What other substitutions could I use to evaluate this integral?

Consider the integral $$\int x^2\sqrt{2 + x} \, dx$$

I need to find the value of this integral, yet all its (seemingly) possible substitutions don't allow me to cancel appropriate terms. Here are three substitutions and their outcomes, all of which cover both terms in the integrand, as well as the composed function inside the root.

$\boxed{\text{Let }u = (2 + x)}$ $$u = 2 + x$$ $$du = 1 \space dx$$ $$dx = du$$ $$\int x^2 \sqrt{u} \space du$$

$\boxed{\text{Let } u = \sqrt{2+x}}$ $$u = \sqrt{2 + x}$$ \begin{align*} du &= \frac{1}{2}(2 + x)^{-\frac{1}{2}} \space dx \\ &= \frac{1}{2\sqrt{2 + x}} \space dx \end{align*} $$dx = 2\sqrt{2 + x} \space du$$ $$2\int ux^2\sqrt{2 + x} \space du$$

$\boxed{\text{Let }u = x^2}$ $$u = x^2$$ $$du = 2x \space dx$$ $$dx = \frac{1}{2x} \space du$$ $$\frac{1}{2} \int \frac{u\sqrt{2 + x}}{x} \space du$$

As you can see, none of these allow me to move forward into integrating with respect to $u$. What other substitutions can I use that would help me move on with this integral?

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@MichaelHardy: Are you sure? Perhaps, you should double check the answer you posted. –  Nana Jun 13 '12 at 1:21
Do you mind if I edit the headings to be not so large? It seems inordinate. Kudos for showing your thoughts by the way. –  Dylan Moreland Jun 13 '12 at 1:55
@DylanMoreland I want to be able to separate different parts so its easier to read, but I get what you're saying. And I appreciate the comments. I work on these questions for quite a while. –  Zolani13 Jun 13 '12 at 3:50
@Zolani13 Well, let me try something, which you should feel free to revert. –  Dylan Moreland Jun 13 '12 at 4:21

Your first substitution should work. That is let $u=2+x$, then $x= u-2$ and thus $$x^2 = (u-2)^2.$$ You should get $$\int \left(u^{5/2}-4u^{3/2}+4u^{1/2}\right)~du.$$

The second substitution also works. Let $u =\sqrt{2+x}$. So $u^2 = 2+x$. Then $x^2 = (u^2-2)^2$. Putting everything together gives you $$\int 2\left(u^6-4u^4+4u^2\right)~du.$$

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@MichaelHardy: Can you point out the error? –  Nana Jun 13 '12 at 0:48
@Nana My apologies! Everything is allright! –  Pedro Tamaroff Jun 13 '12 at 2:08
This seems to check out, to me. $dx = 2u\,du$. –  Dylan Moreland Jun 13 '12 at 2:09
@MichaelHardy You should be more careful when reading answers. –  Pedro Tamaroff Jun 13 '12 at 2:10
The question is (admirably) tagged homework, so perhaps that is why some details were left out. I certainly found it hard to follow (while looking for an error) without writing something down, but perhaps that's the point? –  Dylan Moreland Jun 13 '12 at 4:48

You could even do this problem by parts actually.

Let $u = x^2$ and $dv = \sqrt{x + 2}$. So $du = x$ and $v = \dfrac{2}{3}(x+2)^{3/2}$. So

$$\int x^2\sqrt{x + 2} dx = \dfrac{2}{3}x^2(x+2)^{3/2} - \dfrac{2}{3}\int x (x+2)^{3/2}$$

Now let $s = x$ and $dt = (x+2)^{3/2}$. So $ds = 1$ and $t = \dfrac{2}{5}(x+2)^{5/2}$ Then $$\int x (x+2)^{3/2} = \dfrac{2}{5}x(x+2)^{5/2} - \dfrac{2}{5} \int (x+2)^{5/2} = \dfrac{2}{5}x(x+2)^{5/2} - \dfrac{4}{35} (x+2)^{7/2}.$$

So putting it all together we get $$\int x^2\sqrt{x + 2} dx = \dfrac{2}{3}x^2(x+2)^{3/2} - \dfrac{4}{15} x(x+2)^{5/2} + \dfrac{8}{105} (x+2)^{7/2}.$$

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If $u=\sqrt{2+x}$ then $u^2 = 2+x$, so $2u\,du=dx$, and $x=u^2-2$. Then you have $$\int x^2\sqrt{2+x}\,dx = \int(u^2-2)^2 u\, 2u\,du$$ After expanding the polynomial and getting its antiderivative, put $\sqrt{2+x}$ in place of $u$ wherever it appears.

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This is wrong dude: $$u^2=x+2 \Rightarrow u^2-2=x \Rightarrow (u^2-2)^2=x^2$$ –  Pedro Tamaroff Jun 13 '12 at 2:09