I am very confused by this. I know that the derivative of $\text{arcsec}(x)$ is $\dfrac{1}{|x|\sqrt{x^2-1}}$. However, if you plug in the integral of $\dfrac{1}{x\sqrt{x^2-1}}$ into wolfram alpha it gives some other answer with an inverse tangent: $$ \int \dfrac{1}{x\sqrt{x^2-1}}dx = - \tan^{-1}\Bigg(\frac{1}{\sqrt{x^2-1}} \Bigg) +C $$
I was just wondering why this is, or why wolfram is giving something totally different. Are they equivalent?
The two answers are equivalent. Remember that $\sin$ and $\tan$, from a trigonometric point of view, are just ratios of sides of a right angled triangle. The $\tfrac{1}{\sqrt{x^2-1}}$ in the wolfram result looks suspiciously like an application of Pythagoras's theorem, doesn't it? You can do the calculation yourself.