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 first analyze the expression inside the integral.

### Step 1: Analyze the Absolute Value
The expression \( |x + 2| \) depends on the value of \( x \):
- For \( x < -2 \), \( |x + 2| = -(x + 2) \).
- For \( x \geq -2 \), \( |x + 2| = x + 2 \).

Since our limits of integration are from \(-1\) to \(1\), we note that both \(-1\) and \(1\) are greater than \(-2\). Therefore, within the interval \([-1, 1]\), we have:
\[ |x + 2| = x + 2. \]

### Step 2: Simplify the Integral
Substituting \( |x + 2| \) into the integral, we get:
\[ I = \int_{-1}^{1} \frac{x + 2}{x + 2} \, dx. \]
For \( x \neq -2 \), this simplifies to:
\[ I = \int_{-1}^{1} 1 \, dx. \]

### Step 3: Evaluate the Integral
Now, we can evaluate the integral:
\[ I = \int_{-1}^{1} 1 \, dx = [x]_{-1}^{1} = 1 - (-1) = 1 + 1 = 2. \]

### Final Result
Thus, the value of the integral is:
\[ \boxed{2}. \]