The integral in the image is $\int \frac{\sqrt{x^2-1}}{x} \, dx$. This can be solved using trigonometric substitution, specifically, substituting $x = \sec(\theta)$ because the expression inside the square root, $x^2 - 1$, suggests this substitution would simplify the integral.

Steps:

1. Substitute: $x = \sec(\theta)$ and hence $dx = \sec(\theta)\tan(\theta) d\theta$.
2. Rewrite the integral: $\int \frac{\sqrt{\sec^2(\theta)-1}}{\sec(\theta)} \sec(\theta)\tan(\theta) \, d\theta$
3. Simplify the square root: $\sqrt{\sec^2(\theta) - 1} = \tan(\theta)$.
4. Integral becomes: $\int \tan(\theta) \tan(\theta) \, d\theta = \int \tan^2(\theta) \, d\theta$
5. Simplify further using a trigonometric identity: $\tan^2(\theta) = \sec^2(\theta) - 1$.
6. Integral of $\sec^2(\theta) - 1$: $\int (\sec^2(\theta) - 1) \, d\theta = \tan(\theta) - \theta + C$
7. Back-substitute for $x$: $\tan(\theta) = \sqrt{x^2 - 1}$ and $\theta = \sec^{-1}(x)$.
8. Final answer: $\int \frac{\sqrt{x^2-1}}{x} \, dx = \sqrt{x^2 - 1} - \sec^{-1}(x) + C$

Let me know if you need further details or have any questions! Here are some additional questions to consider:

1. Why is the substitution $x = \sec(\theta)$ appropriate for this integral?
2. How does the trigonometric identity $\tan^2(\theta) = \sec^2(\theta) - 1$ simplify the integral calculation?
3. What are the potential challenges or mistakes one might encounter when using trigonometric substitution?
4. How can one verify that the solution to this integral is correct?
5. Are there alternative methods to solve this integral without using trigonometric substitution?

Tip: Always double-check your back-substitution steps to ensure they match the original variable of integration, which helps avoid common errors in integral calculus solutions.