<s>[INST] 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.
 [/INST] Given that $x=1/y$
Put $x = \dfrac{1}{y},dx = \dfrac{-1}{y^2}.dy$:
$$\int \dfrac{1}{(x\sqrt{(x^2-1)})}dx$$
$$=\int \dfrac{1}{(\frac{1}{y} \sqrt{((\frac{1}{y})^2-1)})}\cdot\dfrac{-1}{y^2}dy$$
$$\int \dfrac{1}{\frac{1}{y} \sqrt{\frac{1-y^2}{y^2}}}\cdot \dfrac{-1}{y^2}dy$$ 
Simplify and you get your answer.
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