Evaluate the integral using the indicated trigonometric substitution

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{1}{x^2\sqrt[\phantom{2}]{x^2-16}}dx$;$,x=4sec$

Select the correct answer.

$\frac{1}{4}\frac{x^2-4}{x^2}+C$

$\frac{1}{9}\frac{x^2-16}{x}+C$

$\frac{1}{16}\frac{\sqrt[\phantom{2}]{x^2-16}}{x^2}+C$

$\frac{1}{16}\frac{\sqrt[\phantom{2}]{x^2-16}}{x}+C$