Given that $x=\dfrac 1y$, show that $∫\frac {dx}{x \sqrt{(x^2-1)}}  = -∫\frac {dy}{\sqrt{1-y^2}}$ Given that $x=\dfrac 1y$, show that $\displaystyle \int \frac 1{x\sqrt{x^2-1}}\,dx = -\int \frac 1{\sqrt{1-y^2}}\,dy$
Have no idea how to prove it. 
here is a link  to wolframalpha showing how to integrate the left side.

Let's treat this like a $u$-substitution problem.
$$\int \frac 1{x\sqrt{x^2-1}}\,dx$$
Let $u = \frac{1}{x}$, so that $du = \frac{-1}{x^2}dx$. Then 
$$\int \frac {x}{x^2\sqrt{x^2-1}}dx = -\int \frac {\frac{1}{u}}{\sqrt{\frac{1}{u^2} - 1}}du = -\int \frac{1}{u\sqrt{1/u^2 - 1}}du = -\int \frac{1}{\sqrt{1 - u^2}}du$$
I just happen to use $u$ because that's how I teach it, but calling it $y$ is just the same. Most importantly, there is no need to actually integrate anything.
