` int_(-1)^(1) (dx) / (x^(2)+2x+5)` is equal to

Options

A. `pi/2`
B. `pi/4`
C. `pi/8`
D. `pi/3`

Explanation

To evaluate the integral

\[ I = \int_{-1}^{1} \frac{dx}{x^2 + 2x + 5}, \]

we will follow these steps:

### Step 1: Complete the square in the denominator

The expression \(x^2 + 2x + 5\) can be rewritten by completing the square.

\[ x^2 + 2x + 5 = (x^2 + 2x + 1) + 4 = (x + 1)^2 + 4. \]

### Step 2: Rewrite the integral

Now we can rewrite the integral \(I\) as:

\[ I = \int_{-1}^{1} \frac{dx}{(x + 1)^2 + 4}. \]

### Step 3: Use substitution

Next, we will use the substitution \(x + 1 = 2 \tan \theta\). This implies:

\[ dx = 2 \sec^2 \theta \, d\theta. \]

We also need to change the limits of integration. When \(x = -1\):

\[ x + 1 = 0 \implies \tan \theta = 0 \implies \theta = 0. \]

When \(x = 1\):

\[ x + 1 = 2 \implies \tan \theta = 1 \implies \theta = \frac{\pi}{4}. \]

### Step 4: Substitute and change limits

Now substituting into the integral, we have:

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{2 \sec^2 \theta \, d\theta}{(2 \tan \theta)^2 + 4}. \]

This simplifies to:

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{2 \sec^2 \theta \, d\theta}{4 \tan^2 \theta + 4} = \int_{0}^{\frac{\pi}{4}} \frac{2 \sec^2 \theta \, d\theta}{4(\tan^2 \theta + 1)}. \]

### Step 5: Simplify the integral

Since \(1 + \tan^2 \theta = \sec^2 \theta\), we can simplify further:

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{2 \sec^2 \theta \, d\theta}{4 \sec^2 \theta} = \int_{0}^{\frac{\pi}{4}} \frac{1}{2} \, d\theta. \]

### Step 6: Evaluate the integral

Now we can evaluate the integral:

\[ I = \frac{1}{2} \int_{0}^{\frac{\pi}{4}} d\theta = \frac{1}{2} \left[ \theta \right]_{0}^{\frac{\pi}{4}} = \frac{1}{2} \left( \frac{\pi}{4} - 0 \right) = \frac{\pi}{8}. \]

### Final Answer

Thus, the value of the integral is:

\[ I = \frac{\pi}{8}. \]

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