. But what about the "Cx+D" part?

Answer 3

You are correct, Micki, that there is supposed to be a (Cx+D) over a quadratic denominator. Luckily, in this case, C=0, so it worked out. Had it been otherwise, I'm sure it would have become apparent.