To solve the integral $\int \frac{3}{\sqrt{4x^2 - 25}} \, dx$, we can use a standard technique for integrals involving expressions of the form $\sqrt{ax^2 - b}$, which is often solved by trigonometric substitution.

Step-by-step solution:

1. Expression Setup:

   The given integral is: $\int \frac{3}{\sqrt{4x^2 - 25}} \, dx$ The denominator suggests the form $\sqrt{a^2x^2 - b^2}$, which is typically solved with a trigonometric substitution.

2. Substitute to simplify:

   First, factor the expression inside the square root: $\sqrt{4x^2 - 25} = \sqrt{(2x)^2 - 5^2}$ This is now in the form $\sqrt{a^2x^2 - b^2}$, where $a = 2$ and $b = 5$.

   We use the substitution: $2x = 5 \sec(\theta)$ This simplifies the square root because: $\sqrt{(2x)^2 - 5^2} = \sqrt{25\sec^2(\theta) - 25} = 5 \tan(\theta)$ Also, differentiate $2x = 5 \sec(\theta)$ with respect to $x$: $2 \, dx = 5 \sec(\theta) \tan(\theta) \, d\theta$ Thus, $dx = \frac{5 \sec(\theta) \tan(\theta)}{2} \, d\theta$.

3. Substitute into the integral:

   Now substitute these into the original integral: $\int \frac{3}{\sqrt{4x^2 - 25}} \, dx = \int \frac{3}{5 \tan(\theta)} \cdot \frac{5 \sec(\theta) \tan(\theta)}{2} \, d\theta$ Simplify the expression: $= \frac{3}{2} \int \sec(\theta) \, d\theta$

4. Integrate:

   The integral of $\sec(\theta)$ is: $\int \sec(\theta) \, d\theta = \ln|\sec(\theta) + \tan(\theta)| + C$ Therefore: $\frac{3}{2} \int \sec(\theta) \, d\theta = \frac{3}{2} \ln|\sec(\theta) + \tan(\theta)| + C$

5. Back-substitute using $\sec(\theta) = \frac{2x}{5}$ and $\tan(\theta) = \frac{\sqrt{4x^2 - 25}}{5}$:

   Finally, we back-substitute to express everything in terms of $x$. Using the relationships: $\sec(\theta) = \frac{2x}{5}, \quad \tan(\theta) = \frac{\sqrt{4x^2 - 25}}{5}$ The solution becomes: $\frac{3}{2} \ln\left| \frac{2x}{5} + \frac{\sqrt{4x^2 - 25}}{5} \right| + C$

Thus, the integral is: $\boxed{\frac{3}{2} \ln\left| \frac{2x + \sqrt{4x^2 - 25}}{5} \right| + C}$

Would you like a more detailed explanation of any of the steps? Or do you have additional questions?

Related questions:

1. How do you handle integrals with square roots in the denominator?
2. What other common trigonometric substitutions can simplify integrals?
3. Can you use substitution to solve integrals with square roots of sums instead of differences?
4. What is the relationship between the inverse trigonometric functions and integrals like these?
5. How do you simplify more complex trigonometric integrals?

Tip: When you encounter square roots in integrals, check if the form matches any standard trigonometric substitution like $\sqrt{a^2x^2 - b^2}$ or $\sqrt{a^2 - x^2}$ to simplify the problem.