The integral ` int_(-1)^(1) (|x+2|)/(x+2)dx` is equal to

Options

A. 1
B. 2
C. 0
D. `-1`

Answer

To solve the integral \( I = \int_{-1}^{1} \frac{|x+2|}{x+2} \, dx \), we will break it down step by step.

### Step 1: Analyze the absolute value
The expression inside the integral contains an absolute value, \( |x + 2| \). To handle this, we need to determine where the expression \( x + 2 \) is positive or negative.

- Set \( x + 2 = 0 \) to find the critical point:
\[ x + 2 = 0 \implies x = -2 \]

### Step 2: Determine the intervals
Now, we analyze the intervals based on the critical point \( x = -2 \):

- For \( x < -2 \): \( x + 2 < 0 \) so \( |x + 2| = -(x + 2) \)
- For \( x \geq -2 \): \( x + 2 \geq 0 \) so \( |x + 2| = x + 2 \)

### Step 3: Evaluate the integral from -1 to 1
Since our limits of integration are from -1 to 1, we note that both limits fall in the interval where \( x + 2 \) is positive (since \( -1 + 2 = 1 \) and \( 1 + 2 = 3 \)). Therefore, we can simplify the integral:

\[ I = \int_{-1}^{1} \frac{|x + 2|}{x + 2} \, dx = \int_{-1}^{1} \frac{x + 2}{x + 2} \, dx = \int_{-1}^{1} 1 \, dx \]

### Step 4: Calculate the integral
Now we compute the integral:

\[ I = \int_{-1}^{1} 1 \, dx \]

The integral of 1 over the interval from -1 to 1 is simply the length of the interval:

\[ I = [x]_{-1}^{1} = 1 - (-1) = 1 + 1 = 2 \]

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

\[ I = 2 \]