Answer
$$ - \frac{{\sqrt {16 - {x^2}} }}{{16x}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{x^2}\sqrt {16 - {x^2}} }}} \cr
& {\text{Substitute }}x = 4\sin \theta ,{\text{ }}dx = 4\cos \theta d\theta \cr
& {\text{Use the trigonometric substitution}} \cr
& \int {\frac{{dx}}{{{x^2}\sqrt {16 - {x^2}} }}} = \int {\frac{{4\cos \theta d\theta }}{{{{\left( {4\sin \theta } \right)}^2}\sqrt {16 - {{\left( {4\sin \theta } \right)}^2}} }}} \cr
& {\text{Simplify}} \cr
& = \int {\frac{{4\cos \theta d\theta }}{{16{{\sin }^2}\theta \sqrt {16 - 16{{\sin }^2}\theta } }}} \cr
& = \int {\frac{{4\cos \theta d\theta }}{{16{{\sin }^2}\theta \left( 4 \right)\sqrt {1 - {{\sin }^2}\theta } }}} \cr
& = \int {\frac{{4\cos \theta d\theta }}{{16{{\sin }^2}\theta \left( 4 \right)\sqrt {{{\cos }^2}\theta } }}} \cr
& = \int {\frac{{d\theta }}{{16{{\sin }^2}\theta }}} \cr
& {\text{Use identity }}\csc \theta = \frac{1}{{\sin \theta }} \cr
& = \frac{1}{{16}}\int {{{\csc }^2}\theta } d\theta \cr
& {\text{Integrate}} \cr
& = - \frac{1}{{16}}\cot \theta + C \cr
& = - \frac{1}{{16}}\left( {\frac{{\cos \theta }}{{\sin \theta }}} \right) + C \cr
& {\text{Writing the answer in terms of }}x,{\text{ we have that }} \cr
& x = 4\sin \theta ,\,\,\,\sin \theta = \frac{x}{4},{\text{ and cos}}\theta = \frac{{\sqrt {16 - {x^2}} }}{4} \cr
& = - \frac{1}{{16}}\left( {\frac{{\left( {\sqrt {16 - {x^2}} } \right)/4}}{{x/4}}} \right) + C \cr
& = - \frac{{\sqrt {16 - {x^2}} }}{{16x}} + C \cr} $$
