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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.

Learn how to solve the integral of 1 / (2x + 1) using the substitution method. This step-by-step solution covers key concepts of integral calculus, logarithmic integration, and substitution. Suitable for advanced high school or early college calculus students.

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