`int_1^2 x^(2x^2+1)(1+2lnx)dx` is equal to

Options

A. 256
B. 255
C. `(255)/(2)`
D. `128`

Answer

To solve the integral \( I = \int_1^2 x^{2x^2 + 1} (1 + 2 \ln x) \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We can rewrite the integral as:
\[ I = \int_1^2 x^{2x^2} \cdot x (1 + 2 \ln x) \, dx \]
This simplifies to:
\[ I = \int_1^2 x^{2x^2} \cdot (x + 2x \ln x) \, dx \]

### Step 2: Substitution
Let \( t = x^{2x^2} \). Taking the natural logarithm of both sides gives:
\[ \ln t = 2x^2 \ln x \]
Differentiating both sides using the product rule:
\[ \frac{1}{t} \frac{dt}{dx} = 2x^2 \cdot \frac{1}{x} + 2 \ln x \cdot 2x = 2x + 4x \ln x \]
Thus, we have:
\[ \frac{dt}{dx} = t(2x + 4x \ln x) \]
This implies:
\[ dx = \frac{dt}{t(2x + 4x \ln x)} \]

### Step 3: Change of Variables
Substituting \( dx \) into the integral:
\[ I = \int_1^2 t \cdot \frac{dt}{t(2x + 4x \ln x)} \]
This simplifies to:
\[ I = \int_1^2 \frac{dt}{2x + 4x \ln x} \]

### Step 4: Evaluate the Integral
Now, we need to evaluate the limits. When \( x = 1 \):
\[ t = 1^{2 \cdot 1^2} = 1 \]
When \( x = 2 \):
\[ t = 2^{2 \cdot 2^2} = 2^8 = 256 \]
Thus, the limits of integration change from \( 1 \) to \( 256 \).

### Step 5: Simplifying the Integral
Now, we can express the integral as:
\[ I = \frac{1}{2} \int_1^{256} dt = \frac{1}{2} [t]_{1}^{256} = \frac{1}{2} (256 - 1) = \frac{255}{2} \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{\frac{255}{2}} \]