Use the given formula ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{du}{u^2\sqrt[\phantom{2}]{u^2-a^2}}=\frac{\sqrt[\phantom{2}]{u^2-a^2}}{a^{2}u}C$ to evaluate the integral ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{1}{y^2\sqrt[\phantom{2}]{9y^2-16}}dy$

A$.\frac{3\sqrt[\phantom{2}]{9y^2-16}}{16y}C$

B$.\frac{\sqrt[\phantom{2}]{9y^2-16}}{4y}C$

C$.$ None of these

D$.\frac{\sqrt[\phantom{2}]{9y^2-16}}{16y}C$

E$.\frac{\sqrt[\phantom{2}]{9y^2-16}}{48y}C$