If n is a positive integer then `int_(0)^(1)(ln x)^(n)dx` is :

Options

A. `n!`
B. `(-1)^(n)n!`
C. `(-1)^(n+2)(n+1)!`
D. None of these

Answer

To solve the integral \( I = \int_0^1 (\ln x)^n \, dx \) where \( n \) is a positive integer, we will use integration by parts.

### Step-by-step Solution:

1. **Set Up the Integral:**
\[ I = \int_0^1 (\ln x)^n \, dx \]

2. **Choose \( u \) and \( dv \):**
We will let:
\[ u = (\ln x)^n \quad \text{and} \quad dv = dx \]
Then, we differentiate \( u \) and integrate \( dv \):
\[ du = n (\ln x)^{n-1} \cdot \frac{1}{x} \, dx \quad \text{and} \quad v = x \]

3. **Apply Integration by Parts:**
Using the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
We have:
\[ I = \left[ x (\ln x)^n \right]_0^1 - \int_0^1 x \cdot n (\ln x)^{n-1} \cdot \frac{1}{x} \, dx \]
Simplifying the integral:
\[ I = \left[ x (\ln x)^n \right]_0^1 - n \int_0^1 (\ln x)^{n-1} \, dx \]

4. **Evaluate the Boundary Terms:**
Evaluating \( \left[ x (\ln x)^n \right]_0^1 \):
- At \( x = 1 \): \( 1 \cdot \ln(1)^n = 0 \)
- At \( x = 0 \): As \( x \to 0 \), \( \ln x \to -\infty \), but \( x (\ln x)^n \to 0 \) (since \( x \) approaches 0 faster than \( (\ln x)^n \) approaches \( -\infty \)).
Thus:
\[ \left[ x (\ln x)^n \right]_0^1 = 0 - 0 = 0 \]

5. **Substituting Back:**
Now substituting back into the equation:
\[ I = 0 - n \int_0^1 (\ln x)^{n-1} \, dx \]
Therefore:
\[ I = -n I_{n-1} \]
where \( I_{n-1} = \int_0^1 (\ln x)^{n-1} \, dx \).

6. **Recursion Relation:**
This gives us a recursive relation:
\[ I_n = -n I_{n-1} \]

7. **Base Case:**
We need to find the base case \( I_0 \):
\[ I_0 = \int_0^1 1 \, dx = 1 \]

8. **Calculate \( I_n \):**
Using the recursive relation:
\[ I_1 = -1 \cdot I_0 = -1 \]
\[ I_2 = -2 \cdot I_1 = -2 \cdot (-1) = 2 \]
\[ I_3 = -3 \cdot I_2 = -3 \cdot 2 = -6 \]
Continuing this pattern, we find:
\[ I_n = (-1)^n n! \]

### Final Result:
Thus, the value of the integral is:
\[ I = \int_0^1 (\ln x)^n \, dx = (-1)^n n! \]