To compute the integral of \( \frac{2z - 1}{4z + 1} \, dz \), we can use the substitution method.

### Step-by-Step Solution:

1. **Substitute**:
   Let \( u = 4z + 1 \), so \( \frac{du}{dz} = 4 \), which gives \( dz = \frac{du}{4} \).

2. **Rewrite the integral**:
   \[
   \int \frac{2z - 1}{4z + 1} \, dz = \int \frac{2z - 1}{u} \cdot \frac{du}{4}.
   \]

3. **Express \( z \) in terms of \( u \)**:
   From \( u = 4z + 1 \), we get \( z = \frac{u - 1}{4} \).
   
   Therefore, \( 2z - 1 = 2 \cdot \frac{u - 1}{4} - 1 = \frac{u - 2}{2} \).

4. **Substitute back into the integral**:
   \[
   \int \frac{\frac{u - 2}{2}}{u} \cdot \frac{du}{4} = \frac{1}{8} \int \frac{u - 2}{u} \, du.
   \]
   
   Simplify the fraction inside the integral:
   \[
   \frac{u - 2}{u} = 1 - \frac{2}{u}.
   \]
   
   So the integral becomes:
   \[
   \frac{1}{8} \int \left(1 - \frac{2}{u}\right) \, du.
   \]

5. **Integrate**:
   \[
   \frac{1}{8} \left( \int 1 \, du - 2 \int \frac{1}{u} \, du \right) = \frac{1}{8} \left( u - 2 \ln |u| \right).
   \]

6. **Substitute \( u = 4z + 1 \) back**:
   \[
   \frac{1}{8} \left( 4z + 1 - 2 \ln |4z + 1| \right) + C.
   \]

7. **Simplify**:
   \[
   \frac{4z + 1}{8} - \frac{1}{4} \ln |4z + 1| + C.
   \]

Thus, the integral of \( \frac{2z - 1}{4z + 1} \, dz \) is:
\[
\frac{4z + 1}{8} - \frac{1}{4} \ln |4z + 1| + C.
\]

Would you like further clarification on any step?

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Here are 5 related questions to explore:
1. How do you perform substitution in integration?
2. What is the difference between indefinite and definite integrals?
3. How would this integral change if the numerator was a higher degree polynomial?
4. How does the natural logarithm appear in integrals like this?
5. Can partial fractions help solve similar integrals?

**Tip**: When facing a rational function in integration, always consider substitution or partial fraction decomposition.

Learn how to integrate the rational function (2z-1)/(4z+1) using the substitution method. This problem demonstrates substitution with natural logarithms and algebraic manipulation. Detailed solution provided, ideal for college-level calculus students.

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