## Calculus: Early Transcendentals (2nd Edition)

$$\frac{1}{2}\ln \left| {\frac{{x + \sqrt {{x^2} + 4} }}{2}} \right| + C$$
\eqalign{ & \int {\frac{{dx}}{{\sqrt {16 + 4{x^2}} }}} \cr & {\text{substitute }}x = 2\tan \theta ,{\text{ }}dx = 2{\sec ^2}\theta d\theta \cr & = \int {\frac{{dx}}{{\sqrt {16 + 4{x^2}} }}} = \int {\frac{{2{{\sec }^2}\theta d\theta }}{{\sqrt {16 + 4{{\left( {2\tan \theta } \right)}^2}} }}} = \int {\frac{{2{{\sec }^2}\theta d\theta }}{{\sqrt {16 + 16{{\tan }^2}\theta } }}} \cr & = \int {\frac{{2{{\sec }^2}\theta d\theta }}{{4\sqrt {1 + {{\tan }^2}\theta } }}} = \int {\frac{{2{{\sec }^2}\theta d\theta }}{{4\sec \theta }}} \cr & = \frac{1}{2}\int {\sec \theta } d\theta \cr & {\text{integrating}} \cr & = \frac{1}{2}\ln \left| {\sec \theta + \tan \theta } \right| + C \cr & {\text{with sec}}\theta = \frac{{\sqrt {4{x^2} + 16} }}{4}{\text{ and tan}}\theta = \frac{{2x}}{4} \cr & = \frac{1}{2}\ln \left| {\frac{{\sqrt {4{x^2} + 16} }}{4} + \frac{{2x}}{4}} \right| + C \cr & {\text{simplify}} \cr & = \frac{1}{2}\ln \left| {\frac{{\sqrt {{x^2} + 4} }}{2} + \frac{x}{2}} \right| + C \cr & = \frac{1}{2}\ln \left| {\frac{{x + \sqrt {{x^2} + 4} }}{2}} \right| + C \cr}