Answer
$$\frac{1}{{2000}}{\sec ^{ - 1}}\left( {\frac{x}{{10}}} \right) - \frac{{\sqrt {{x^2} - 100} }}{{200{x^2}}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{x^3}\sqrt {{x^2} - 100} }}} \cr
& {\text{substitute }}x = 10\sec \theta ,{\text{ }}dx = 10\sec \theta \tan \theta d\theta \cr
& \int {\frac{{dx}}{{{x^3}\sqrt {{x^2} - 100} }}} = \int {\frac{{10\sec \theta \tan \theta d\theta }}{{{{\left( {10\sec \theta } \right)}^3}\sqrt {{{\left( {10\sec \theta } \right)}^2} - 100} }}} \cr
& {\text{simplify}} \cr
& = \int {\frac{{10\sec \theta \tan \theta d\theta }}{{1000{{\sec }^3}\theta \left( {10} \right)\sqrt {{{\sec }^2}\theta - 1} }}} \cr
& {\text{pythagorean identity}} \cr
& = \int {\frac{{10\sec \theta \tan \theta d\theta }}{{1000{{\sec }^3}\theta \left( {10} \right)\sqrt {{{\tan }^2}\theta } }}} \cr
& {\text{simplify}} \cr
& = \frac{1}{{1000}}\int {\frac{{d\theta }}{{{{\sec }^2}\theta }}} \cr
& = \frac{1}{{1000}}\int {{{\cos }^2}\theta } d\theta \cr
& {\text{half - angle formula}} \cr
& = \frac{1}{{1000}}\int {\left( {\frac{{1 + \cos 2\theta }}{2}} \right)} d\theta \cr
& {\text{integrating}} \cr
& = \frac{1}{{1000}}\left( {\frac{\theta }{2} - \frac{1}{4}\sin 2\theta } \right) + C \cr
& = \frac{1}{{1000}}\left( {\frac{\theta }{2} - \frac{1}{2}\sin \theta \cos \theta } \right) + C \cr
& = \frac{1}{{2000}}\theta - \frac{1}{{2000}}\sin \theta \cos \theta + C \cr
& = \frac{1}{{2000}}{\sec ^{ - 1}}\left( {\frac{x}{{10}}} \right) - \frac{1}{{2000}}\left( {\frac{{\sqrt {{x^2} - 100} }}{x}} \right)\left( {\frac{{10}}{x}} \right) + C \cr
& {\text{simplify}} \cr
& = \frac{1}{{2000}}{\sec ^{ - 1}}\left( {\frac{x}{{10}}} \right) - \frac{{\sqrt {{x^2} - 100} }}{{200{x^2}}} + C \cr} $$
