Let $t\in \mathbb R, n\in \mathbb N.$
How should I choose $A, B$ and $C$ so
$$\frac{(t-n)^2t^2}{[1+(t-n)^2)]^3} = \frac{A}{[1+(t-n)^2)]^3}+ \frac{B}{[1+(t-n)^2)]^2} + \frac{C}{[1+(t-n)^2)]}$$
My Try:
We may rewrite, $$t^{4}-2nt^3+n^2t^2= A+ B(n^2-2nt+t^2+1)^2 + C (n^2-2nt+t^2+1)$$
If we take $t=0,$ we have $$0=A+B(n^2+1)^2 + C(n^2+1)$$
If we take $t=n,$ then we have $$0=A+B+C$$
If we take $t=1,$ we have $$(1-n)^2=A+ B(1+ (1-n)^2)^2 + C (1+ (1-n)^2)$$
But I am struggling to decide $A, B, C$ from these informations...
HINT : Change of variable $x=t-n$ $$\frac{(t-n)^2t^2}{(1+(t-n)^2)^3}=\frac{x^2(x+n)^2}{(1+x^2)^3}=n^2\frac{x^2}{(1+x^2)^3}+2n\frac{x^3}{(1+x^2)^3}+\frac{x^4}{(1+x^2)^3}$$ isn't it easier now to decompose each of the three terms ?