The integral shown in the image is:

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

Solution:

1. Substitute: To simplify the square root term, let’s use the substitution:

   $x = \frac{5}{2} \sec(\theta)$ This choice is motivated by the fact that $\sec^2(\theta) - 1 = \tan^2(\theta)$, which will simplify the expression under the square root.

2. Differentiate $x$: $dx = \frac{5}{2} \sec(\theta) \tan(\theta) d\theta$

3. Substitute into the integral: Now, we substitute $x = \frac{5}{2} \sec(\theta)$ and $dx = \frac{5}{2} \sec(\theta) \tan(\theta) d\theta$ into the integral.

   First, compute the term under the square root:

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

   Therefore, $\sqrt{4x^2 - 25} = 5 \tan(\theta)$.

   Now the integral becomes:

   $\int \frac{\frac{5}{2} \sec(\theta) \tan(\theta) d\theta}{\frac{5}{2} \sec(\theta) \cdot 5 \tan(\theta)} = \int \frac{d\theta}{5}$

4. Integrate: $\int \frac{d\theta}{5} = \frac{\theta}{5} + C$

5. Re-substitute $x = \frac{5}{2} \sec(\theta)$: From the substitution $x = \frac{5}{2} \sec(\theta)$, we have $\sec(\theta) = \frac{2x}{5}$. Therefore, $\theta = \sec^{-1}\left(\frac{2x}{5}\right)$.

6. Final answer: The final result is:

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

Would you like further details on any step or have questions about a specific part of the solution?

Here are 5 related questions to further expand your understanding:

1. How do trigonometric substitutions simplify integrals with square roots?
2. What is the general method for choosing a substitution in integrals involving $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, or $\sqrt{x^2 - a^2}$?
3. What is the derivative of $\sec^{-1}(x)$ and how is it derived?
4. Can you explain how the limits of integration would change in a definite integral using the same substitution?
5. How do hyperbolic trigonometric substitutions compare to standard trigonometric substitutions in integration?

Tip: When dealing with integrals involving square roots of quadratic expressions, using trigonometric substitutions based on common identities ($\sec^2(\theta) - 1 = \tan^2(\theta)$) often simplifies the problem dramatically.