<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] write $$\frac{x-1}{(x^2-2x-3)^2}$$ as $$-1/8\, \left( x+1 \right) ^{-2}+1/8\, \left( x-3 \right) ^{-2}$$
use that $$\frac{(x-1)}{(x^2-2x-3)^2}=\frac{x-1}{(x+1)^2(x-3)^2}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x-3}+\frac{D}{(x-3)^2}$$
it is 
$$\frac{x-1}{(x^2-2x-3)^2}=\frac{A(x+1)(x-3)^2+B(x-3)^2+C(x-3)(x+1)^2+D(x+1)^2}{(x+1)^2(x-3)^2}$$
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