Answer
$$\frac{\pi }{8}$$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^1 {\frac{{dx}}{{{x^2} + 2x + 5}}} \cr
& {\text{Complete the square }} \cr
& \int_{ - 1}^1 {\frac{{dx}}{{{x^2} + 2x + 5}}} = \int_{ - 1}^1 {\frac{{dx}}{{{x^2} + 2x + 1 + 4}}} = \int_{ - 1}^1 {\frac{{dx}}{{{{\left( {x + 1} \right)}^2} + {2^2}}}} \cr
& {\text{Integrate using }}\int {\frac{{du}}{{{u^2} + {a^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{u}{a}} \right) + C} \cr
& \int_{ - 1}^1 {\frac{{dx}}{{{{\left( {x + 1} \right)}^2} + {2^2}}}} = \frac{1}{2}\left[ {{{\tan }^{ - 1}}\left( {\frac{{x + 1}}{2}} \right)} \right]_{ - 1}^1 \cr
& {\text{Evaluate}} \cr
& \frac{1}{2}\left[ {{{\tan }^{ - 1}}\left( {\frac{{x + 1}}{2}} \right)} \right]_{ - 1}^1 = \frac{1}{2}\left[ {{{\tan }^{ - 1}}\left( {\frac{{1 + 1}}{2}} \right) - {{\tan }^{ - 1}}\left( {\frac{{ - 1 + 1}}{2}} \right)} \right] \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\left[ {{{\tan }^{ - 1}}\left( 1 \right) - {{\tan }^{ - 1}}\left( 0 \right)} \right] \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\left( {\frac{\pi }{4}} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{\pi }{8} \cr} $$
