Answer
$$\int^{2}_{\sqrt2}\frac{\sec^2(\sec^{-1}x)dx}{x\sqrt{x^2-1}}=\sqrt3-1$$
Work Step by Step
$$A=\int^{2}_{\sqrt2}\frac{\sec^2(\sec^{-1}x)dx}{x\sqrt{x^2-1}}$$
We set $u=\sec^{-1}x$, which means $$du=\frac{1}{x\sqrt{x^2-1}}dx$$
- For $x=\sqrt2$, we have $u=\sec^{-1}\sqrt2=\cos^{-1}\frac{\sqrt2}{2}=\frac{\pi}{4}$
- For $x=2$, we have $u=\sec^{-1}2=\cos^{-1}\frac{1}{2}=\frac{\pi}{3}$
Therefore, $$A=\int^{\pi/3}_{\pi/4}\sec^2udu$$ $$A=\tan u\Big]^{\pi/3}_{\pi/4}$$ $$A=\tan\frac{\pi}{3}-\tan\frac{\pi}{4}$$ $$A=\sqrt3-1$$