According to Arfken et al. Mathematical Methods p.277 $$[x,p^2]=xp^2 - pxp + pxp -p^2x =[x,p]p + p[x,p]= 2ip \, .$$ According to the text this follows solely from $[x,p]=i$.
I'm not understanding how the squared operator is being applied.
How is the commutator on squared operators supposed to work to get the middle terms $pxp$? Why would it not be $[x,p]p - p[x,p]$.
You can either think of the middle term as "adding zero" to an equality, or you can use the commutator relationship to replace $xp^2$ and $p^2 x$ with
$$xp^2 = (xp)p = (px + [x,p])p$$
and
$$p^2 x = p(px)= p(xp - [x,p]).$$
Using commutivity and substituting these into the original expression, you'll see the $pxp$ terms cancel:
$$xp^2 - p^2x = pxp + [x,p]p - pxp + p[x,p] = [x,p]p + p[x,p].$$
Since $[x,p]$ is a scalar, namely $i$, you can move $p$ to either side freely, and
$$[x,p]p + [x,p]p = 2[x,p]p = 2ip.$$
So you see, all we needed to prove this was the commutator $[x,p]$, and that $p$ commutes with it. You can follow a structurally identical line of reasoning to show by induction that
$$[x,p^n] = inp^{n-1}.$$
I think you're missing an $\hbar$, but it is due to the useful formula,
$$ [x, p^{n}] = i \hbar \frac{\partial}{\partial p} (p^{n})$$
for which in your case, $n = 2$.
This formula I provided above is a special case of the more general formula, (in one dimension)
$$ [x, F(p)] = i \hbar \frac{\partial}{\partial p} (F)$$
