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

Question

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!

Answer 1

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 $$\int x^3\sqrt{x^2 +2}\,dx = \int x^2 \sqrt{x^2 + 2}(x\,dx)$$ $$ = \frac 12 \int (u-2)\sqrt u \,du = \frac 12\int \left(u^{3/2} -2u^{1/2}\right)\,du$$

Answer 2

Hint: integrate by parts using

$$u=x^2$$

and

$$dv=x\sqrt{x^2+2}\,dx$$

instead.

Answer 3

Hint: $$\int x^3 \sqrt{x^2 + 1} \ \mathrm{d}x = \frac{1}{2}\int x^2 (2x) \sqrt{x^2 + 1} \ \mathrm{d}x$$ and $u = x^2 + 1$, along with $\mathrm{d}u = 2x\ \mathrm{d}x$ gives: $$\int x^3 \sqrt{x^2 + 1} \ \mathrm{d}x = \frac{1}{2} \int (u - 1) \sqrt{u} \ \mathrm{d}u.$$