We are given the equation:

$\int (2x - 1) e^{2x + 3} \, dx = yz - \int z \, dy$

We are tasked with finding the value of $\int z \, dy$, and we are provided with some answer choices.

Step 1: Solve the given integral

We begin by evaluating the integral on the left-hand side:

$\int (2x - 1) e^{2x + 3} \, dx$

Using substitution, let: $u = 2x + 3 \quad \Rightarrow \quad du = 2 \, dx$ Thus, the integral becomes:

$\frac{1}{2} \int (u - 1) e^u \, du$

Now, break this into two parts:

$\frac{1}{2} \left[ \int u e^u \, du - \int e^u \, du \right]$

The first integral can be solved using integration by parts. Let: $v = u \quad \text{and} \quad dw = e^u \, du$

Then:

$\int u e^u \, du = u e^u - \int e^u \, du = u e^u - e^u$

Thus, we have:

[ \frac{1}{2} \left[ (u e^u - e^u) - e^u \right] = \frac{1}{2} \left[ u e^u - 2 e^u \right] ]

Substituting back $u = 2x + 3$:

[ \frac{1}{2} \left[ (2x + 3) e^{2x + 3} - 2 e^{2x + 3} \right] = \frac{1}{2} e^{2x + 3} \left[ (2x + 3) - 2 \right] ]

Simplifying:

$\frac{1}{2} e^{2x + 3} (2x + 1)$

Step 2: Compare with the right-hand side

We are given that this is equal to $yz - \int z \, dy$, so we can now solve for $\int z \, dy$:

$\int z \, dy = yz - \frac{1}{2} e^{2x + 3} (2x + 1)$

Looking at the answer choices, we conclude that $\int z \, dy = \boxed{\frac{1}{2} e^{2x + 3}} + c$, which matches option (b).

Would you like any further details on the steps?

Related questions:

1. What is the technique of integration by parts and when is it used?
2. How is substitution used in solving integrals like the one in the problem?
3. What are some common integrals involving exponential functions?
4. How can you verify if an integral solution is correct?
5. How does the constant of integration ($+ c$) affect the solution?

Tip:

When solving integrals involving products of functions, always check if integration by parts or substitution can simplify the process.