Answer
$$\frac{1}{{16}}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + \frac{x}{{8\left( {{x^2} + 4} \right)}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{{\left( {4 + {x^2}} \right)}^2}}}} \cr
& {\text{substitute }}x = 2\tan \theta ,{\text{ }}dx = 2{\sec ^2}\theta d\theta \cr
& = \int {\frac{{dx}}{{{{\left( {4 + {x^2}} \right)}^2}}}} = \int {\frac{{2{{\sec }^2}\theta d\theta }}{{{{\left( {4 + 4{{\tan }^2}\theta } \right)}^2}}}} \cr
& = \int {\frac{{2{{\sec }^2}\theta d\theta }}{{16{{\left( {1 + {{\tan }^2}\theta } \right)}^2}}}} \cr
& {\text{pythagorean identity}} \cr
& = \int {\frac{{2{{\sec }^2}\theta d\theta }}{{16{{\left( {{{\sec }^2}\theta } \right)}^2}}}} \cr
& = \int {\frac{{2{{\sec }^2}\theta d\theta }}{{16{{\sec }^4}\theta }}} \cr
& = \frac{1}{8}\int {\frac{{d\theta }}{{{{\sec }^2}\theta }}} \cr
& = \frac{1}{8}\int {{{\cos }^2}\theta } d\theta \cr
& {\text{identity co}}{{\text{s}}^2}\theta = \frac{{1 + 2\cos \theta }}{2} \cr
& = \frac{1}{8}\int {\frac{{1 + 2\cos \theta }}{2}} d\theta \cr
& = \frac{1}{{16}}\int {\left( {1 + 2\cos \theta } \right)} d\theta \cr
& {\text{integrating}} \cr
& = \frac{1}{{16}}\left( {\theta + 2\sin \theta } \right) + C \cr
& {\text{write in terms of }}x \cr
& = \frac{1}{{16}}\left( {{{\tan }^{ - 1}}\left( {\frac{x}{2}} \right) + 2\left( {\frac{x}{{{x^2} + 4}}} \right)} \right) + C \cr
& {\text{simplify}} \cr
& = \frac{1}{{16}}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + \frac{x}{{8\left( {{x^2} + 4} \right)}} + C \cr} $$
