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!
 A: 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:
$$\dfrac{2}{x(x+1)(x-1)} \sim \dfrac{2}{x^3} \ \text{as}\ x \to \infty $$
$$\int_2^\infty \dfrac{2}{x^3}\; dx\  \text{converges}$$
A: The integral function is non-negative over the integration range and behaves like $\frac{1}{x^3}$ for large values of $x$, hence it is integrable for sure. Partial fraction decomposition gives:
$$\frac{2}{(x-1)x(x+1)} = \frac{1}{x(x-1)}-\frac{1}{x(x+1)}=\frac{1}{2}\left(\frac{1}{x-1}-\frac{2}{x}+\frac{1}{x+1}\right)\tag{1}$$
hence:
$$ \int_{2}^{M}\frac{2\,dx}{(x-1)x(x+1)}=\log\left(\frac{4}{3}\right)+\log\left(1-\frac{1}{M^2}\right)\tag{2} $$
and by letting $M\to +\infty$:

$$ \int_{2}^{+\infty}\frac{2\,dx}{(x-1)x(x+1)}=\color{red}{\log\left(\frac{4}{3}\right)}.\tag{3}$$

A: I know you're asking if it converges, but unless I'm interpreting your answer incorrectly I think it's wrong. As others have pointed out before me you can see it converges via a limit comparison test to see $x\rightarrow\infty$. So here's how I get $\ln(4/3)$:
$$\int\dfrac{2}{x(x+1)(x-1)}dx$$
$$=2\int\dfrac{1}{x(x+1)(x-1)}dx$$
$$=2\int-\frac{1}{x}+\dfrac{1}{2(x+1)}+\dfrac{1}{2(x-1)}dx$$
$$=2\left(-\ln|x|+\frac{1}{2}\ln|x+1|+\frac{1}{2}\ln|x-1|\right)+C$$
Now the bounds:
$$2\left(-\ln|2|+\frac{1}{2}\ln|2+1|+\frac{1}{2}\ln|2-1|\right)\Rightarrow-\ln\left(\frac{4}{3}\right)$$
$$2\left(-\ln|\infty|+\frac{1}{2}\ln|\infty+1|+\frac{1}{2}\ln|\infty-1|\right)\Rightarrow0$$
$$=0-\left(-\ln\left(\frac{4}{3}\right)\right)$$
$$=\ln\left(\frac{4}{3}\right)$$
A: Short answer: yes, it does: the denominator doesn't cancel and the asymptotic behavior is $\sim x^{-3}$.
