Upper approximation of $\mathrm{atanh}(x)$? Is there are nice upper approximation of $\mathrm{atanh(x)}$?
For example, $\ln(x)$ is nicely approximated by $x-1$ for $x$ around $1$.

As @Rahul said in the comment above we can write inverse hyperbolic functions as logarithms. With this approach we get for all $\left| x \right| < 1$:
$$\operatorname{artanh}(x) = \frac{1}{2}\ln \left( \frac{1 + x}{1 - x} \right).$$
No we can use logarithmic inequalities. We have that for all $x>0$:
$$1-\frac{1}{x} \leq \ln x \leq x-1.$$
Use of this we can get bounds for $\operatorname{artanh}$ for all $0<x<1$:
$$\frac{1}{2}\left(1-\frac{1-x}{1+x}\right) \leq \operatorname{artanh}(x) \leq \frac{1}{2}\left(\frac{1+x}{1-x}-1\right).$$
You can have more accurate bounds with Padé approximant. For example you can get an upper bound with second order Padé approximant like this. We know that for all $x>0$:
$$\ln x \leq \frac{(x-1)(x+5)}{2+4x}.$$
Using this we get for $\operatorname{atanh}$ function for all $0<x<1$ the following.
$$\operatorname{artanh}(x) \leq \frac{x(2x-3)}{(x+3)(x-1)} = \psi_2(x).$$
I give you result using this method.
$
\begin{align}
\operatorname{artanh}(0.1) = 0.1003353477 & \leq 0.1003584228 = \psi_2(0.1), \\
\operatorname{artanh}(0.3) = 0.3095196042 & \leq 0.3116883116 = \psi_2(0.3), \\
\operatorname{artanh}(0.5) = 0.5493061443 & \leq 0.5714285715 = \psi_2(0.5), \\
\operatorname{artanh}(0.7) = 0.8673005277 & \leq 1.009009010\phantom{0} = \psi_2(0.7), \\
\operatorname{artanh}(0.9) = 1.472219490\phantom{0} & \leq 2.769230769\phantom{0} = \psi_2(0.9).
\end{align}
$
With this approach you can get also lower bounds and of course for more precise bounds you can use higher order Padé approximants. In this really good paper you find hiher order Padé approximants for $\ln$, from them you can construct bounds for $\operatorname{artanh}$.
I have written more details about Padé approximants and approximation of the logarithm function in this answer. Feel free to read it.
