Evaluate the integral $\int_2^\infty \frac{1}{\sqrt x}log(\frac{x+1}{x-1})\,dx$ 
I actually need to check whether the following series is convergent
  $\sum_{n=2}^\infty \frac{1}{\sqrt n}log(\frac{n+1}{n-1})$.

The question specifically asks to use Cauchy's integral method, but I am not able to evaluate the integral nor able to find the answer through comparison test.
Any help is appreciated.

$$\int_2^\infty \log\left(\frac{x+1}{x-1}\right)\frac{dx}{\sqrt{x}}=2\sqrt{x}\log\left(\frac{x+1}{x-1} \right)|_2^\infty-2\int_2^\infty\sqrt{x}\left(\frac{1}{x+1}-\frac{1}{x-1}\right)dx\\=-\log(9)\sqrt{2}-2\int_2^\infty \frac{\sqrt{x}}{x+1}-\frac{\sqrt{x}}{x-1}dx$$ Let $u^2=x$, $2udu=dx$.
$$-2\int_2^\infty \frac{\sqrt{x}}{x+1}-\frac{\sqrt{x}}{x-1}dx=-4\int_\sqrt{2}^\infty \frac{u^2}{u^2+1}-\frac{u^2}{u^2-1}du$$ We can evaluate this integral in two pieces due to its linearity and compute the indefinite integrals.
$$\int \frac{u^2}{u^2+1}du=\int\frac{u^2 +1 -1}{u^2+1}du=u-\tan^{-1}(u)$$
$$\int \frac{u^2}{u^2-1}du=\int\frac{u^2 -1 +1}{u^2 -1}du=u+\frac{1}{2}\int \frac{1}{u-1}-\frac{1}{u+1}du=u+\frac{1}{2}\log\left(\frac{u-1}{u+1}\right)$$ Combining and taking the respective limits, we get $$-\log(9)\sqrt{2}-4\left(u-\tan^{-1}(u)-u-\frac{1}{2}\log\left(\frac{u-1}{u+1}\right)\right)\bigg|_\sqrt{2}^\infty\\=-\log(9)\sqrt{2}+2\pi-4\tan^{-1}(\sqrt{2})+2\log\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)$$
