## Thomas' Calculus 13th Edition

$$\frac{1}{3}{\sec ^{ - 1}}\left| {3t + 1} \right| + C$$
\eqalign{ & \int {\frac{{dt}}{{\left( {3t + 1} \right)\sqrt {9{t^2} + 6t} }}} \cr & = \int {\frac{{dt}}{{\left( {3t + 1} \right)\sqrt {9\left( {{t^2} + \frac{2}{3}t} \right)} }}} \cr & = \int {\frac{{dt}}{{3\left( {3t + 1} \right)\sqrt {{t^2} + \frac{2}{3}t} }}} \cr & {\text{complete the square for }}{t^2} + \frac{2}{3}t \cr & {t^2} + \frac{2}{3}t = {t^2} + \frac{2}{3}t + \frac{1}{9} - \frac{1}{9} \cr & {t^2} + \frac{2}{3}t = {\left( {t + \frac{1}{3}} \right)^2} - {\left( {\frac{1}{3}} \right)^2} \cr & {\text{then}}{\text{,}} \cr & \int {\frac{{dt}}{{3\left( {3t + 1} \right)\sqrt {{t^2} + \frac{2}{3}t} }}} = \int {\frac{{dt}}{{9\left( {t + \frac{1}{3}} \right)\sqrt {{{\left( {t + \frac{1}{3}} \right)}^2} - {{\left( {\frac{1}{3}} \right)}^2}} }}} \cr & {\text{use the substitution method}} \cr & u = t + \frac{1}{3},\,\,\,\,\,\,du = dt \cr & \int {\frac{{dt}}{{9\left( {t + \frac{1}{3}} \right)\sqrt {{{\left( {t + \frac{1}{3}} \right)}^2} - {{\left( {\frac{1}{3}} \right)}^2}} }}} = \int {\frac{{dt}}{{9u\sqrt {{u^2} - {{\left( {\frac{1}{3}} \right)}^2}} }}} \cr & {\text{integrate by the formula }}\int {\frac{{du}}{{\left| u \right|\sqrt {{u^2} - {a^2}} }} = \frac{1}{a}{{\sec }^{ - 1}}\left| {\frac{u}{a}} \right| + C} \cr & = \frac{1}{9}\left( {\frac{1}{{1/3}}{{\sec }^{ - 1}}\left| {\frac{u}{{1/3}}} \right|} \right) + C \cr & = \frac{1}{3}{\sec ^{ - 1}}\left| {3u} \right| + C \cr & {\text{write in terms of }}t{\text{; replace }}u = t + \frac{1}{3} \cr & = \frac{1}{3}{\sec ^{ - 1}}\left| {3t + 1} \right| + C \cr}