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.