evaluate if integral converge: $ \int_2^\infty \frac{2}{x(x+1)(x-1)} dx $ Evaluate if the following integral converges:
$$
\int_2^\infty \frac{2}{x(x+1)(x-1)} dx
$$
Here I go:
$$ 2\int_2^\infty \frac{1}{x(x+1)(x-1)} dx $$
Partial fractions of $\dfrac{1}{x(x+1)(x-1)}$ :
$$\frac{1}{x(x+1)(x-1)} =\frac{A}{x}+\frac{B}{(x+1)} + \frac{C}{(x-1)} = \frac{-1}{x}+\frac{1/2}{(x+1)} + \frac{1/2}{(x-1)} $$
Back to the integral:
$$2(-\int_2^\infty\frac{1}{x}+\frac{1}{2}\int_2^\infty\frac{1}{(x+1)} + \frac{1}{2}\int_2^\infty\frac{1}{(x-1)})dx$$
$$\lim_{a\to2..{b\to\infty}}2[(-\ln(x)+\frac{\ln(x+1)}{2} + \frac{\ln(x-1)}{2}]_a^b$$
$$2[(-\ln(b)+\frac{\ln(b+1)}{2} + \frac{\ln(b-1)}{2}]-2[(-\ln(2)+\frac{\ln(3)}{2} + \frac{\ln(1)}{2}] $$
$$(-\infty+\infty+\infty) + (\ln(3/4)) $$
So, the integral does not converge? I am unsure about the last part, if anyone can confirm me the answer or explain why it's wrong... thank you guys!

No: the "infinities" cancel.
$$ \eqalign{-\ln(b) &+ \dfrac{\ln(b+1)}{2} + \dfrac{\ln(b-1)}{2} = \ln \left(\frac{\sqrt{b+1}\sqrt{b-1}}{b}\right) = \ln\left( \sqrt{1+1/b}\sqrt{1-1/b}\right)\cr & \to 0\ \text{as}\ b \to \infty}$$
But the simplest way to see this converges is by a limit comparison test:
$$\dfrac{2}{x(x+1)(x-1)} \sim \dfrac{2}{x^3} \ \text{as}\ x \to \infty $$
$$\int_2^\infty \dfrac{2}{x^3}\; dx\  \text{converges}$$
