Find the indefinite integral using trig substitution. $,{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{1}{x^2\sqrt[\phantom{2}]{16-x^2}}dx$

a$)\frac{1}{16x(16-x^2{)}^{\frac{3}{2}}}+C$

b$)-\frac{1}{16x(16-x^2{)}^{\frac{3}{2}}}+C$

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

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

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