## Calculus: Early Transcendentals (2nd Edition)

$$\ln \left[ {\sec \left( {{{\tan }^{ - 1}}\left( {\frac{1}{4}} \right)} \right) + \frac{1}{4}} \right]$$
\eqalign{ & \int_0^1 {\frac{{dx}}{{\sqrt {{x^2} + 16} }}} \cr & x = 4\tan \theta ,{\text{ }}dx = 4{\sec ^2}\theta d\theta \cr & {\text{New limits of integration}} \cr & \theta = {\tan ^{ - 1}}\left( {x/4} \right) \cr & x = 0,{\text{ }}\theta = 0 \cr & x = 1,{\text{ }}\theta = {\tan ^{ - 1}}\left( {1/4} \right) \cr & {\text{we have}} \cr & \int_0^1 {\frac{{dx}}{{\sqrt {{x^2} + 16} }}} = \int_0^{{{\tan }^{ - 1}}\left( {1/4} \right)} {\frac{{4{{\sec }^2}\theta d\theta }}{{\sqrt {{{\left( {4\tan \theta } \right)}^2} + 16} }}} \cr & = \int_0^{{{\tan }^{ - 1}}\left( {1/4} \right)} {\frac{{{{\sec }^2}\theta d\theta }}{{\sqrt {{{\left( {4\tan \theta } \right)}^2} + 16} }}} = \int_0^{{{\tan }^{ - 1}}\left( {1/4} \right)} {\frac{{4{{\sec }^2}\theta d\theta }}{{4\sqrt {{{\tan }^2}\theta + 1} }}} \cr & {\text{pythagorean identity}} \cr & = \int_0^{{{\tan }^{ - 1}}\left( {1/4} \right)} {\frac{{4{{\sec }^2}\theta d\theta }}{{4\sqrt {{{\sec }^2}\theta } }} = } \int_0^{{{\tan }^{ - 1}}\left( {1/4} \right)} {\frac{{4{{\sec }^2}\theta d\theta }}{{4\sec \theta }} = } \int_0^{{{\tan }^{ - 1}}\left( {1/4} \right)} {\sec \theta d\theta } \cr & {\text{integrating}} \cr & = \left[ {\ln \left| {\sec \theta + \tan \theta } \right|} \right]_0^{{{\tan }^{ - 1}}\left( {1/4} \right)} \cr & {\text{evaluate limits}} \cr & = \left[ {\ln \left| {\sec \left( {{{\tan }^{ - 1}}\left( {1/4} \right)} \right) + \tan \left( {{{\tan }^{ - 1}}\left( {1/4} \right)} \right)} \right|} \right] - \left[ {\ln \left| {\sec \left( 0 \right) + \tan \left( 0 \right)} \right|} \right] \cr & {\text{simplify}} \cr & = \left[ {\ln \left| {\sec \left( {{{\tan }^{ - 1}}\left( {1/4} \right)} \right) + \frac{1}{4}} \right|} \right] - \left[ {\ln \left| 1 \right|} \right] \cr & = \ln \left[ {\sec \left( {{{\tan }^{ - 1}}\left( {\frac{1}{4}} \right)} \right) + \frac{1}{4}} \right] \cr}