Where to place the operator? I believe it's standard to place the operator in between the conjugate of the wavefunction and the wavefunction itself. For instance,
$$\langle p\rangle = \int_{-\infty}^{\infty}\Psi * \frac{\hbar}{i}\frac{d}{dx}\Psi dx$$
Is it wrong to do this?
$$\langle p\rangle = \int_{-\infty}^{\infty}\frac{\hbar}{i}\frac{d}{dx}|\Psi|^2 dx$$
 A: The 2nd way 
$$\langle p\rangle = \int_{-\infty}^{\infty}\frac{\hbar}{i}\frac{d}{dx}|\Psi|^2 dx$$
will produce a complex result in general (in the example above it is will simply be zero), not having a physical measurement analog. The operator operates on some vector (either $\Psi$ or $\bar{\Psi}$), whereas the $|\Psi|^2$ is a simple real number.
A: 
So I believe it's standard to place the operator inbetween the conjugate of the wavefunction and the wavefunction itself. For instance,
$$\langle p\rangle = \int_{-\infty}^{\infty}\Psi * \frac{\hbar}{i}\frac{d}{dx}\Psi dx$$

Yes, that is correct, and

Is it wrong to do this?
$$\langle p\rangle = \int_{-\infty}^{\infty}\frac{\hbar}{i}\frac{d}{dx}|\Psi|^2 dx$$

yes, that is wrong.
One easy way to see why the latter must be wrong is that the integral of an exact derivative is always doable:
$$\int_{-\infty}^{\infty}\frac{d}{dx}|\Psi|^2 dx=\left.|\Psi(x)|^2\right|_{-\infty}^{\infty}=0,$$
for all $\Psi$, which can't be right, not least because you require $⟨p⟩$ to depend on $\Psi$ in some meaningful fashion.
