The value of the integral `inte^(x^(2)+(1)/(2))(2x^(2)-(1)/(x)+1)dx` is equal to (where C is the constant of integration)

Options

A. `e^(x^(2)+(1)/(x))+C`
B. `x^(2)(x^(2)+(1)/(x))+C`
C. `xe^(x^(2)+(1)/(x))+C`
D. `x.e^(x)+C`

Answer

To solve the integral \( \int e^{x^2 + \frac{1}{2}} (2x^2 - \frac{1}{x} + 1) \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We can factor out \( e^{\frac{1}{2}} \) from the integral since it is a constant:
\[ \int e^{x^2 + \frac{1}{2}} (2x^2 - \frac{1}{x} + 1) \, dx = e^{\frac{1}{2}} \int e^{x^2} (2x^2 - \frac{1}{x} + 1) \, dx \]

### Step 2: Distribute the Exponential
Next, we can distribute \( e^{x^2} \) across the terms inside the integral:
\[ = e^{\frac{1}{2}} \left( \int 2x^2 e^{x^2} \, dx - \int \frac{1}{x} e^{x^2} \, dx + \int e^{x^2} \, dx \right) \]

### Step 3: Solve Each Integral
Now we will solve each integral separately.

1. **Integral of \( 2x^2 e^{x^2} \)**:
Using integration by parts, let \( u = e^{x^2} \) and \( dv = 2x \, dx \). Then, \( du = 2x e^{x^2} \, dx \) and \( v = x^2 \).
\[ \int 2x^2 e^{x^2} \, dx = x e^{x^2} - \int e^{x^2} \, dx \]

2. **Integral of \( \frac{1}{x} e^{x^2} \)**:
This integral does not have a simple elementary form, but we can denote it as \( I_1 \).

3. **Integral of \( e^{x^2} \)**:
This integral is known and can be denoted as \( I_2 \).

### Step 4: Combine the Results
Now we can combine the results:
\[ \int e^{x^2 + \frac{1}{2}} (2x^2 - \frac{1}{x} + 1) \, dx = e^{\frac{1}{2}} \left( x e^{x^2} - I_1 + I_2 + C \right) \]

### Step 5: Simplify
Finally, we can express the result in a simpler form:
\[ = e^{\frac{1}{2}} \left( x e^{x^2} + C \right) \]

### Final Result
Thus, the value of the integral is:
\[ \int e^{x^2 + \frac{1}{2}} (2x^2 - \frac{1}{x} + 1) \, dx = e^{\frac{1}{2}} (x e^{x^2} + C) \]