Evaluate the integral $\int_{-1}^{1} \frac{dx}{x^{2}+2x+5}$.

(D) We are given the integral $I = \int_{-1}^{1} \frac{dx}{x^{2}+2x+5}$.
First,complete the square in the denominator: $x^{2}+2x+5 = (x^{2}+2x+1) + 4 = (x+1)^{2} + 2^{2}$.
So,$I = \int_{-1}^{1} \frac{dx}{(x+1)^{2} + 2^{2}}$.
Let $u = x+1$,then $du = dx$.
Change the limits of integration:
When $x = -1$,$u = -1+1 = 0$.
When $x = 1$,$u = 1+1 = 2$.
Now,the integral becomes $I = \int_{0}^{2} \frac{du}{u^{2} + 2^{2}}$.
Using the standard formula $\int \frac{dx}{x^{2}+a^{2}} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$,we get:
$I = \left[ \frac{1}{2} \tan^{-1}(\frac{u}{2}) \right]_{0}^{2}$.
Evaluating at the limits:
$I = \frac{1}{2} \tan^{-1}(\frac{2}{2}) - \frac{1}{2} \tan^{-1}(\frac{0}{2})$
$I = \frac{1}{2} \tan^{-1}(1) - \frac{1}{2} \tan^{-1}(0)$
$I = \frac{1}{2} (\frac{\pi}{4}) - \frac{1}{2} (0) = \frac{\pi}{8}$.