We need to compute the integral:

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

This integral suggests the use of a trigonometric substitution because of the form $\sqrt{x^2 - 1}$.

Step 1: Use the substitution $x = \sec(\theta)$

We know that: $\sec^2(\theta) - 1 = \tan^2(\theta)$ Thus, let: $x = \sec(\theta) \quad \text{so} \quad dx = \sec(\theta) \tan(\theta) \, d\theta$ and: $\sqrt{x^2 - 1} = \sqrt{\sec^2(\theta) - 1} = \tan(\theta)$

Step 2: Rewrite the integral

Substitute $x = \sec(\theta)$ and