Question

Solve: $\int \frac{dx}{x\sqrt{x^4-1}}$

Answer 1

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

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

Let $\frac{1}{2}=t$

$=\Rightarrow \frac{1}{x^2}dx=-dt$

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

$==-\int \frac{tdt}{\sqrt{1-t^4}}$

Let $t^2=u$

$=\Rightarrow tdt=\frac{1}{2}du$

$=\therefore I=-\frac{1}{2}\int \frac{du}{\sqrt{1-u^2}}$

$=\{\because \int \frac{1}{1-x^2}dx-{\sin }^{-1}x+e\}$

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

$=\{\because {\sin }^{-1}x+{\cos }^{-1}x=\frac{x}{2}\}$

$=-\frac{1}{2}(\frac{x}{2}-{\cos }^{-1})+c$

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

$=\frac{1}{2}{\cos }^{-1}\left(t^2\right)+c^{\prime }$ $\{\because {\cos }^{-1}(\frac{1}{x})-{\sec }^{-1}x\}$

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

$I=\frac{1}{2}{\sec }^{-1}\left(x^2\right)+c^{\prime }$