If `I_(m,n)= int_(0)^(1) x^(m) (ln x)^(n)`dx then `I_(m,n)` is also equal to

Options

A. `(n)/(n+1) I_(m,n-1)`
B. `(-m)/(n+1) I_(m,n-1)`
C. `(-n)/(m+1) I_(m,n-1)`
D. `(m)/(n+1) I_(m,n-1)`

Answer

To solve the integral \( I_{m,n} = \int_0^1 x^m (\ln x)^n \, dx \), we will use integration by parts.

### Step-by-Step Solution:

1. **Identify the parts for integration by parts**:
- Let \( u = (\ln x)^n \) and \( dv = x^m \, dx \).
- Then, we need to find \( du \) and \( v \):
- \( du = n (\ln x)^{n-1} \cdot \frac{1}{x} \, dx \)
- \( v = \frac{x^{m+1}}{m+1} \)

2. **Apply the integration by parts formula**:
The formula for integration by parts is:
\[ \int u \, dv = uv - \int v \, du \]
Applying this, we have:
\[ I_{m,n} = \left[ (\ln x)^n \cdot \frac{x^{m+1}}{m+1} \right]_0^1 - \int_0^1 \frac{x^{m+1}}{m+1} \cdot n (\ln x)^{n-1} \cdot \frac{1}{x} \, dx \]

3. **Evaluate the boundary term**:
- At \( x = 1 \), \( \ln(1) = 0 \), so the first term becomes \( 0 \).
- At \( x = 0 \), \( \ln(0) \) approaches \( -\infty \), but \( x^{m+1} \) approaches \( 0 \) faster than \( (\ln x)^n \) approaches \( -\infty \). Therefore, this term also evaluates to \( 0 \).
- Thus, the boundary term is \( 0 - 0 = 0 \).

4. **Simplify the integral**:
Now we have:
\[ I_{m,n} = 0 - \frac{n}{m+1} \int_0^1 x^m (\ln x)^{n-1} \, dx \]
This simplifies to:
\[ I_{m,n} = -\frac{n}{m+1} I_{m,n-1} \]

### Final Result:
Thus, we have:
\[ I_{m,n} = -\frac{n}{m+1} I_{m,n-1} \]