## Thomas' Calculus 13th Edition

$$\frac{1}{3}{\sec ^{ - 1}}\left| {\frac{{t + 1}}{3}} \right| + C$$
\eqalign{ & \int {\frac{{dt}}{{\left( {t + 1} \right)\sqrt {{t^2} + 2t - 8} }}} \cr & {\text{Complete the square for }}{t^2} + 2t - 8 \cr & {t^2} + 2t - 8 = {t^2} + 2t + 1 - 9 \cr & {t^2} + 2t - 8 = {\left( {t + 1} \right)^2} - 9 \cr & {\text{Then}}{\text{,}} \cr & = \int {\frac{{dt}}{{\left( {t + 1} \right)\sqrt {{{\left( {t + 1} \right)}^2} - 9} }}} \cr & {\text{Use the substitution method}} \cr & u = t + 1,\,\,\,\,\,\,du = dt \cr & \int {\frac{{dt}}{{\left( {t + 1} \right)\sqrt {{{\left( {t + 1} \right)}^2} - 9} }}} = \int {\frac{{du}}{{u\sqrt {{u^2} - 9} }}} \cr & {\text{integrating 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}{3}{\sec ^{ - 1}}\left| {\frac{u}{3}} \right| + C \cr & {\text{write in terms of }}t{\text{; replace }}u = t + 1 \cr & = \frac{1}{3}{\sec ^{ - 1}}\left| {\frac{{t + 1}}{3}} \right| + C \cr}