Partial fractions with an irreducible quadratic factor in the denominator

So I'm supposed to decompose this expression into partial fractions

[Note T_A is a constant in this case]

Although I can get the right coefficients, I'm just wondering why it is C and not CT+D - should it not be the latter considering it has a quadratic in its denominator?

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It's even worse. It should actually be CT+D, if all possibilities were given. The question, stated as such, actually already tells you, however that the linear term here vanishes. You're quite right to remark this. Otherwise you'd be in enormous trouble when looking at, for example $\frac{t^2}{t^3-t^2+t-1} = \frac{1}{t-1} + \frac{t+1}{t^2+1}$. –  HSN Jan 17 '13 at 17:20
Sorry, I meant CT+D. That's very reassuring - I think the clue is trying to help tell you that there is no linear term, as you said, but it is very confusing. –  Mathlete Jan 17 '13 at 17:22
If you were using partial fractions on $\frac{aT^3+bT^2+cT+d}{T_A^4-T^4}$, then the partial fraction decomposition would have a term of the shape $\frac{mT+n}{T_A^2+T^2}$, more or less like you describe.
However, in this special case where $a=b=c=0$, it turns out that $m=0$. An easy way to see this is to let $U=T^2$. Then $$\frac{1}{T_A^4-U^2}=\frac{1}{2T_A^2}\left(\frac{1}{T_A^2-U}+\frac{1}{T_A^2+U}\right).$$
Replace $U$ by $T^2$. The term $\frac{1}{T_A^2-T^2}$ decomposes further, but $\frac{1}{T_A^2+T^2}$ does not.