Evaluate The Indefinite Integral With U-Sub

Question

I can't get beyond the u-sub in this problem, here's what i've tried:

$\int{x^2\sqrt{2+x}}dx$

Let $u=\sqrt{2+x}$

Let $z = 2 + x$

$du = (z)'(\sqrt{z})'dx$ <- Chain Rule

$du = \frac{1}{2\sqrt{2+x}}dx$

How do I solve from here?

Answer 1

Let $u = \sqrt{2+x}$. Then notice that $u^2 = 2+x$. Differentiating this (implicitly via chain rule) yields $2u \, du = dx$. Furthermore, observe that $x = u^2 - 2$. Hence, substituting all of this yields: $$ \int (x)^2(\sqrt{2+x}) \, (dx) = \int (u^2 - 2)^2(u)(2u \, du) = 2\int u^2(u^2 - 2)^2 \, du $$ From there, you can expand and integrate with some tedious but simple arithmetic.

Answer 2

Hint: Just expand $(u^2-2)^2$ and multiply through by $u^2$, then take the integral. All the terms should be elementary integrands.