We know that the antiderivative of $\frac{1}{x^{2} - 1}$ on $(-1,1)$ is $\operatorname{artanh}(x)$. Is there a nice way of writing the antiderivative on the intervals $(-\infty,-1)$ and $(1,\infty)$?
I realize we could write $$\int_{a}^{x}\frac{1}{y^{2} - 1}\,dy$$ for $a>1$, for the antiderivative on $(1,\infty)$, for example, but I'm curious if there's an expression that doesn't include an integral, and WolframAlpha chokes when I ask it to compute this.
For the sake of completeness, here are the two suggested ways of writing the antiderivative:
First Way: $$\boxed{\int \frac{dx}{x^2-1} = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C.}$$ This holds because: \begin{align} \frac{d}{dx}\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| &=\frac{1}{2}\frac{d}{dx}\bigl(\ln|x-1| - \ln|x+1|\bigr)\\[5pt] &=\frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right)\\[5pt] &=\frac{1}{x^2-1}. \end{align}
Second Way: $$\boxed{\int \frac{dx}{x^2-1} = \begin{cases} -\operatorname{artanh}(x)+C, &|x|<1\\ -\operatorname{arcoth}(x)+C, &|x|>1. \end{cases}}$$
This holds because: $$-\operatorname{artahn}(x) = \frac{1}{2}\ln\left(\frac{1-x}{x+1}\right) = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| \quad\text{for } |x|<1$$ and \begin{align} -\operatorname{arcoth}(x) &= -\operatorname{artanh}\left(\frac{1}{x}\right)\\[5pt] &= \frac{1}{2}\ln\left(\frac{1-\frac{1}{x}}{1+\frac{1}{x}}\right)\\[5pt] &= \frac{1}{2}\ln\left(\frac{x-1}{x+1}\right)\\[5pt] &=\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| \quad\text{for }|x|>1. \end{align}