## Thomas' Calculus 13th Edition

$$\frac{1}{4}\sqrt {4{t^2} - 1} + C$$
\eqalign{ & \int {\frac{{tdt}}{{\sqrt {4{t^2} - 1} }}} \cr & \cr & \left( {\bf{a}} \right){\text{using the substitution method}} \cr & \,\,\,\,{\text{Let }}u = 4{t^2} - 1,\,\,\,\,\,du = 8tdt,\,\,\,\,dt = \frac{{du}}{{8t}} \cr & \int {\frac{{tdt}}{{\sqrt {4{t^2} - 1} }}} = \int {\frac{t}{{\sqrt u }}\left( {\frac{{du}}{{8t}}} \right)} \cr & = \frac{1}{8}\int {\frac{1}{{\sqrt u }}du} \cr & = \frac{1}{8}\int {{u^{ - 1/2}}du} \cr & {\text{integrate}} \cr & = \frac{1}{8}\left( {\frac{{{u^{1/2}}}}{{1/2}}} \right) + C \cr & = \frac{1}{4}\sqrt u + C \cr & {\text{write in terms of }}t,{\text{ substitute }}4{t^2} - 1{\text{ for }}u \cr & = \frac{1}{4}\sqrt {4{t^2} - 1} + C \cr & \cr & \left( {\bf{b}} \right){\text{using a trigonometric substitution}} \cr & \,\,\,\,\,{\text{Let }}t = \frac{1}{2}\sec \theta ,\,\,\,dt = \frac{1}{2}\sec \theta \tan \theta d\theta \cr & \,\,\,\,\int {\frac{{tdt}}{{\sqrt {4{t^2} - 1} }}} = \int {\frac{{\left( {1/2} \right)\sec \theta }}{{\sqrt {4{{\left( {\frac{1}{2}\sec \theta } \right)}^2} - 1} }}} \left( {\frac{1}{2}\sec \theta \tan \theta d\theta } \right) \cr & = \frac{1}{4}\int {\frac{{{{\sec }^2}\theta \tan \theta }}{{\sqrt {4\left( {\frac{1}{4}{{\sec }^2}\theta } \right) - 1} }}} d\theta \cr & = \frac{1}{4}\int {\frac{{{{\sec }^2}\theta \tan \theta }}{{\sqrt {{{\sec }^2}\theta - 1} }}} d\theta \cr & = \frac{1}{4}\int {\frac{{{{\sec }^2}\theta \tan \theta }}{{\sqrt {{{\tan }^2}\theta } }}} d\theta \cr & = \frac{1}{4}\int {{{\sec }^2}\theta } d\theta \cr & {\text{Integrate}} \cr & = \frac{1}{4}\tan \theta + C \cr & {\text{where }}\tan \theta = \sqrt {4{t^2} - 1} \cr & = \frac{1}{4}\sqrt {4{t^2} - 1} + C \cr}