Definite integral of a rational fraction

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 1

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 $$\frac{x-1}{(x^2-2x-3)^2}=\frac{A(x+1)(x-3)^2+B(x-3)^2+C(x-3)(x+1)^2+D(x+1)^2}{(x+1)^2(x-3)^2}$$

Answer 2

Use partial fraction decomposition, that is, find $A,B,C,D\in\mathbb{R}$ such that$$\frac{x-1}{(x+1)^2(x-3)^2}=\frac A{x+1}+\frac B{(x+1)^2}+\frac C{x-3}+\frac D{(x-3)^2}.$$

Answer 3

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