How do we calculate the primitive of $r^{3}\sqrt{r^{2} + 2}$ without guessing?

Question

I encountered the following problem:

$\int r^3 \sqrt{r^2+2}\;dr$

I found the answer to be:

$1/5(r^2+2)^{5/2}−2/3(r^2+2)^{3/2}$

Although this is the correct answer, I am not entirely sure how this works, more specifically on how you find line of equation such as in this case:

$r^3=r^2r=[(r^2+2)−2]r=(r^2+2)r−2r$

Can anyone explain how you find this value other than trial and error through testing everything you can think of?

Answer 1

To follow from your line of thinking, you may consider using u-substitution. If we let $u=r^2 + 2$, then $du = 2r \ dr$. Note that $$u = r^2 + 2 \iff u-2=r^2$$ Therefore, \begin{align*} \int r^3 \sqrt{r^2+2}\ dr &= \int \underbrace{r^2}_{=u-2} \underbrace{\sqrt{r^2+2}}_{=\sqrt{u}} \ \underbrace{r \ dr}_{=du/2} \\ &= \int (u-2) \sqrt{u} \ \frac{1}{2}du \\ &= \frac{1}{2}\int \left(u^{3/2} - 2u^{1/2}\right) du \\ &= \frac{1}{5} u^{5/2} - \frac{2}{3} u^{3/2} + C \\ &= \frac{1}{5} (r^2+2)^{5/2} - \frac{2}{3} (r^2 + 2)^{3/2} +C \end{align*}

This approach avoids "guessing" some way to rewrite $r^3$.

Answer 2

Integration by parts gives:

\begin{align*} \int r^{3}\sqrt{r^{2} + 2}\mathrm{d}r & = \int r^{2}\left(r\sqrt{r^{2} + 2}\right)\mathrm{d}r\\\\ & = \frac{1}{3}\int r^{2}\left[(r^{2} + 2)^{3/2}\right]'\mathrm{d}r\\\\ & = \frac{1}{3}r^{2}(r^{2} + 2)^{3/2} - \frac{2}{3}\int r(r^{2} + 2)^{3/2}\mathrm{d}r\\\\ & = \frac{1}{3}(r^{2} + 2)^{5/2} - \frac{2}{3}(r^{2} + 2)^{3/2} - \frac{2}{3}\int r(r^{2} + 2)^{3/2}\mathrm{d}r\\\\ & = \left(\frac{1}{3} - \frac{2}{15}\right)(r^{2} + 2)^{5/2} - \frac{2}{3}(r^{2} + 2)^{3/2}\\\\ & = \frac{1}{5}(r^{2} + 2)^{5/2} - \frac{2}{3}(r^{2} + 2)^{3/2} + C \end{align*}

EDIT

To be more precise, notice that \begin{align*} \int r\sqrt{r^{2} + 2}\mathrm{d}r & = \frac{1}{2}\int 2r\sqrt{r^{2} + 2}\mathrm{d}r\\\\ & = \frac{1}{2}\int\sqrt{r^{2} + 2}\,\mathrm{d}(r^{2} + 2)\\\\ & = \frac{1}{2}\int \sqrt{u}\,\mathrm{d}u\\\\ & = \frac{1}{3}u^{3/2} + C\\\\ & = \frac{1}{3}(r^{2} + 2)^{3/2} + C \end{align*}

Answer 3

By Substitution

Let $x=\sqrt{r^2+2}$, then $x^2=r^2+2$ and $xdx=rdr$ transforms the integral into $$ \begin{aligned} I & =\int\left(x^2-2\right) x \cdot x d x \\ & =\frac{x^5}{5}-\frac{2 x^3}{3}+c\\&=\frac{\left(r^2+2\right)^{\frac{5}{2}}}{5}-\frac{2\left(r^2+2\right)^{\frac{3}{2}}}{3}+C\\&= \frac{1}{15}\left(r^2+2\right)^{\frac{3}{2}}\left(3 r^2-4\right)+C \end{aligned} $$