Integrating $\int \frac{r^3}{\sqrt{16 + r^2}} dr$ I need to calculate: $$\int \frac{r^3}{\sqrt{16 + r^2}} dr$$
I am currently in the midst of learning AP Calculus BC and the lesson from which this problem came from goes over integration by parts.
Basically, it utilizes that : $$\frac{d}{dx} \left[ f(x)g(x) \right] = f'(x)g(x) + f(x)g'(x)$$ in order to get: $$\int f'(x)g(x) dx = f(x)g(x) - \int f(x)g'(x) dx.$$
Sometimes, we have to apply this process multiple times.
However, for this question, I cannot seem to figure out what to set $f(x)$ and $g(x)$ to. So far, I have tried setting $f(x) = x \rightarrow f'(x) = 1$ and $g(x) = \frac{x^3}{\sqrt{16 + x^2}}$.
This left me needing to calculate $$\int x \cdot \frac{2x^2 (x^2 + 24)}{ (16 + x^2)^{\frac{3}{2}}}$$ which I have no idea how to do.
For my second attempt, I tried letting $f(x) = \frac{1}{4} x^4 \rightarrow f'(x) = x^3$ and $g(x) = \frac{1}{\sqrt{16 + x^2}}.$
This led to me needing to calculate: $$\int \frac{1}{4} x^4 \cdot \bigg(- \frac{x}{(16 + x^2)^{\frac{3}{2}}} \bigg)$$ which I have also been unable to compute.
What should I set $f(x)$ and $g(x)$ to? Thank you in advance for any help.

The fastest way to solve this integral (IMO), does not use IBP.  We let
\begin{align*}
u &= \sqrt{16 + r^2} &&\implies \boxed{r^2 = u^2 - 16}\\
\frac{du}{dr} &= \frac{2r}{2\sqrt{16+r^2}}&&\implies\boxed{du = {\frac{r}{\sqrt{16+r^2}}}\,dr.}
\end{align*}
Making this substitution:
\begin{align*}\int \frac{r^{3}}{\sqrt{16+r^2}}\,dr &= \int r^2\,du\\
&= \int (u^2-16)\,du\\
&= \frac{u^{3}}{3} - 16u + C\\
&= \boxed{\frac{(16+r^2)^{3/2}}{3} - 16\sqrt{16+r^2} + C.}
\end{align*}
