The bottom factors to (x^2-4)(x^2+2)
(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
= -13/(24(x+2)) + 13/(24(x-2)) - 7/(6(x^2+2))
Thank you. But what about the "Cx+D" part?
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.
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