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 find the expression for \( I_n = \int (\ln x)^n \, dx \) and then compute \( I_{10} + 10I_{9} \).

### Step 1: Define the integral \( I_n \)
We start with the integral:
\[ I_n = \int (\ln x)^n \, dx \]

### Step 2: Use integration by parts
We can use integration by parts where we let:
- \( u = (\ln x)^n \) (first function)
- \( dv = dx \) (second function)

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

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

### Step 3: Simplify the integral
The integral simplifies to:
\[ I_n = x (\ln x)^n - n \int (\ln x)^{n-1} \, dx \]
\[ I_n = x (\ln x)^n - n I_{n-1} \]

### Step 4: Rearranging the equation
Now we can rearrange the equation:
\[ I_n + n I_{n-1} = x (\ln x)^n \]

### Step 5: Substitute \( n = 10 \)
Now, substituting \( n = 10 \):
\[ I_{10} + 10 I_{9} = x (\ln x)^{10} \]

### Step 6: Final result
Thus, we have:
\[ I_{10} + 10 I_{9} = x (\ln x)^{10} + C \]

### Conclusion
The final answer is:
\[ I_{10} + 10 I_{9} = x (\ln x)^{10} + C \]