`int_(-1)^(0) (dx)/( x^(2) + 2x+ 2)` is equal to

Options

A. 0
B. `(pi)/(2)`
C. `(pi)/(4)`
D. `-(pi)/(4)`

Answer

To solve the integral \( I = \int_{-1}^{0} \frac{dx}{x^2 + 2x + 2} \), we will follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ I = \int_{-1}^{0} \frac{dx}{x^2 + 2x + 2} \]

### Step 2: Complete the square in the denominator
We need to rewrite the quadratic expression in the denominator in a completed square form. The expression \( x^2 + 2x + 2 \) can be rewritten as:
\[ x^2 + 2x + 2 = (x^2 + 2x + 1) + 1 = (x + 1)^2 + 1 \]
Thus, we have:
\[ I = \int_{-1}^{0} \frac{dx}{(x + 1)^2 + 1} \]

### Step 3: Make a substitution
Let \( t = x + 1 \). Then, \( dt = dx \). When \( x = -1 \), \( t = 0 \), and when \( x = 0 \), \( t = 1 \). Therefore, the limits of integration change from \( -1 \) to \( 0 \) for \( x \) to \( 0 \) to \( 1 \) for \( t \):
\[ I = \int_{0}^{1} \frac{dt}{t^2 + 1} \]

### Step 4: Recognize the standard integral
The integral \( \int \frac{dt}{t^2 + 1} \) is a standard integral that evaluates to:
\[ \int \frac{dt}{t^2 + 1} = \tan^{-1}(t) \]
Thus, we can write:
\[ I = \left[ \tan^{-1}(t) \right]_{0}^{1} \]

### Step 5: Evaluate the integral at the limits
Now we evaluate the definite integral:
\[ I = \tan^{-1}(1) - \tan^{-1}(0) \]
We know that:
\[ \tan^{-1}(1) = \frac{\pi}{4} \quad \text{and} \quad \tan^{-1}(0) = 0 \]
So, we have:
\[ I = \frac{\pi}{4} - 0 = \frac{\pi}{4} \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{\frac{\pi}{4}} \]