`intx^(x)(1+log_(e)x)`dx is equal to

Options

A. `x^(x)log_(e)x+C`
B. `ex^(x)+C`
C. `x^(x)+C`
D. none of these

Explanation

To solve the integral \( \int x^x (1 + \log_e x) \, dx \), we can follow these steps:

### Step 1: Substitution
Let \( t = x^x \). To differentiate \( t \), we take the natural logarithm of both sides:
\[ \log t = \log(x^x) = x \log x \]

### Step 2: Differentiate
Now we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(\log t) = \frac{d}{dx}(x \log x) \]
Using the product rule on the right side:
\[ \frac{1}{t} \frac{dt}{dx} = \log x + 1 \]
Thus,
\[ \frac{dt}{dx} = t (\log x + 1) \]

### Step 3: Express \( dx \)
Now we can express \( dx \) in terms of \( dt \):
\[ dx = \frac{dt}{t (\log x + 1)} \]

### Step 4: Substitute in the Integral
Substituting \( t = x^x \) and \( dx \) into the integral:
\[ \int x^x (1 + \log_e x) \, dx = \int t (1 + \log_e x) \cdot \frac{dt}{t (\log x + 1)} \]
The \( t \) cancels out:
\[ = \int dt \]

### Step 5: Integrate
The integral of \( dt \) is:
\[ t + C \]

### Step 6: Substitute Back
Now substitute back \( t = x^x \):
\[ = x^x + C \]

### Conclusion
Thus, the final answer is:
\[ \int x^x (1 + \log_e x) \, dx = x^x + C \]