Evaluate the integral using the given substitution

Evaluate the integral using the given substitution?
the integral of
12s^3*ds / √(6-s^4) u= 6-s^4

∫12s³ds/√(6-s⁴)
We want to get everything in terms of u:

Since u = 6-s⁴, we can differentiate both sides with respect to u:
1 = -4s³ * ds/du
du = -4s³ds
-3du = 12s³ds

Now we can substitute in -3du for 12s³ds and u for 6-s⁴:
∫12s³ds/√(6-s⁴)
= ∫-3du/√u
= ∫-3u^(-1/2)du
= -6u^(1/2)
= -6√u + C

Now we substitute back in for s (since we started out with s). Since u = 6-s⁴,
-6√u + C
= -6√(6-s⁴) + C

∫ 12s^3 ds/√(6 - s^4)

Let u = 6 - s^4
then du = -4s^3 ds
so (-1/4) du = s^3 ds
and -3 du = 12s^3 ds

The integral becomes -3 ∫ du/√u = -3 ∫ u^(-1/2) du = -3 u^(1/2)/(1/2) = -6 u^(1/2) = -6 √(6 - s^4) + C