Great work, EB!
It seems to invite Trig Substitution,
. . so I tried it . . . and found yet another solution.
$\displaystyle \int\frac{dx}{x\sqrt{x^2-1}}$
Let $\displaystyle x = \sec\theta\quad\Rightarrow\quad dx = \sec\theta\tan\theta\,d\theta\quad\Rightarrow\quad \sqrt{x^2 - 1} = \tan\theta$
Substitute: .$\displaystyle \int\frac{\sec\theta\tan\theta\,d\theta}{\sec \theta\tan\theta}\;=\;\int d\theta \;=\;\theta + C$
Back-substitute: .$\displaystyle \boxed{\text{arcsec }x + C}$
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Our three answers are equivalent.
I had: $\displaystyle \sec\theta = \frac{x}{1}$
$\displaystyle \theta$ is an angle is a right triangle with: $\displaystyle hyp = x,\;adj = 1$ Code:
*
x * | ______
* | √x² - 1
* θ |
* - - - - - *
1
Your answer .$\displaystyle \arctan\sqrt{x^2-1}$ .is the same angle.
The other acute angle is: $\displaystyle \frac{\pi}{2} - \theta$ . and $\displaystyle \sin\left(\frac{\pi}{2} - \theta\right) = \frac{1}{x}$
Hence: .$\displaystyle \frac{\pi}{2} - \theta\:=\:\arcsin\left(\frac{1}{x}\right)\quad \Rightarrow\quad\theta \:= \:\frac{\pi}{2} - \arcsin\left(\frac{1}{x}\right)$
. . which verifies margaritas' solution.