<s>[INST] Integral $\int \frac{x^2}{\sqrt[] {3-x^3} } \operatorname d \! x$ $$\int \frac{x^2}{\sqrt[] {3-x^3} } \operatorname d \! x$$
I have the above indefinite integral problem with substitution and I wanted to see if I could get a double check from the masses just in case I am missing anything crucial to making the final solution that much more explanatory!  Thank you for your help and I apologize for the chicken scratch again!
Here are the steps I have taken the time to complete and am requesting a double check/critique of:

$$\int \frac{x^2}{\sqrt{3-x^3}} \,dx =
\begin{vmatrix}
u=3-x^3\\
du=-3x^2 dx
\end{vmatrix} =
\int -\frac13 \int \frac1{\sqrt u} du \overset{(1)}=
-\frac{2\sqrt u}3+C \overset{(2)}=
-\frac{2\sqrt{3-x^3}}3+C$$
(1) $\int\frac1{\sqrt u}=2\sqrt u+C$
(2) susbstitute $u$ 
 [/INST] we have $$\int \frac{x^2}{\sqrt{3-x^3}}dx$$
let $$u = 3 - x^3$$
$$-\frac{du}{3} = x^2dx$$
$$-\frac{1}{3}\int \frac{1}{\sqrt{u}}du = -\frac{2}{3}\sqrt{u} +C = -\frac{2}{3}\sqrt{3 - x^3} +C$$
</s>