Answer
$\dfrac{\pi}{2}$
Work Step by Step
We are given the integral $\int_1^{\infty} \dfrac{dx}{x\sqrt {x^2-1}}$
Let us substitute $a=\sqrt {x^2-1}$ and $da=\dfrac{x dx}{\sqrt {x^2-1}}$
Now, consider $I = \int \dfrac{da}{a^2+1}= tan^{-1} (a) +C\\= \tan^{-1} \sqrt {x^2-1}+C$
Since, $ \lim\limits_{x \to 1} \tan^{-1} \sqrt {x^2-1} =0$ and $ \lim\limits_{x \to \infty} \tan^{-1} \sqrt {x^2-1} =\dfrac{\pi}{2}$
So, our given integral becomes:
$I=\int_1^{\infty} \dfrac{dx}{x\sqrt {x^2-1}}=\dfrac{\pi}{2}-0=\dfrac{\pi}{2}$
