Question: Can I find the value of $$\int\_3^{\infty}\frac{x-1}{(x^2-2x-3)^2}dx$$ by just factoring the fraction?

I tried to wrote:
$$\frac{x-1}{(x^2-2x-3)^2}=\frac{x-1}{(x^2-2x+1-4)^2}=\frac{x-1}{[(x-1)^2-2^2]^2}=\frac{x-1}{(x+1)^2\cdot(x-3)^2}$$ but didn't work out. Any ideas?

Answer: 
With $u=x^2-2x-3$ we have $du = 2(x-1)\,dx$, so that
$$\int \frac{x-1}{(x^2-2x-3)^2}\,dx = \frac{1}{2}\int\frac{1}{u^2}\,du
 = -\frac{1}{2u} + c= -\frac{1}{2(x^2-2x-3)}+c.$$
Use partial fraction decomposition, that is, find $A,B,C,D\in\mathbb{R}$ such that$$\frac{x-1}{(x+1)^2(x-3)^2}=\frac A{x+1}+\frac B{(x+1)^2}+\frac C{x-3}+\frac D{(x-3)^2}.$$