Evaluate :
`int x log 2x dx`

Answer

To evaluate the integral \( \int x \log(2x) \, dx \), we will use the method of integration by parts.

### Step-by-Step Solution:

1. **Identify the functions for integration by parts**:
We will use the formula for integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
Here, we can choose:
- \( u = \log(2x) \) (logarithmic function)
- \( dv = x \, dx \) (algebraic function)

2. **Differentiate \( u \) and integrate \( dv \)**:
- Differentiate \( u \):
\[ du = \frac{d}{dx}(\log(2x)) \, dx = \frac{1}{2x} \cdot 2 \, dx = \frac{1}{x} \, dx \]
- Integrate \( dv \):
\[ v = \int x \, dx = \frac{x^2}{2} \]

3. **Apply the integration by parts formula**:
Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula:
\[ \int x \log(2x) \, dx = uv - \int v \, du \]
This gives us:
\[ \int x \log(2x) \, dx = \log(2x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx \]

4. **Simplify the integral**:
The integral simplifies to:
\[ \int x \log(2x) \, dx = \frac{x^2}{2} \log(2x) - \int \frac{x^2}{2x} \, dx = \frac{x^2}{2} \log(2x) - \int \frac{x}{2} \, dx \]

5. **Evaluate the remaining integral**:
\[ \int \frac{x}{2} \, dx = \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{4} \]

6. **Combine the results**:
Now, substituting back, we have:
\[ \int x \log(2x) \, dx = \frac{x^2}{2} \log(2x) - \frac{x^2}{4} + C \]

### Final Answer:
\[ \int x \log(2x) \, dx = \frac{x^2}{2} \log(2x) - \frac{x^2}{4} + C \]