The integral shown in the image is:

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

This is a standard integral and can be solved using a trigonometric substitution. Let's go through the steps to solve it.

Step-by-step solution:

1. Substitute: Use the trigonometric identity $x = \sec(\theta)$, which implies: $dx = \sec(\theta) \tan(\theta) \, d\theta$ and, $\sqrt{x^2 - 1} = \sqrt{\sec^2(\theta) - 1} = \tan(\theta)$

2. Rewrite the integral: Substituting these values into the integral: $\int \frac{\sqrt{x^2 - 1}}{x} \, dx = \int \frac{\tan(\theta)}{\sec(\theta)} \sec(\theta) \tan(\theta) \, d\theta$

   This simplifies to: $\int \tan^2(\theta) \, d\theta$

3. Use the trigonometric identity: Recall that $\tan^2(\theta) = \sec^2(\theta) - 1$. So the integral becomes: $\int (\sec^2(\theta) - 1) \, d\theta$

4. Integrate: $\int \sec^2(\theta) \, d\theta - \int 1 \, d\theta = \tan(\theta) - \theta + C$

5. Substitute back $\theta$: Since $x = \sec(\theta)$, we know: $\tan(\theta) = \sqrt{x^2 - 1}$ and $\theta = \sec^{-1}(x)$

   Therefore, the solution is: $\sqrt{x^2 - 1} - \sec^{-1}(x) + C$

Final Answer:

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

Would you like further details or explanations on any of the steps?

Here are 5 related questions that expand on this topic:

1. How can other trigonometric substitutions be used for different types of integrals?
2. What is the significance of using trigonometric identities in solving integrals?
3. How can the inverse secant function be interpreted geometrically?
4. What are other methods to solve integrals involving square roots of quadratic expressions?
5. How does the method of partial fractions apply to integrals with rational expressions?

Tip: Always consider using trigonometric identities when encountering square roots involving quadratic expressions in integrals. They often simplify the problem significantly.