Answer
$$\frac{1}{3}{\tan ^{ - 1}}\left( {\frac{4}{3}} \right)$$
Work Step by Step
$$\eqalign{
& \int_{ - 2}^2 {\frac{{dx}}{{{x^2} + 4x + 13}}} \cr
& {\text{Completing the square}} \cr
& \int_{ - 2}^2 {\frac{{dx}}{{{x^2} + 4x + 13}}} = \int_{ - 2}^2 {\frac{{dx}}{{{x^2} + 4x + 4 + 9}}} \cr
& = \int_{ - 2}^2 {\frac{{dx}}{{{{\left( {x + 2} \right)}^2} + 9}}} \cr
& {\text{Let }}u = x + 2,{\text{ }}du = dx,{\text{ }} \cr
& {\text{The new limits of integration are}} \cr
& x = 2,{\text{ }}u = 2 + 2 = 4 \cr
& x = - 2,{\text{ }}u = - 2 + 2 = 0 \cr
& {\text{Substituting}} \cr
& \int_{ - 2}^2 {\frac{{dx}}{{{{\left( {x + 2} \right)}^2} + 9}}} = \int_0^4 {\frac{{du}}{{{u^2} + 9}}} \cr
& {\text{Integrate using basic integration rules}} \cr
& \int_0^4 {\frac{{du}}{{{u^2} + 9}}} = \frac{1}{3}\left[ {{{\tan }^{ - 1}}\left( {\frac{u}{3}} \right)} \right]_0^4 \cr
& = \frac{1}{3}\left[ {{{\tan }^{ - 1}}\left( {\frac{4}{3}} \right) - {{\tan }^{ - 1}}\left( {\frac{0}{3}} \right)} \right] \cr
& = \frac{1}{3}\left[ {{{\tan }^{ - 1}}\left( {\frac{4}{3}} \right) - {{\tan }^{ - 1}}\left( 0 \right)} \right] \cr
& = \frac{1}{3}{\tan ^{ - 1}}\left( {\frac{4}{3}} \right) \cr} $$
