`int(x^x)^x(2xlog_ex+x)dx` is equal to

Options

A. `x^((x^(x)))+C`
B. `(x^(x))^(x)+C`
C. `x^(2).log_(e)x+C`
D. None of these

Answer

To solve the integral \( \int (x^x)^x (2x \log_e x + x) \, dx \), we can follow these steps:

### Step 1: Set up the integral
Let
\[ I = \int (x^x)^x (2x \log_e x + x) \, dx \]

### Step 2: Simplify the expression
We can rewrite \( (x^x)^x \) as \( x^{x^2} \). Therefore, we have:
\[ I = \int x^{x^2} (2x \log_e x + x) \, dx \]

### Step 3: Factor out \( x \)
Now, we can factor out \( x \) from the expression inside the integral:
\[ I = \int x^{x^2} \left(2x \log_e x + x\right) \, dx = \int x^{x^2} x (2 \log_e x + 1) \, dx \]
This simplifies to:
\[ I = \int x^{x^2 + 1} (2 \log_e x + 1) \, dx \]

### Step 4: Use substitution
Let
\[ t = x^{x^2} \]
Now, we take the natural logarithm of both sides:
\[ \log_e t = x^2 \log_e x \]

### Step 5: Differentiate both sides
Differentiating both sides with respect to \( x \):
\[ \frac{1}{t} \frac{dt}{dx} = 2x \log_e x + x \]
Thus,
\[ dt = t (2x \log_e x + x) \, dx \]

### Step 6: Substitute back into the integral
Substituting \( dt \) into the integral, we have:
\[ I = \int dt \]

### Step 7: Integrate
The integral of \( dt \) is:
\[ I = t + C \]

### Step 8: Substitute back for \( t \)
Recalling that \( t = x^{x^2} \), we get:
\[ I = x^{x^2} + C \]

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