If `I_(n)-int(lnx)^(n)dx,` then `I_(10)+10I_(9)` is equal to (where C is the constant of integration)

Options

A. `x(lnx)^(10)+C`
B. `10(lnx)^(9)+C`
C. `9(lnx)^(10)+C`
D. `x(lnx)^(9)+C`

Answer

To solve the problem, we need to evaluate \( I_n = \int (\ln x)^n \, dx \) and find the expression for \( I_{10} + 10 I_{9} \).

### Step-by-Step Solution:

1. **Integration by Parts Setup**:
We will use integration by parts, where we let:
- \( u = (\ln x)^n \) (first function)
- \( dv = dx \) (second function)

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

2. **Applying Integration by Parts**:
Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ I_n = x (\ln x)^n - \int x \cdot n (\ln x)^{n-1} \cdot \frac{1}{x} \, dx \]
Simplifying the integral:
\[ I_n = x (\ln x)^n - n \int (\ln x)^{n-1} \, dx \]
Thus, we can express this as:
\[ I_n = x (\ln x)^n - n I_{n-1} \]

3. **Finding \( I_{10} + 10 I_{9} \)**:
Now, substituting \( n = 10 \) into the equation:
\[ I_{10} = x (\ln x)^{10} - 10 I_{9} \]
Rearranging gives:
\[ I_{10} + 10 I_{9} = x (\ln x)^{10} \]

4. **Final Expression**:
Therefore, we have:
\[ I_{10} + 10 I_{9} = x (\ln x)^{10} + C \]
where \( C \) is the constant of integration.

### Final Answer:
\[ I_{10} + 10 I_{9} = x (\ln x)^{10} + C \]