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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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up vote 5 down vote accepted

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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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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