To find the integral of $\frac{1}{2x+1} \, dx$, we can use the method of substitution.

Let: $u = 2x + 1$ Then, differentiate $u$ with respect to $x$: $\frac{du}{dx} = 2 \quad \text{or} \quad du = 2 \, dx$ This implies: $dx = \frac{du}{2}$

Now, substitute these into the integral: $\int \frac{1}{2x+1} \, dx = \int \frac{1}{u} \cdot \frac{du}{2}$ $= \frac{1}{2} \int \frac{1}{u} \, du$

The integral of $\frac{1}{u}$ with respect to $u$ is: $\frac{1}{2} \ln |u| + C$

Finally, substitute back $u = 2x + 1$ to get: $\int \frac{1}{2x+1} \, dx = \frac{1}{2} \ln |2x+1| + C$ where $C$ is the constant of integration.

Would you like more details or have any questions?

Here are some related questions:

1. How do you perform integration using substitution in more complex cases?
2. What are other methods for integrating rational functions?
3. How does the integration of natural logarithms differ from other functions?
4. Can integration be used to solve differential equations?
5. How do you determine when to use substitution versus partial fractions?

Tip: Always check if the function you're integrating can be simplified before applying substitution or other techniques.