To solve the integral $\int \frac{1}{\sqrt{9 + 64x^2}} \, dx$, we can use a trigonometric substitution method.

Steps to Solve:

1. Identify the substitution: The integral has the form $\int \frac{1}{\sqrt{a^2 + b^2x^2}} \, dx$, which suggests the substitution $x = \frac{a}{b} \tan(\theta)$, where $a = 3$ and $b = 8$.

   Set $x = \frac{3}{8} \tan(\theta)$.

2. Compute the differential: $dx = \frac{3}{8} \sec^2(\theta) \, d\theta$

3. Substitute into the integral: Substitute $x = \frac{3}{8} \tan(\theta)$ and $dx = \frac{3}{8} \sec^2(\theta) \, d\theta$ into the integral: $\int \frac{1}{\sqrt{9 + 64\left(\frac{3}{8} \tan(\theta)\right)^2}} \cdot \frac{3}{8} \sec^2(\theta) \, d\theta$

   Simplify the expression inside the square root: $\sqrt{9 + 64\left(\frac{9}{64} \tan^2(\theta)\right)} = \sqrt{9 + 9 \tan^2(\theta)} = 3\sqrt{1 + \tan^2(\theta)} = 3\sec(\theta)$ So the integral becomes: $\int \frac{1}{3\sec(\theta)} \cdot \frac{3}{8} \sec^2(\theta) \, d\theta = \int \frac{1}{8} \sec(\theta) \, d\theta$

4. Integrate: The integral of $\sec(\theta)$ is known: $\int \sec(\theta) \, d\theta = \ln|\sec(\theta) + \tan(\theta)| + C$ Thus, the integral becomes: $\frac{1}{8} \ln|\sec(\theta) + \tan(\theta)| + C$

5. Substitute back: Since $x = \frac{3}{8} \tan(\theta)$, $\tan(\theta) = \frac{8x}{3}$, and $\sec(\theta) = \sqrt{1 + \tan^2(\theta)} = \sqrt{1 + \left(\frac{8x}{3}\right)^2} = \frac{\sqrt{9 + 64x^2}}{3}$, the final answer is: $\frac{1}{8} \ln\left|\frac{\sqrt{9 + 64x^2}}{3} + \frac{8x}{3}\right| + C$

This is the solution to the integral.

Would you like more details or have any questions about the steps? Here are some related questions you might find interesting:

1. How do we solve integrals using trigonometric substitution in general?
2. What is the general form of integrals involving $\sqrt{a^2 + b^2x^2}$?
3. What are the common trigonometric identities used in calculus?
4. Can you explain the integration of $\sec(\theta)$ in more detail?
5. How do we handle integrals that don't easily fit common forms?

Tip: When dealing with square roots in integrals, look out for patterns that suggest trigonometric substitution. This can simplify the problem significantly.