Integral $\int \frac{x^2}{\sqrt[] {3-x^3} } \operatorname d \! x$

Question

$$\int \frac{x^2}{\sqrt[] {3-x^3} } \operatorname d \! x$$

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 $u$

Answer 1

All ok, but a small thing. When you wrote

$$ - \frac23 \sqrt u + C $$

you mentally imagined the first part, but did not put it down. Just say " integral =" or some such thing on the left hand side so those who may read sequentially cannot detect the omission.

Answer 2

we have $$\int \frac{x^2}{\sqrt{3-x^3}}dx$$ let $$u = 3 - x^3$$ $$-\frac{du}{3} = x^2dx$$ $$-\frac{1}{3}\int \frac{1}{\sqrt{u}}du = -\frac{2}{3}\sqrt{u} +C = -\frac{2}{3}\sqrt{3 - x^3} +C$$