We are tasked with finding the integral:

$\int \frac{\sqrt{25x^2 - 4}}{x} \, dx$

Step 1: Substitution $x = \frac{2}{5} \sec \theta$

Given the hint, we'll substitute $x = \frac{2}{5} \sec \theta$. From this, we can compute the differential:

$dx = \frac{2}{5} \sec \theta \tan \theta \, d\theta$

Step 2: Simplify the square root expression

Now, let's simplify the square root part of the integrand:

$25x^2 - 4 = 25\left(\frac{2}{5} \sec \theta\right)^2 - 4 = 25 \cdot \frac{4}{25} \sec^2 \theta - 4 = 4\sec^2 \theta - 4$

This simplifies to:

$4(\sec^2 \theta - 1) = 4\tan^2 \theta$

Thus, the square root becomes:

$\sqrt{25x^2 - 4} = \sqrt{4\tan^2 \theta} = 2\tan \theta$

Step 3: Substituting everything back into the integral

Substitute $x = \frac{2}{5} \sec \theta$, $dx = \frac{2}{5} \sec \theta \tan \theta \, d\theta$, and $\sqrt{25x^2 - 4} = 2 \tan \theta$:

$\int \frac{2 \tan \theta}{\frac{2}{5} \sec \theta} \cdot \frac{2}{5} \sec \theta \tan \theta \, d\theta$

Simplify:

$\int 2 \tan^2 \theta \, d\theta$

Step 4: Use the identity for $\tan^2 \theta$

We can now use the trigonometric identity $\tan^2 \theta = \sec^2 \theta - 1$:

$\int 2 (\sec^2 \theta - 1) \, d\theta = 2 \int (\sec^2 \theta - 1) \, d\theta$

Step 5: Integrate

The integral of $\sec^2 \theta$ is $\tan \theta$, and the integral of $1$ is $\theta$:

$2 \left( \tan \theta - \theta \right) + C$

Step 6: Substitute back in terms of $x$

Since $x = \frac{2}{5} \sec \theta$, we can find:

$\sec \theta = \frac{5x}{2} \quad \Rightarrow \quad \tan \theta = \sqrt{\sec^2 \theta - 1} = \sqrt{\left(\frac{5x}{2}\right)^2 - 1} = \sqrt{\frac{25x^2}{4} - 1} = \frac{\sqrt{25x^2 - 4}}{2}$

Also, from $x = \frac{2}{5} \sec \theta$, we can solve for $\theta$:

$\theta = \sec^{-1} \left( \frac{5x}{2} \right)$

Thus, the final answer is:

$2 \left( \frac{\sqrt{25x^2 - 4}}{2} - \sec^{-1} \left( \frac{5x}{2} \right) \right) + C$

Which simplifies to:

$\sqrt{25x^2 - 4} - 2 \sec^{-1} \left( \frac{5x}{2} \right) + C$

Summary of Solution:

$\int \frac{\sqrt{25x^2 - 4}}{x} \, dx = \sqrt{25x^2 - 4} - 2 \sec^{-1} \left( \frac{5x}{2} \right) + C$

Would you like any further details or clarification on any of these steps?

Here are 5 follow-up questions you might find useful:

1. How do trigonometric identities like $\tan^2 \theta = \sec^2 \theta - 1$ help simplify integrals?
2. What is the significance of the inverse secant function in integrals?
3. How do we choose the right substitution for integrals involving square roots?
4. Why do integrals often involve trigonometric substitution when dealing with quadratic expressions?
5. Can we solve this integral using any other substitution method?

Tip: When dealing with integrals involving $\sqrt{a^2x^2 - b^2}$, consider using trigonometric substitutions involving secant, tangent, or sine to simplify the problem.