Antiderivative of $\frac{1}{x^2-1}$ for $x>1$ or $x<-1$ 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.
 A: 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}
