`int x^(x) (1+ log x) dx`

Answer

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

### Step 1: Substitution
Let \( t = x^x \). To differentiate this, we can use the property of logarithms:
\[ \log t = x \log x \]
Now, differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(\log t) = \frac{1}{t} \frac{dt}{dx} = \log x + 1 \]
Thus, we have:
\[ \frac{dt}{dx} = t (\log x + 1) \]

### Step 2: Express \( dx \) in terms of \( dt \)
From the equation above, we can express \( dx \) as:
\[ dx = \frac{dt}{t (\log x + 1)} \]

### Step 3: Substitute back into the integral
Now, substitute \( t = x^x \) and \( dx \) into the integral:
\[ \int x^x (1 + \log x) \, dx = \int t \cdot \frac{dt}{t (\log x + 1)} \]
This simplifies to:
\[ \int dt = t + C \]

### Step 4: Substitute back for \( t \)
Now, substitute back \( t = x^x \):
\[ \int x^x (1 + \log x) \, dx = x^x + C \]

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

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