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