Answer
$$\frac{1}{5}{\sec ^{ - 1}}\left| {\frac{x}{5}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{x\sqrt {{x^2} - 25} }}dx} \cr
& {\text{from the table 4}}{\text{.10 }}\int {\frac{{dx}}{{x\sqrt {{x^2} - {a^2}} }}} = \frac{1}{a}{\sec ^{ - 1}}\left| {\frac{x}{a}} \right| + C \cr
& {\text{letting }}a = 5,{\text{ }} \cr
& = \frac{1}{5}{\sec ^{ - 1}}\left| {\frac{x}{5}} \right| + C \cr
& {\text{check by differentiation}} \cr
& {\text{ = }}\frac{d}{{dx}}\left( {\frac{1}{5}{{\sec }^{ - 1}}\left| {\frac{x}{5}} \right| + C} \right) \cr
& {\text{ = }}\frac{1}{5}\frac{d}{{dx}}\left( {{{\sec }^{ - 1}}\left| {\frac{x}{5}} \right|} \right) + \frac{d}{{dx}}\left( C \right) \cr
& {\text{ = }}\frac{1}{5}\left( {\frac{5}{{x\sqrt {{x^2} - {{\left( 5 \right)}^2}} }}} \right) \cr
& {\text{ = }}\frac{1}{{x\sqrt {{x^2} - 25} }} \cr} $$
