Find the asymptote and the upper bound for $1\big/ \ln\left(\frac{x+1}{x-1}\right)$ as $x\to\infty$. I need to show that $\left[\ln\left(\frac{x+1}{x-1}\right)\right]^{-1}$ asymptotically approaches to a line from below that I should determine, and hence, show that $\frac{x}{2} > \left[\ln\left(\frac{x+1}{x-1}\right)\right]^{-1}$ for $x>1$.  I would be very happy if someone helps. Thank you.

Fix $x > 1$ and write $\varphi(t) = \frac{1}{x+t}$. Since $\varphi_x$ is strictly convex, it follows from the Jensen's inequality that
\begin{align*}
\frac{1}{2} \log\left(\frac{x+1}{x-1}\right)
= \frac{1}{2} \int_{-1}^{1} \varphi_x(t) \, \mathrm{d}t
> \varphi_x \left( \frac{1}{2} \int_{-1}^{1} t \, \mathrm{d}t \right)
= \frac{1}{x}
\end{align*}
On the other hand, again by the strict convexity of $\varphi_x$,
\begin{align*}
\frac{1}{2} \log\left(\frac{x+1}{x-1}\right)
= \frac{1}{2} \int_{-1}^{1} \varphi_x(t) \, \mathrm{d}t
< \frac{\varphi_x(-1) + \varphi_x(1)}{2}
= \frac{x}{x^2-1}.
\end{align*}
Altogether, it follows that
$$ \frac{x}{2} - \frac{1}{2x} < \left[\log\left(\frac{x+1}{x-1}\right)\right]^{-1} < \frac{x}{2} $$
for all $x > 1$.

Addendum. It is easy to check that the above function admits the Laurent expansion of the form
$$ \left[\log\left(\frac{x+1}{x-1}\right)\right]^{-1} = \frac{x}{2} - \sum_{n=0}^{\infty} \frac{a_{2n+1}}{x^{2n+1}} $$
as $x\to\infty$. Now here comes a hard part: $\texttt{Mathematica}$ seems to suggests $a_{2n+1} > 0$ for all $n \geq 0$. For instance,
$$
a_1 = \frac{1}{6}, \quad
a_3 = \frac{2}{45}, \quad
a_5 = \frac{22}{945}, \quad
a_7 = \frac{214}{14175}, \quad
a_9 = \frac{5098}{467775}, \quad \cdots. $$
It will be interesting to be able to prove it, although I have no good idea to begin with.
