<s>[INST] Integrating $\;\int x^3\sqrt{x^2 + 2}\,dx$ Integrate the following: $$\int x^3\sqrt{x^2 + 2}\,dx$$
I understand how to do basic integration by parts but I don't know what to do with 
$\sqrt{x^2+2}$.  
Do I divide the $\sqrt{x^2+2}$ by 2 first to make  it becomes to $\sqrt{2}\sqrt{\frac{x^2}{2}+1}$ ?  If so, how do I keep going?
Thank you for any help!
 [/INST] If you let $u = x^2 + 2\implies du = 2x\,dx \iff \dfrac {du}{2} = x\,dx$,
And note that $x^3 = x\cdot x^2,$ where $x^2 = u-2$,
Then your integral is equivalent to $$\int x^3\sqrt{x^2 +2}\,dx = \int x^2 \sqrt{x^2 + 2}(x\,dx)$$ $$ = \frac 12 \int (u-2)\sqrt u \,du = \frac 12\int \left(u^{3/2} -2u^{1/2}\right)\,du$$
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