Question on Indefinite Integrals (Calculus II) hope you are doing well. I am working on a problem solving the following indefinite integral :
$$\int 6x \sqrt{2x^3+8}\, dx$$
I have gotten up to the point where :
$$\int \sqrt{u}\frac{du}{x}$$
But am stuck as I am not sure where to place the $x$, not sure if I have done an incorrect method or am missing something. Thank you for your assistance in advance :)

What you did is fine. Let $u=2x^3+8$, $\frac{du}{dx}=6x^2$.
Hence $$\int 6x\sqrt{2x^3+8}\, dx = \int \frac{\sqrt{u}}{x}\, du$$
However, I do not expect this integral to have a nice form. Wolfram alpha expressed the solution in hypergeometric function.
