## Thomas' Calculus 13th Edition

$${\sec ^{ - 1}}\left( {\frac{{x - 1}}{7}} \right)$$
\eqalign{ & \int {\frac{{7dx}}{{\left( {x - 1} \right)\sqrt {{x^2} - 2x - 48} }}} \cr & {\text{complete the square for }}{x^2} - 2x - 48 \cr & {x^2} - 2x - 48 = {x^2} - 2x + 1 - 49 \cr & {x^2} - 2x - 48 = {\left( {x - 1} \right)^2} - 49 \cr & {\text{then}}{\text{,}} \cr & \int {\frac{{7dx}}{{\left( {x - 1} \right)\sqrt {{x^2} - 2x - 48} }}} = \int {\frac{{7dx}}{{\left( {x - 1} \right)\sqrt {{{\left( {x - 1} \right)}^2} - 49} }}} \cr & {\text{integrate by the substitution method:}} \cr & {\text{set }}u = {e^z}{\text{; then }}du = {e^z}dz \cr & {\text{write the integrand in terms of }}u \cr & \int {\frac{{7dx}}{{\left( {x - 1} \right)\sqrt {{{\left( {x - 1} \right)}^2} - 49} }}} = 7\int {\frac{{du}}{{u\sqrt {{u^2} - {7^2}} }}} \cr & {\text{integrate by using the formula }}\int {\frac{{du}}{{u\sqrt {{u^2} - {a^2}} }} = \frac{1}{a}{{\sec }^{ - 1}}\left( {\frac{u}{a}} \right) + C},.{\text{ }}a = 7 \cr & 7\int {\frac{{du}}{{u\sqrt {{u^2} - {7^2}} }}} = 7\left( {\frac{1}{7}{{\sec }^{ - 1}}\left( {\frac{u}{7}} \right)} \right) + C \cr & = {\sec ^{ - 1}}\left( {\frac{u}{7}} \right) \cr & {\text{replace }}x - 1{\text{ for }}u \cr & = {\sec ^{ - 1}}\left( {\frac{{x - 1}}{7}} \right) \cr}