Integral of $\frac{1}{\sqrt{x^2-1}}$: Solving & Understanding

[PFuser1232]

What exactly is the integral of $\frac{1}{\sqrt{x^2 - 1}}$?
I know that the derivative of $\cosh^{-1}{x}$ is $\frac{1}{\sqrt{x^2 - 1}}$, but $\cosh^{-1}{x}$ is only defined for $x \geq 1$, whereas $\frac{1}{\sqrt{x^2 - 1}}$ is defined for all $|x| \geq 1$. How do I take that into account? Do I write:
$$\int \frac{1}{\sqrt{x^2 - 1}} dx = \cosh^{-1}{|x|} + c$$
UPDATE: I tried differentiating $\cosh^{-1}{|x|}$ as a piecewise function and I ended up with $-\frac{1}{\sqrt{x^2 - 1}}$ for $x \leq 1$. Now I'm more confused.
Should I write the integral as:
$$\int \frac{1}{\sqrt{x^2 - 1}} dx = S(x) \cosh^{-1}{|x|}$$
where $S$ is the signum function?

 

[Matternot]

Consider a comparison to ln(x)

notice how:
$\frac{d}{dx}\ln(x)$ is the same as $\frac{d}{dx}\ln(-x)$

This is inconsistent with your $\cosh^{-1}{|x|}$

The idea that resolves this is that you can use the idea that $\frac{1}{\sqrt{x^2 - 1}}$ is an even function.
This leads to your last line which sounds right to me.

There could be some quirks using complex numbers... I'll try it and get back.

What exactly is the integral of $\frac{1}{\sqrt{x^2 - 1}}$?
I know that the derivative of $\cosh^{-1}{x}$ is $\frac{1}{\sqrt{x^2 - 1}}$, but $\cosh^{-1}{x}$ is only defined for $x \geq 1$, whereas $\frac{1}{\sqrt{x^2 - 1}}$ is defined for all $|x| \geq 1$. How do I take that into account? Do I write:
$$\int \frac{1}{\sqrt{x^2 - 1}} dx = \cosh^{-1}{|x|} + c$$
UPDATE: I tried differentiating $\cosh^{-1}{|x|}$ as a piecewise function and I ended up with $-\frac{1}{\sqrt{x^2 - 1}}$ for $x \leq 1$. Now I'm more confused.
Should I write the integral as:
$$\int \frac{1}{\sqrt{x^2 - 1}} dx = S(x) \cosh^{-1}{|x|}$$
where $S$ is the signum function?

 

[Matternot]

one idea is to consider $\cosh(x) = cos(ix)$

i.e.
$\cosh^{-1}(x) = -i cos^{-1}(x)$
this leads to simply $±\frac{1}{\sqrt{x^{2}-1}}$

Most obvious solution would be your last line

 

[PFuser1232]

one idea is to consider $\cosh(x) = cos(ix)$

i.e.
$\cosh^{-1}(x) = -i cos^{-1}(x)$
this leads to simply $±\frac{1}{\sqrt{x^{2}-1}}$

Most obvious solution would be your last line

Consider a comparison to ln(x)

notice how:
$\frac{d}{dx}\ln(x)$ is the same as $\frac{d}{dx}\ln(-x)$

This is inconsistent with your $\cosh^{-1}{|x|}$

The idea that resolves this is that you can use the idea that $\frac{1}{\sqrt{x^2 - 1}}$ is an even function.
This leads to your last line which sounds right to me.

There could be some quirks using complex numbers... I'll try it and get back.

I think I figured it out. It's $\ln|x + \sqrt{x^2 - 1}|$.
Which is identical to the inverse hyperbolic cosine, except it's defined for both positive and negative $x + \sqrt{x^2 - 1}$.

 

[Matternot]

yeah, that's a much nicer way of writing it.