The problem asks to compute the integral:

$\int \frac{1}{2x + 1} \, dx$

To solve this, we can use the method of substitution.

Let: $u = 2x + 1$ Then, differentiate both sides with respect to $x$: $du = 2 \, dx \quad \Rightarrow \quad dx = \frac{du}{2}$

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 $\frac{1}{u}$ is $\ln |u|$, so we have:

$\frac{1}{2} \ln |u| + C$

Finally, substitute $u = 2x + 1$ back into the result:

$\frac{1}{2} \ln |2x + 1| + C$

Thus, the solution is:

$\int \frac{1}{2x + 1} \, dx = \frac{1}{2} \ln |2x + 1| + C$

Would you like more details on any step?

Here are 5 related questions for further exploration:

1. How does substitution help simplify integrals like this?
2. What is the integral of $\frac{1}{ax + b}$ in general?
3. How do we apply logarithmic properties to integrals involving $\ln |u|$?
4. Can this integral be solved using partial fractions or other methods?
5. What happens if the denominator is a quadratic expression instead of linear?

Tip: Always check if substitution is a suitable method for simplifying an integral. It can be very helpful for linear expressions in the denominator.