`int x ^(x)((ln x )^(2) +lnx+1/x) ` dx is equal to:

Options

A. `x ^(2) ((ln x)^(2) -1/x +C`
B. `x ^(x) (ln x-x) +C`
C. `x ^(x) ((ln x )^(2))/(2 )+C`
D. `x ^(x)ln x +C`

Answer

To solve the integral \( I = \int x^x \left( (\ln x)^2 + \ln x + \frac{1}{x} \right) dx \), we can follow these steps:

### Step 1: Define the Integral
Let
\[ I = \int x^x \left( (\ln x)^2 + \ln x + \frac{1}{x} \right) dx \]

### Step 2: Introduce a Substitution
Let
\[ t = x^x \ln x \]
Now, we will differentiate \( t \) with respect to \( x \).

### Step 3: Differentiate \( t \)
Using the product rule and properties of logarithms:
\[ \ln t = \ln(x^x) + \ln(\ln x) = x \ln x + \ln(\ln x) \]
Differentiating both sides with respect to \( x \):
\[ \frac{1}{t} \frac{dt}{dx} = \frac{d}{dx}(x \ln x) + \frac{d}{dx}(\ln(\ln x)) \]
Using the product rule on \( x \ln x \):
\[ \frac{d}{dx}(x \ln x) = \ln x + 1 \]
And for \( \ln(\ln x) \):
\[ \frac{d}{dx}(\ln(\ln x)) = \frac{1}{\ln x} \cdot \frac{1}{x} \]
Thus, we have:
\[ \frac{1}{t} \frac{dt}{dx} = \ln x + 1 + \frac{1}{x \ln x} \]

### Step 4: Solve for \( dt \)
Multiplying both sides by \( t \):
\[ dt = t \left( \ln x + 1 + \frac{1}{x \ln x} \right) dx \]
Substituting \( t = x^x \ln x \):
\[ dt = x^x \ln x \left( \ln x + 1 + \frac{1}{x \ln x} \right) dx \]

### Step 5: Substitute Back into the Integral
From the expression for \( dt \), we can see that:
\[ I = \int dt \]

### Step 6: Integrate
The integral of \( dt \) is simply:
\[ I = t + C = x^x \ln x + C \]

### Final Answer
Thus, the integral evaluates to:
\[ \int x^x \left( (\ln x)^2 + \ln x + \frac{1}{x} \right) dx = x^x \ln x + C \]