## Calculus: Early Transcendentals (2nd Edition)

$$\frac{1}{6}{\tan ^{ - 1}}\left( {\frac{{u - 3}}{3}} \right) + C$$
\eqalign{ & \int {\frac{{du}}{{2{u^2} - 12u + 36}}} \cr & {\text{factor}} \cr & = \int {\frac{{du}}{{2\left( {{u^2} - 6u + 18} \right)}}} \cr & {\text{completing the square}} \cr & = \int {\frac{{du}}{{2\left( {{u^2} - 6u + 9} \right) + 18}}} = \int {\frac{{du}}{{2{{\left( {u - 3} \right)}^2} + 18}}} \cr & = \frac{1}{2}\int {\frac{{du}}{{{{\left( {u - 3} \right)}^2} + 9}}} \cr & z = u - 3,{\text{ }}dz = du \cr & = \frac{1}{2}\int {\frac{{dz}}{{{z^2} + 9}}} \cr & {\text{substitute }}z = 3\tan \theta ,{\text{ }}dz = 3{\sec ^2}\theta d\theta \cr & \frac{1}{2}\int {\frac{{dz}}{{{z^2} + 9}}} = \frac{1}{2}\int {\frac{{3{{\sec }^2}\theta d\theta }}{{{{\left( {3\tan \theta } \right)}^2} + 9}}} \cr & = \frac{1}{2}\int {\frac{{3{{\sec }^2}\theta d\theta }}{{9{{\tan }^2}\theta + 9}}} = \frac{3}{{18}}\int {\frac{{{{\sec }^2}\theta d\theta }}{{{{\tan }^2}\theta + 1}}} \cr & {\text{pythagorean identity}} \cr & = \frac{1}{6}\int {\frac{{{{\sec }^2}\theta d\theta }}{{{{\sec }^2}\theta }}} \cr & {\text{integrating}} \cr & = \frac{1}{6}\theta + C \cr & z = 3\tan \theta \cr & = \frac{1}{6}{\tan ^{ - 1}}\left( {\frac{u}{3}} \right) + C \cr & z = u - 3 \cr & = \frac{1}{6}{\tan ^{ - 1}}\left( {\frac{{u - 3}}{3}} \right) + C \cr}