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

$\frac{1}{2} \sec^{-1}(x^2) + C$

The correct answer is Option (1) → $\frac{1}{2} \sec^{-1}(x^2) + C$

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

Put $x^2 = \sec \theta \Rightarrow \theta = \sec^{-1} x^2$

$\Rightarrow 2x \, dx = \sec \theta \cdot \tan \theta \, d\theta$

$I = \int \frac{\sec \theta \tan \theta \, d\theta}{2 \sec \theta \cdot \sqrt{\sec^2 \theta - 1}}$

$∴I = \frac{1}{2} \int \frac{\sec \theta \cdot \tan \theta}{\sec \theta \tan \theta} \, d\theta = \frac{1}{2} \int d\theta = \frac{1}{2} \theta + C \quad [∵ \sec^2 \theta - 1 = \tan^2 \theta]$

$= \frac{1}{2} \sec^{-1}(x^2) + C$