- #1

PFuser1232

- 479

- 20

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?

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?

Last edited: