Heisenberg EOM for $\langle x \rangle$ in momentum eigenstate - where is my error? Equation of motion for expectation value of a quantum particle in a momentum eigenstate: 
$$\frac{d}{dt} \langle x \rangle = \frac{1}{i h} \langle [x,H] \rangle$$
and since it's in a momentum eigenstate, 
$$\frac{1}{i h} \langle [x,H] \rangle = \frac{1}{i h} \langle p \vert [x,H] \vert p \rangle$$
Expanding this, 
$$\frac{1}{i h} \langle p \vert (x \frac{p^2}{2m} + xV(x)) -  (\frac{p^2}{2m}x + V(x)x)\vert p \rangle = \frac{1}{i h} (\frac{p^2}{2m} - \frac{p^2}{2m}) \langle p \vert x \vert p \rangle +  \frac{1}{i h} \langle p \vert[x,V(x)] \vert p \rangle  = 0$$
since $V$ is $V(x)$.  But $\frac{d}{dt}\langle x \rangle = \frac{p}{m}$.  Where is my mistake?
 A: Well, you know that the eigenfunction of $\hat p$ is $\exp(ipx)$, so let's try to find what the expectation value of $x$ is, to begin with: $$\langle p | x | p\rangle = \int \exp(-ipx) x \exp(ipx) \, dx = \int x\,  dx$$
and this integral doesn't exist. Then it shouldn't come as a surprise that trying to take the time derivative gives nonsense. The underlying reason is that $\exp(ipx)$ isn't normalizable. This answer to a similar question gives more details on how to resolve this.
A: Update: Comments by Robin pointed out my confusion.
Consider the following:
$$ [x, p^2] = x p p - p p x = x p p - p x p + p x p - p p x = [x,p] p + p [x,p] = 2 i h p $$
If you plug this in the initial expression, you will get exactly what you would expect from this observable: $p/m$.
But formally correct manipulations that you performed provide a different answer, showing that something is not right. For the conclusion, please see Robin's answer.

Original mistaken answer:
You cannot take the momentum term outside the average - momentum p is an operator that does not commute with x.
