<s>[INST] 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!
 [/INST] 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 $M\to +\infty$:

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

</s>