Answer
$$\int_2^4\frac{1}{\sqrt{x^2-1}}dx=\cosh^{-1}x|_2^4=\cosh^{-1}4-\cosh^{-1}2.$$
Work Step by Step
Since $\frac{d}{dx} \cosh^{-1}x=\frac{1}{\sqrt{x^2-1}}$, then we have
$$\int_2^4\frac{1}{\sqrt{x^2-1}}dx=\cosh^{-1}x|_2^4=\cosh^{-1}4-\cosh^{-1}2.$$
