If =`int_(1)^(e) (logx)^(n) dx, "then"I_(n)+nI_(n-1)` is equal to

Options

A. `1//e`
B. e
C. `e-1`
D. none of these

Explanation

To solve the problem, we need to evaluate the expression \( I_n + n I_{n-1} \), where \( I_n = \int_1^e (\log x)^n \, dx \).

### Step-by-Step Solution:

1. **Define the Integral**:
\[ I_n = \int_1^e (\log x)^n \, dx \]

2. **Apply Integration by Parts**:
We will use integration by parts, where we let:
- \( u = (\log x)^n \) (first part)
- \( dv = dx \) (second part)

Then, we differentiate and integrate:
- \( du = n (\log x)^{n-1} \cdot \frac{1}{x} \, dx \)
- \( v = x \)

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ I_n = \left[ x (\log x)^n \right]_1^e - \int_1^e x \cdot n (\log x)^{n-1} \cdot \frac{1}{x} \, dx \]

3. **Evaluate the Boundary Terms**:
Calculate the boundary terms:
- At \( x = e \): \( e (\log e)^n = e \cdot 1^n = e \)
- At \( x = 1 \): \( 1 (\log 1)^n = 1 \cdot 0^n = 0 \)

Thus,
\[ I_n = e - n \int_1^e (\log x)^{n-1} \, dx \]
This gives us:
\[ I_n = e - n I_{n-1} \]

4. **Rearranging the Equation**:
We can rearrange the equation to find \( I_n + n I_{n-1} \):
\[ I_n + n I_{n-1} = e \]

5. **Final Result**:
Therefore, the value of \( I_n + n I_{n-1} \) is:
\[ \boxed{e} \]