Answer
$$ - \frac{1}{{16}}\ln \left| {\frac{{x + 4}}{{\sqrt {{x^2} - 16} }}} \right| + \frac{1}{{4x}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{x^2}\sqrt {{x^2} - 16} }}} \cr
& {\text{substitute }}x = 4\sec \theta ,{\text{ }}dx = 4\sec \theta \tan \theta d\theta \cr
& = \int {\frac{{4\sec \theta \tan \theta d\theta }}{{{{\left( {4\sec \theta } \right)}^2}\sqrt {{{\left( {4\sec \theta } \right)}^2} - 16} }}} \cr
& = \int {\frac{{4\sec \theta \tan \theta d\theta }}{{{{\left( {4\sec \theta } \right)}^2}\sqrt {16{{\sec }^2}\theta - 16} }}} \cr
& {\text{simplifying}} \cr
& = \int {\frac{{d\theta }}{{4\sec \theta \sqrt {16{{\sec }^2}\theta - 16} }}} \cr
& = \int {\frac{{d\theta }}{{16\sec \theta \sqrt {{{\sec }^2}\theta - 1} }}} \cr
& = \int {\frac{{d\theta }}{{16\sec \theta \sqrt {{{\tan }^2}\theta } }}} \cr
& = \int {\frac{{d\theta }}{{16\sec \theta \tan \theta }}} \cr
& = \frac{1}{{16}}\int {\frac{{{{\cos }^2}\theta d\theta }}{{sin\theta }}} \cr
& = \frac{1}{{16}}\int {\left( {\frac{{1 - {{\sin }^2}\theta }}{{\sin \theta }}} \right)d\theta } \cr
& = \frac{1}{{16}}\int {\left( {\csc \theta - \sin \theta } \right)d\theta } \cr
& {\text{integrating}} \cr
& = \frac{1}{{16}}\left( { - \ln \left| {\csc \theta + \cot \theta } \right| + \cos \theta } \right) + C \cr
& = - \frac{1}{{16}}\ln \left| {\csc \theta + \cot \theta } \right| + \frac{1}{{16}}\cos \theta + C \cr
& {\text{write in terms of }}x \cr
& = - \frac{1}{{16}}\ln \left| {\frac{x}{{\sqrt {{x^2} - 16} }} + \frac{4}{{\sqrt {{x^2} - 16} }}} \right| + \frac{1}{{16}}\left( {\frac{4}{x}} \right) + C \cr
& {\text{simplify}} \cr
& = - \frac{1}{{16}}\ln \left| {\frac{{x + 4}}{{\sqrt {{x^2} - 16} }}} \right| + \frac{1}{{4x}} + C \cr} $$
