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We are given the equation:

\[
\int (2x + 3) \ln x \, dx = y z - \int z \, dy
\]

The objective is to find the expression for \( yz \) and match it with one of the given options.

Let's break down the process:

### Step 1: Solve the Integral
We need to evaluate:

\[
\int (2x + 3) \ln x \, dx
\]

This can be split into two parts using the distributive property of integrals:

\[
\int (2x + 3) \ln x \, dx = \int 2x \ln x \, dx + \int 3 \ln x \, dx
\]

Now, we handle each integral separately:

1. **For** \(\int 2x \ln x \, dx\):  
   This is an integration by parts problem. Let:
   - \( u = \ln x \), so \( du = \frac{1}{x} dx \)
   - \( dv = 2x \, dx \), so \( v = x^2 \)

   Using integration by parts formula \( \int u \, dv = uv - \int v \, du \), we get:
   \[
   \int 2x \ln x \, dx = x^2 \ln x - \int x^2 \cdot \frac{1}{x} dx = x^2 \ln x - \int x \, dx
   \]
   \[
   = x^2 \ln x - \frac{x^2}{2}
   \]

2. **For** \(\int 3 \ln x \, dx\):  
   Again, using integration by parts, let:
   - \( u = \ln x \), so \( du = \frac{1}{x} dx \)
   - \( dv = 3 \, dx \), so \( v = 3x \)

   Applying the formula, we get:
   \[
   \int 3 \ln x \, dx = 3x \ln x - \int 3x \cdot \frac{1}{x} dx = 3x \ln x - \int 3 \, dx
   \]
   \[
   = 3x \ln x - 3x
   \]

### Step 2: Combine the Results
Now, combine both integrals:

\[
\int (2x + 3) \ln x \, dx = \left( x^2 \ln x - \frac{x^2}{2} \right) + (3x \ln x - 3x)
\]

Simplifying:

\[
= x^2 \ln x + 3x \ln x - \frac{x^2}{2} - 3x
\]

### Step 3: Find \( yz \)
From the equation \( \int (2x + 3) \ln x \, dx = yz \), we equate the above expression to \( yz \):

\[
yz = x^2 \ln x + 3x \ln x - \frac{x^2}{2} - 3x
\]

### Step 4: Compare with Options
The expression most closely resembles **option (b)**:

\[
(2x + 3) \ln x
\]

### Final Answer:
The correct answer is **(b)**.

---

Do you want further details or have any questions? Here are 5 related questions to deepen your understanding:

1. How does integration by parts work, and when is it useful?
2. Why do we split the integral when integrating functions like \( (2x + 3) \ln x \)?
3. What are other methods to approach integrals involving logarithmic functions?
4. Could substitution methods have been used here instead of integration by parts?
5. How can you verify your integral solutions by differentiation?

**Tip:** When solving integrals involving products of functions, always check if integration by parts or substitution is applicable to simplify the expression.

If ∫(2x + 3)lnx dx = yz - ∫z dy, then yz = ....

Explore how to solve the integral ∫(2x + 3)lnx dx using the integration by parts method. This problem applies advanced calculus concepts and theorems to solve integrals with logarithmic functions. Suitable for college-level calculus students.

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