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

Answer

To solve the integral \( \int \log_e x \, dx \), we will use the method of integration by parts. Here’s a step-by-step solution:

### Step 1: Identify the parts for integration by parts
We will use the formula for integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
Let:
- \( u = \log_e x \) (which we will differentiate)
- \( dv = dx \) (which we will integrate)

### Step 2: Differentiate \( u \) and integrate \( dv \)
Now, we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = \frac{1}{x} \, dx \]
- Integrate \( dv \):
\[ v = x \]

### Step 3: Apply the integration by parts formula
Substituting into the integration by parts formula:
\[ \int \log_e x \, dx = x \log_e x - \int x \cdot \frac{1}{x} \, dx \]
This simplifies to:
\[ \int \log_e x \, dx = x \log_e x - \int 1 \, dx \]

### Step 4: Integrate the remaining integral
Now, we can integrate \( \int 1 \, dx \):
\[ \int 1 \, dx = x \]

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

### Final Answer
Thus, the final result is:
\[ \int \log_e x \, dx = x \log_e x - x + C \]

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