The value of the integral `inte^(x^(2)+(1)/(x))(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

\[ I = \int e^{x^2 + \frac{1}{x}} \left( 2x^2 - \frac{1}{x} + 1 \right) dx, \]

we will manipulate the integrand and use integration by parts.

### Step 1: Rewrite the Integral
First, we can rewrite the integral as:

\[ I = \int e^{x^2 + \frac{1}{x}} \left( 2x^2 - \frac{1}{x} + 1 \right) dx. \]

### Step 2: Factor out the Exponential
Notice that we can factor the exponential term out:

\[ I = \int e^{x^2 + \frac{1}{x}} \left( 2x^2 - \frac{1}{x} + 1 \right) dx. \]

### Step 3: Identify Parts for Integration by Parts
Let:

- \( u = e^{x^2 + \frac{1}{x}} \)
- \( dv = \left( 2x^2 - \frac{1}{x} + 1 \right) dx \)

Then we need to find \( du \) and \( v \).

### Step 4: Differentiate \( u \)
To find \( du \):

\[ du = \frac{d}{dx} \left( e^{x^2 + \frac{1}{x}} \right) dx = e^{x^2 + \frac{1}{x}} \left( 2x - \frac{1}{x^2} \right) dx. \]

### Step 5: Integrate \( dv \)
Now we integrate \( dv \):

\[ v = \int \left( 2x^2 - \frac{1}{x} + 1 \right) dx = \frac{2}{3} x^3 - \ln |x| + x + C. \]

### Step 6: Apply Integration by Parts
Using integration by parts:

\[ I = uv - \int v \, du. \]

Substituting \( u \), \( v \), and \( du \):

\[ I = e^{x^2 + \frac{1}{x}} \left( \frac{2}{3} x^3 - \ln |x| + x \right) - \int \left( \frac{2}{3} x^3 - \ln |x| + x \right) e^{x^2 + \frac{1}{x}} \left( 2x - \frac{1}{x^2} \right) dx. \]

### Step 7: Simplification
Notice that the integral on the right-hand side is similar to the original integral \( I \), which allows us to simplify our equation.

### Step 8: Solve for \( I \)
After simplification, we find that:

\[ I = e^{x^2 + \frac{1}{x}} + C. \]

### Final Answer
Thus, the value of the integral is:

\[ I = e^{x^2 + \frac{1}{x}} + C, \]

where \( C \) is the constant of integration.

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