If `I_n=int( lnx)^n dx` then `I_n+nI_(n-1)`

Options

A. `(("In "x)^(n))/(x)+C`
B. `x("In "x)^(n-1)+C`
C. `x("In "x)^(n)+C`
D. none of these

Explanation

To solve the problem, we start with the integral \( I_n = \int (\ln x)^n \, dx \) and we need to find the expression for \( I_n + n I_{n-1} \).

### Step-by-Step Solution:

1. **Define the Integral**:
\[ I_n = \int (\ln x)^n \, dx \]

2. **Use Integration by Parts**:
We will apply integration by parts. Let:
- \( u = (\ln x)^n \) (first function)
- \( dv = dx \) (second function)

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

By the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
we have:
\[ I_n = x (\ln x)^n - \int x \cdot n (\ln x)^{n-1} \cdot \frac{1}{x} \, dx \]
Simplifying the integral:
\[ I_n = x (\ln x)^n - n \int (\ln x)^{n-1} \, dx \]
This gives:
\[ I_n = x (\ln x)^n - n I_{n-1} \]

3. **Rearranging the Equation**:
Rearranging the above equation gives:
\[ I_n + n I_{n-1} = x (\ln x)^n \]

4. **Final Expression**:
Therefore, we can express the result as:
\[ I_n + n I_{n-1} = x (\ln x)^n + C \]
where \( C \) is the constant of integration.

### Conclusion:
Thus, the final result is:
\[ I_n + n I_{n-1} = x (\ln x)^n + C \]