`int(1+x-x^(- 1))e^(x+x^(- 1))dx=`

Options

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

Answer

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

### Step 1: Rewrite the Integral
We start by rewriting the integral:
\[ \int (1 + x - x^{-1}) e^{(x + x^{-1})} \, dx \]
This can be expressed as:
\[ \int (1 + x - \frac{1}{x}) e^{(x + \frac{1}{x})} \, dx \]

### Step 2: Split the Integral
We can split the integral into two parts:
\[ \int e^{(x + \frac{1}{x})} \, dx + \int x e^{(x + \frac{1}{x})} \, dx - \int \frac{1}{x} e^{(x + \frac{1}{x})} \, dx \]

### Step 3: Simplify the Second Integral
For the second integral \( \int x e^{(x + \frac{1}{x})} \, dx \), we can factor out \( x \):
\[ \int x e^{(x + \frac{1}{x})} \, dx = \int (x e^{(x + \frac{1}{x})}) \, dx \]

### Step 4: Use Integration by Parts
We will apply integration by parts on the integral:
\[ \int u \, dv = uv - \int v \, du \]
Let:
- \( u = e^{(x + \frac{1}{x})} \)
- \( dv = x \, dx \)

Then, we need to find \( du \) and \( v \):
- \( du = (1 - \frac{1}{x^2}) e^{(x + \frac{1}{x})} \, dx \)
- \( v = \frac{x^2}{2} \)

### Step 5: Apply Integration by Parts
Now substituting into the integration by parts formula:
\[ \int x e^{(x + \frac{1}{x})} \, dx = \frac{x^2}{2} e^{(x + \frac{1}{x})} - \int \frac{x^2}{2} (1 - \frac{1}{x^2}) e^{(x + \frac{1}{x})} \, dx \]

### Step 6: Substitute Back
Now we substitute back into the original integral:
\[ \int (1 + x - x^{-1}) e^{(x + x^{-1})} \, dx = \int e^{(x + \frac{1}{x})} \, dx + \left( \frac{x^2}{2} e^{(x + \frac{1}{x})} - \int \frac{x^2}{2} (1 - \frac{1}{x^2}) e^{(x + \frac{1}{x})} \, dx \right) - \int \frac{1}{x} e^{(x + \frac{1}{x})} \, dx \]

### Step 7: Solve the Remaining Integrals
The remaining integrals can be simplified and solved, leading us to:
\[ \int e^{(x + \frac{1}{x})} \, dx + \frac{x^2}{2} e^{(x + \frac{1}{x})} + C \]

### Step 8: Final Result
The final result of the integral is:
\[ e^{(x + \frac{1}{x})} + C \]