`int " log"_(e) " x dx "`

Answer

To solve the integral \( \int \log_e x \, dx \), we can use integration by parts. The formula for integration by parts is given by:

\[ \int u \, dv = uv - \int v \, du \]

### Step-by-Step Solution:

1. **Choose \( u \) and \( dv \)**:
Let:
\[ u = \log_e x \quad \text{and} \quad dv = dx \]

2. **Differentiate \( u \) and integrate \( dv \)**:
Now, we differentiate \( u \) and integrate \( dv \):
\[ du = \frac{1}{x} \, dx \quad \text{and} \quad v = x \]

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

4. **Simplify the integral**:
The integral simplifies as follows:
\[ \int x \cdot \frac{1}{x} \, dx = \int 1 \, dx = x \]

5. **Combine the results**:
Now substitute back into the equation:
\[ \int \log_e x \, dx = x \log_e x - x + C \]
where \( C \) is the constant of integration.

6. **Factor out \( x \)**:
To make the expression neater, we can factor out \( x \):
\[ \int \log_e x \, dx = x \left( \log_e x - 1 \right) + C \]

### Final Answer:
\[ \int \log_e x \, dx = x \left( \log_e x - 1 \right) + C \]