`int_(e )^(e^(2))log x dx =`

Options

A. 1
B. `e^(2)`
C. `e^(2)-1`
D. `1+e^(2)`

Answer

To solve the integral \( \int_{e}^{e^2} \log x \, dx \), we can use integration by parts.

### Step-by-Step Solution:

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:
- \( u = \log x \) (which implies \( du = \frac{1}{x} \, dx \))
- \( dv = dx \) (which implies \( v = x \))

2. **Apply integration by parts**:
Using the integration by parts formula:
\[ \int \log x \, dx = x \log x - \int x \cdot \frac{1}{x} \, dx = x \log x - \int dx \]
\[ = x \log x - x + C \]

3. **Evaluate the definite integral from \( e \) to \( e^2 \)**:
We need to evaluate:
\[ \int_{e}^{e^2} \log x \, dx = \left[ x \log x - x \right]_{e}^{e^2} \]

4. **Calculate the upper limit \( e^2 \)**:
\[ \text{At } x = e^2: \quad e^2 \log(e^2) - e^2 = e^2 \cdot 2 - e^2 = 2e^2 - e^2 = e^2 \]

5. **Calculate the lower limit \( e \)**:
\[ \text{At } x = e: \quad e \log e - e = e \cdot 1 - e = e - e = 0 \]

6. **Combine the results**:
Now, we subtract the lower limit from the upper limit:
\[ \int_{e}^{e^2} \log x \, dx = \left( e^2 \right) - \left( 0 \right) = e^2 \]

### Final Answer:
\[ \int_{e}^{e^2} \log x \, dx = e^2 \]