Question

\[\int\frac{1}{x\sqrt{x^4 - 1}} dx\]

Answer 1

\[\int\frac{dx}{x\sqrt{x^4 - 1}}\]
 ` =  ∫  {x dx}/{x^2 \sqrt{( x^2 )^2 - 1} `

`  Let   x^2 = t `
\[ \Rightarrow 2x = \frac{dt}{dx}\]
\[ \Rightarrow \text{x  dx} = \frac{dt}{2}\]
Now,  ` =  ∫  {x    dx}/{x^2 \sqrt{( x^2 )^2 - 1} `
\[ = \frac{1}{2}\int\frac{dt}{t\sqrt{t^2 - 1}}\]
\[ = \frac{1}{2} \sec^{- 1} \left( t \right) + C\]
\[ = \frac{1}{2} \sec^{- 1} \left( x^2 \right) + C\]