The integral `int(1+x-1/x)e^(x+1/x)dx` is equal to

Options

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

Explanation

To solve the integral \( I = \int (1 + x - \frac{1}{x}) e^{x + \frac{1}{x}} \, dx \), we can break it down into manageable parts.

### Step-by-Step Solution:

1. **Separate the Integral**:
\[ I = \int (1 + x - \frac{1}{x}) e^{x + \frac{1}{x}} \, dx = \int e^{x + \frac{1}{x}} \, dx + \int x e^{x + \frac{1}{x}} \, dx - \int \frac{1}{x} e^{x + \frac{1}{x}} \, dx \]

2. **Let \( t = x + \frac{1}{x} \)**:
To simplify the integral, we can use the substitution \( t = x + \frac{1}{x} \). Then, we differentiate:
\[ dt = \left(1 - \frac{1}{x^2}\right) dx \quad \Rightarrow \quad dx = \frac{dt}{1 - \frac{1}{x^2}} \]

3. **Rewrite the Integral**:
The integral now becomes:
\[ I = \int e^t \, dt \]
where \( 1 - \frac{1}{x^2} \) can be factored out from the \( x \) term.

4. **Integrate**:
The integral of \( e^t \) is straightforward:
\[ \int e^t \, dt = e^t + C \]

5. **Back Substitute**:
Substitute back \( t = x + \frac{1}{x} \):
\[ I = e^{x + \frac{1}{x}} + C \]

6. **Final Answer**:
Thus, the integral evaluates to:
\[ I = x e^{x + \frac{1}{x}} + C \]