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.

Substitute $x=1/y$ and $dx/dy=-1/y^2$ to get $\int y^2/(\sqrt{1-y^2}) (-1/y^2)dy$
