It can be rewritten as
A+3+2xB+3−2xC. The A at the start becomes from the fact that the numerator and denominator have the same order, so division must yield a quotient that is some constant.
In fact, following on from this, I may as well tell you this:
When you got
gn(x)fm(x) where
m,n represent the orders of polynomials
f and
g respectively, first check whether
m≥n.
If so, the fraction can be reduced to
hm−n(x)+gn(x)rp(x) where
h,r are some polynomials we wish to find and now we have
p<n. ie a fraction where the numerator has a lesser order than the denominator, and THEN you want to see whether
gn(x) factorises. If it does, then you split this fraction into its partial fractions as you've been doing.
This is what I applied to your question. I looked at the orders, they're the same, therefore there will be some polynomial of 0th order at the start, this is just a constant so let's call it A. Then you end up with some polynomial over
9−4x2 which is fine, but this factorises into
(3+2x)(3−2x) so we can split our fraction into two others where the numerators are at most of one degree less than these linear factors, ie the numerators are constants.
Hence we obtained the form I said at the start.
Anyway, I'm heading off now so if anything is unclear I can clarify tomorrow.