$\int \frac{x^2}{\sqrt{x^3+2}}\, dx$

Question

Find $$\int \frac{x^2}{\sqrt{x^3+2}}\, dx$$

Answer 1

Let

\begin{align*} u &= \sqrt{x^3 + 2}\\ du &= \frac{3x^{2}}{2\sqrt{x^{3} + 2}}\,dx \end{align*}

Then the integral reduces to $$\frac{2}{3} \int du = \frac{2}{3}u + C = \frac{2}{3}\sqrt{x^{3} +2}+C.$$

Answer 2

Alternatively, let $u=x^3+2$, so that $du=3x^2\,dx$, and then $$\int \frac{x^2}{\sqrt{x^3+2}}\,dx=\int\frac{1}{3u^{1/2}}\,du=\frac{2}{3}\sqrt{u}+C=\frac{2}{3}\sqrt{x^3+2}+C,$$ so a $u$-substitution of course works...