Answer
$$\int\frac{1}{\sqrt{x^2-4}}dx =\frac{1}{2}\cosh^{-1}(x/2)+c.$$
Work Step by Step
Since $\frac{d}{dx} \cosh^{-1}x=\frac{1}{\sqrt{x^2-1}}$, then we have
$$\int\frac{1}{\sqrt{x^2-4}}dx=\int\frac{1}{2\sqrt{(x/2)^2-1}}dx=\frac{1}{2}\cosh^{-1}(x/2)+c.$$
