To solve the integral of $\frac{3}{x\sqrt{x^2 - 1}}$, we can use a trigonometric substitution. Here's the step-by-step process:

Step 1: Substitution

We can use the substitution: $x = \sec(\theta)$ Then, $dx = \sec(\theta)\tan(\theta) d\theta$, and the expression $\sqrt{x^2 - 1}$ simplifies to $\sqrt{\sec^2(\theta) - 1} = \tan(\theta)$.

Step 2: Substituting into the Integral

Substitute $x = \sec(\theta)$ and $dx = \sec(\theta)\tan(\theta) d\theta$ into the integral:

$\int \frac{3}{x\sqrt{x^2 - 1}} dx = \int \frac{3}{\sec(\theta)\tan(\theta)} \cdot \sec(\theta)\tan(\theta) d\theta$

This simplifies to: $\int 3 d\theta$

Step 3: Solve the Integral

The integral of $3 \, d\theta$ is straightforward: $3\theta + C$

Step 4: Back-Substitute $\theta$

Since $x = \sec(\theta)$, we have $\theta = \sec^{-1}(x)$. Therefore, the solution is:

$3\sec^{-1}(x) + C$

Final Answer:

$\int \frac{3}{x\sqrt{x^2 - 1}} dx = 3\sec^{-1}(x) + C$

Would you like more details on trigonometric substitution or any other step?

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Here are 5 related questions you might find useful:

1. What are other integrals that involve trigonometric substitution?
2. How does one choose the correct substitution for integrals involving square roots?
3. What is the derivative of $\sec^{-1}(x)$?
4. How can you apply hyperbolic substitution for similar integrals?
5. How would you integrate $\frac{1}{x^2\sqrt{x^2 - 1}}$?

Tip: Trigonometric substitution is especially useful for integrals involving expressions like $\sqrt{x^2 - 1}$, $\sqrt{x^2 + 1}$, or $\sqrt{1 - x^2}$.