Question

If $I=\int \frac{dx}{x^3\sqrt{x^2-1}}$, then $I$ equals

Answer 1

$I=\int \frac{dx}{x^3\sqrt{x^2-1}}$

$=\int \frac{dx}{x^3.x\sqrt{1-1/2^2}}$

$=\int \frac{dx}{x^4\sqrt{1-x^2}}$

let $1-x^{-2}=t^2$

$\frac{2}{x^3}dx=\alpha tdt$

$=\int \frac{tdt}{x.t}$

$=\int \frac{dt}{\sqrt{1-t^2}}$

$={\sin }^{-1}\left(t\right)+c$

$={\sin }^{-1}\left(\sqrt{1-\frac{1}{x^2}}\right)+c$

$={\sin }^{-1}\left[\frac{\sqrt{x^2-1}}{x}\right]+c$