To verify that
\[
\int \frac{2x-1}{2x+3} \, dx = x - \log|(2x+3)^{2}| + C,
\]
we will perform the integration step by step.
### Step 1: Rewrite the integrand
We start with the integral:
\[
\int \frac{2x-1}{2x+3} \, dx.
\]
We can rewrite the numerator \(2x - 1\) as follows:
\[
2x - 1 = (2x + 3) - 4.
\]
Thus, we can express the integral as:
\[
\int \frac{(2x + 3) - 4}{2x + 3} \, dx = \int \left(1 - \frac{4}{2x + 3}\right) \, dx.
\]
### Step 2: Separate the integral
Now we can separate the integral into two parts:
\[
\int \left(1 - \frac{4}{2x + 3}\right) \, dx = \int 1 \, dx - \int \frac{4}{2x + 3} \, dx.
\]
### Step 3: Integrate each part
1. The integral of \(1\) is:
\[
\int 1 \, dx = x.
\]
2. For the second integral, we have:
\[
\int \frac{4}{2x + 3} \, dx.
\]
Using the substitution \(u = 2x + 3\), we find \(du = 2 \, dx\) or \(dx = \frac{du}{2}\). Therefore, we can rewrite the integral as:
\[
\int \frac{4}{u} \cdot \frac{du}{2} = 2 \int \frac{1}{u} \, du = 2 \log |u| + C = 2 \log |2x + 3| + C.
\]
### Step 4: Combine the results
Now substituting back, we have:
\[
\int \frac{2x - 1}{2x + 3} \, dx = x - 2 \log |2x + 3| + C.
\]
### Step 5: Simplify the logarithm
Using the property of logarithms, \(a \log b = \log b^a\), we can rewrite \(2 \log |2x + 3|\) as:
\[
\log |(2x + 3)^2|.
\]
Thus, we have:
\[
\int \frac{2x - 1}{2x + 3} \, dx = x - \log |(2x + 3)^2| + C.
\]
### Conclusion
This matches the right-hand side of the original equation:
\[
x - \log |(2x + 3)^2| + C.
\]
Therefore, we have verified that
\[
\int \frac{2x-1}{2x+3} \, dx = x - \log|(2x+3)^{2}| + C.
\]
