<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] With $u=x^2-2x-3$ we have $du = 2(x-1)\,dx$, so that
$$\int \frac{x-1}{(x^2-2x-3)^2}\,dx = \frac{1}{2}\int\frac{1}{u^2}\,du
     = -\frac{1}{2u} + c= -\frac{1}{2(x^2-2x-3)}+c.$$
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