The value of $x$ such that $\displaystyle \int_{\sqrt{2}}^{\mathrm{x}}\frac{1}{x\sqrt{x^{2}-1}}dx=\frac{\pi}{12}$ is

Let $x=\sec{\theta} \implies dx=\sec{\theta}\tan{\theta}d\theta$

When $x=\sqrt{2}, \theta=\dfrac{\pi}{4}$ and when $x=x, \theta=\sec^{-1} {x}$

Now, $\displaystyle\int_{\pi/4}^{\sec^{-1} {x}} \dfrac{1}{\sec{\theta} \sqrt{\sec^2{\theta} -1}}\sec{\theta}\tan{\theta}d\theta=\dfrac{\pi}{12}$

$\implies\displaystyle\int_{\pi/4}^{\sec^{-1} {x}} \dfrac{1}{\sec{\theta} \tan{\theta}}\sec{\theta}\tan{\theta}d\theta=\dfrac{\pi}{12}$

$\implies\displaystyle\int_{\pi/4}^{\sec^{-1} {x}} d\theta=\dfrac{\pi}{12}$

$\implies\sec^{-1} {x} - \dfrac{\pi}{4}=\dfrac{\pi}{4}$

$\implies\sec^{-1} {x}=\dfrac{\pi}{4}+\dfrac{\pi}{12}=\dfrac{\pi}{3}$

$\implies x=\sec{\dfrac{\pi}{3}}=2$

$\implies x=2$