`int( x-1)e^(-x) dx` is equal to : a) `( x- 2)e^(x) + C` b) ` x e^(-x) + C` c) ` - x e^(x) + C` d) ` ( x+1)e^(-x) +C`

Options

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

Answer

To solve the integral \( \int (x - 1)e^{-x} \, dx \), we will use integration by parts. The formula for integration by parts is:

\[ \int u \, dv = uv - \int v \, du \]

### Step 1: Choose \( u \) and \( dv \)

Let:
- \( u = x - 1 \) (which means \( du = dx \))
- \( dv = e^{-x} \, dx \) (which means \( v = \int e^{-x} \, dx = -e^{-x} \))

### Step 2: Apply the integration by parts formula

Using the integration by parts formula, we have:

\[ \int (x - 1)e^{-x} \, dx = (x - 1)(-e^{-x}) - \int (-e^{-x}) \, dx \]

### Step 3: Simplify the expression

This simplifies to:

\[ -(x - 1)e^{-x} + \int e^{-x} \, dx \]

### Step 4: Integrate \( e^{-x} \)

Now, we need to integrate \( e^{-x} \):

\[ \int e^{-x} \, dx = -e^{-x} \]

### Step 5: Substitute back into the equation

Substituting this back, we get:

\[ -(x - 1)e^{-x} - e^{-x} + C \]

### Step 6: Combine like terms

Now, we can combine the terms:

\[ -(x - 1)e^{-x} - e^{-x} = -xe^{-x} + e^{-x} - e^{-x} = -xe^{-x} \]

Thus, the final result is:

\[ \int (x - 1)e^{-x} \, dx = -xe^{-x} + C \]

### Final Answer

The correct answer is:

\[ \boxed{-xe^{-x} + C} \]