Decomposing $\frac{(\omega-1)^2}{(1+\omega^2)^2}$ into partial fractions

Question

How can I decompose $$\frac{(\omega-1)^2}{(1+\omega^2)^2}$$ into partial fractions?

Should I solve $$\frac{A\omega + B}{1+\omega^2} + \frac{C\omega^3 + D\omega^2 + E\omega + D}{(1+\omega^2)^2} = \frac{(\omega-1)^2}{(1+\omega^2)^2}$$

It seems a bit complicated, and if that's indeed the case what is the rationale behind it?

I tried with a CAS and got $$\frac1{1+\omega^2} - \frac{2\omega}{(1+\omega^2)^2}$$ but it's beyond me how that can be achieved!

Answer 1

$$\frac{ (1 + w^2) - 2w}{(1 + w^2)^2} = \frac{1}{(1 + w^2)} - \frac{2w}{(1 + w^2)^2}$$

Answer 2

the right ansatz is $$\frac{A\omega+B}{1+\omega^2}+\frac{C\omega+D}{(1+\omega^2)^2}$$