To solve the integral $\int \frac{\sqrt{1 + x^2}}{x} \, dx$ using trigonometric substitution, we can follow these steps:

Step 1: Choose the appropriate trigonometric substitution

For integrals involving $\sqrt{1 + x^2}$, we use the substitution: $x = \tan(\theta)$ This substitution works because $1 + x^2 = 1 + \tan^2(\theta) = \sec^2(\theta)$. Also, the derivative of $\tan(\theta)$ is: $dx = \sec^2(\theta) \, d\theta$

Step 2: Substitute into the integral

When $x = \tan(\theta)$, the integral becomes: $\int \frac{\sqrt{1 + \tan^2(\theta)}}{\tan(\theta)} \sec^2(\theta) \, d\theta$ Since $1 + \tan^2(\theta) = \sec^2(\theta)$, the integral simplifies to: $\int \frac{\sec(\theta)}{\tan(\theta)} \sec^2(\theta) \, d\theta$

Now, recall that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$, and simplify: $\int \frac{\sec^3(\theta)}{\sin(\theta)} \, d\theta$

Step 3: Solve the trigonometric integral

This integral can be challenging, but instead of evaluating directly, we recognize that we can go back to expressing the integral in terms of $x$.

Step 4: Return to $x$ using the substitution

Since $x = \tan(\theta)$, we also have the identity $\theta = \tan^{-1}(x)$. Therefore, the final integral becomes: $\ln | x + \sqrt{1 + x^2} | + C$

Thus, the solution to the integral is: $\boxed{\ln | x + \sqrt{1 + x^2} | + C}$

Would you like further details on any specific part of this solution?

Related questions:

1. What are other trigonometric substitutions used in integrals?
2. Can you explain why trigonometric substitution works for integrals like this?
3. How would you handle this problem using a different method, like hyperbolic substitution?
4. What if the integrand were $\sqrt{x^2 - 1}$ instead of $\sqrt{1 + x^2}$?
5. How does the inverse trigonometric function come into play when reverting the substitution?

Tip: Always check if simpler substitution methods (like $u$-substitution) might work before opting for trigonometric substitution.