<s>[INST] Definite integral of a rational fraction 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?
 [/INST] Use partial fraction decomposition, that is, find $A,B,C,D\in\mathbb{R}$ such that$$\frac{x-1}{(x+1)^2(x-3)^2}=\frac A{x+1}+\frac B{(x+1)^2}+\frac C{x-3}+\frac D{(x-3)^2}.$$
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