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?
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.