Solve using trig substitution:
$\int\left(\frac{dx}{x\sqrt{x^{2}-9}}\right)$
- $u = x = \left({3\sec\theta}\right)^{2} = 9\sec^{2}\theta$ and $ dx = \sec\theta\tan\theta$
- $\theta = \sec^{-1}\left(\frac{x}{3}\right)$ and $ \sqrt{x^2-9} = 3\tan\theta$
- $\int \frac{\sec\theta\tan\theta}{9\sec\theta\tan\theta}d\theta$
- $\frac{1}{9}\int d\theta = \frac{1}{9}\theta = \frac{1}{9}\sec^{-1}\left(\frac{x}{3}\right) + C$
As you can see, we end up at $$ \bbox[5px,border:2px solid #282] {\frac{1}{9}\sec^{-1}\left(\frac{x}{3}\right) + C} $$
However, when I use calculators to solve the same problem, I end up getting $$ \bbox[5px,border:2px solid red] {\frac{1}{3}\tan^{-1}\left(\frac{\sqrt{x^2-9}}{3}\right) + C} $$ or $$ \bbox[5px,border:2px solid red] {\sqrt{x^2-9} + 3tan^{-1}\left(\frac{3}{\sqrt{x^2-9}}\right) + C} $$
WolframAlpha does not explain why (obviously I don't know) $ \sec^{-1}x$ ended up being simplified to $\tan^{-1}x$, the $ x $ parameter being whatever with substitute in for.
My guess is it has to do with the trig substitution rules.
I haven't found any trig identity that looks like that or why the answer is simplified this way, and differently if that matters at all.
Can someone explain why?Thanks.