Improper integral $\int_{1}^{\infty}\frac{dx}{x\sqrt{x^2 - 1}}$

Question

I wonder how to figure it out this integral: $$\int_{1}^{\infty}\dfrac{dx}{x \sqrt{x^2 - 1}}$$. What I know is: $$\int_{1}^{\infty}\dfrac{dx}{x\sqrt{(x+1)(x-1)}}$$, but what does this help me to figure it out the entire integral? Please help ...

Answer 1

By making the substitution $x\to1/x$, we have $$ \int_1^\infty \frac{dx}{x\sqrt{x^2-1}} =\int_0^1 \frac{dx}{\sqrt{1-x^2}}=\bigg[\arcsin x\bigg]_0^1=\frac{\pi}{2} $$

Answer 2

But, if you take $u = \sqrt{x^2 -1}dx$ and make a substitution, you get the integral of $\tan^{-1}(u)\Big|_{0}^{\infty} = \tan^{-1}(\infty) - \tan^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$ which means, that this integral is convergent. ;)