Evaluate the integral using the given substitution.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{12s^{3}ds}{\sqrt[\phantom{2}]{2-s^4}}$;$u=2*s^4$

$-6\sqrt[\phantom{2}]{2-s^4}C$

$\frac{6s^4}{\sqrt[\phantom{2}]{2-s^4}}$

$\frac{-3}{2\sqrt[\phantom{2}]{2-n^4}}C$

$-6s^3\sqrt[\phantom{2}]{2-s^4}C$