To solve this integral:

$\int \frac{1 + \sin(3x)}{3x - \cos(3x)} \, dx$

we need to choose an appropriate substitution. Observing the integrand, let's try a substitution based on the term $u = 3x - \cos(3x)$, as this could simplify the integral.

Step-by-Step Solution

1. Substitute: Set $u = 3x - \cos(3x)$.
2. Differentiate: Find $du$ in terms of $dx$: $\frac{du}{dx} = 3 + 3\sin(3x)$ which simplifies to: $du = (3 + 3\sin(3x)) \, dx$ or $du = 3(1 + \sin(3x)) \, dx$
3. Rewrite the Integral: Since $1 + \sin(3x) = \frac{du}{3}$, substitute into the integral: $\int \frac{1 + \sin(3x)}{3x - \cos(3x)} \, dx = \int \frac{\frac{du}{3}}{u}$ Simplify this expression to: $= \frac{1}{3} \int \frac{1}{u} \, du$
4. Integrate: $= \frac{1}{3} \ln |u| + C$
5. Substitute Back: Substitute $u = 3x - \cos(3x)$ back into the result: $\frac{1}{3} \ln |3x - \cos(3x)| + C$

Final Answer

$\int \frac{1 + \sin(3x)}{3x - \cos(3x)} \, dx = \frac{1}{3} \ln |3x - \cos(3x)| + C$

Would you like further clarification on any steps?

Here are five related questions that expand on this topic:

1. How does one choose an appropriate substitution for different types of integrals?
2. What are other techniques for solving integrals involving trigonometric functions?
3. How can one verify the correctness of an integral solution?
4. What is the importance of the absolute value in the logarithmic result of integrals?
5. How does substitution help simplify complex integrals involving trigonometric functions?

Tip: Always check if the derivative of your substitution expression matches part of the integrand, as it often indicates a good substitution choice.