A Derivation of Ehrenfest's Theorem in a particular case What are the missing lines in the integration?
$$\frac{\text d \langle {p} \rangle}{ \text{d} t} $$ 
$$= \frac{\text d}{\text d t} \int\limits_{-\infty}^{\infty} \Psi^* \left( \frac{\hbar}{i}\frac{\partial}{\partial x} \right) \Psi ~\text d x$$
$$= \frac{\hbar}{i}\int\frac{\partial }{\partial t} \left(\Psi^*\frac{\partial\Psi}{\partial x}\right)~\text d x $$
$$= \frac{\hbar}{i}\int \frac{\partial \Psi^*}{\partial t}\frac{\partial \Psi}{\partial x} +\Psi^*\frac{\partial }{\partial x}\frac{\partial \Psi }{ \partial t} ~\text {d} x$$
$$= \frac{\hbar}{i}\int\left( -\frac{i\hbar}{2m}\frac{\partial^2 \Psi^*}{\partial x^2}+\frac{i}{\hbar} V\Psi^*\right)\frac{\partial \Psi}{\partial x}+\Psi^* \frac{\partial}{\partial x} \left(\frac{i\hbar}{2m} \frac{\partial^2 \Psi }{\partial x^2} -\frac{i}{\hbar}V\Psi \right)~\text d x$$
$$=\int\left(V\Psi^*-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi^*}{\partial x^2}\right) \frac{\partial \Psi}{\partial x}+\Psi^*\frac{\partial}{\partial x}\left(\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2}-V\Psi\right)\text d x$$
$$=\left. \left(V\Psi^*-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi^*}{ \partial x^2} \right) \Psi \right|_{-\infty}^{\infty}-\int\left(\frac{\partial}{\partial x} (V\Psi^*)-\frac{\hbar^2}{2m}\frac{\partial^3 \Psi^*}{ \partial x^3}\right)\Psi  \text d x+ \left.\Psi^*\left(\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} - V\Psi\right)\right|_{-\infty}^\infty-\int\frac{\partial\Psi^*}{\partial x} \left( \frac{\hbar^2}{2m}\frac{\partial ^2 \Psi}{\partial x^2} - V \Psi \right)\text d x$$
$$=0 + \int\left(\frac{\hbar^2}{2m}\frac{\partial^3 \Psi^*}{ \partial x^3}-\frac{\partial}{\partial x} (V\Psi^*)\right) \Psi \text d x+0+\int\frac{\partial\Psi^*}{\partial x}\left(V\Psi - \frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2}\right) \text d x$$
$$=\int\frac{\hbar^2}{2m} \left(\frac{\partial^3\Psi^*}{\partial x^3}\Psi -\frac{\partial\Psi^*}{\partial x} \frac{\partial^2\Psi}{\partial x^2}\right)+\frac{\partial\Psi^*}{\partial x}(V\Psi )-\frac{\partial}{\partial x}(V\Psi^*)\Psi\text d x $$
$$\vdots$$
$$=\int \frac{\hbar^2}{2m}\left( \Psi^* \frac{\partial^3 \Psi}{\partial x^3} -\frac {\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x} \right)+\left( V \Psi^* \frac{\partial \Psi}{\partial x}-\Psi^* \frac{\partial}{\partial x} (V \Psi) \right)~\text d x$$
$$\vdots  $$
$$= \int \left( V \Psi^* \frac{\partial \Psi}{\partial x}- \Psi^* \frac{\partial V}{\partial x} \Psi-\Psi^*V \frac{\partial \Psi}{\partial x}  \right)~\text d x$$
$$= \int\limits_{-\infty}^{\infty} -\Psi^* \frac{\partial V}{\partial x} \Psi ~\text {d} x$$
$$=\left\langle - \frac{ \partial V }{\partial x} \right\rangle $$
 A: That's kind of horrible. But in the first three dots, the changed quantity is at least not wrong, as
$$\Psi^*_{,x}(V\Psi) - (V\Psi^*)_{,x}\Psi = V\Psi\Psi^*_{.x} - (\Psi^*V_{,x} + V\Psi^*_{,x})\Psi
 = -\Psi^* V_{,x} \Psi$$
is definitely the same as
$$V\Psi^*\Psi_{,x} - \Psi^*(V\Psi)_{,x} = \Psi^*V\Psi_{,x} - \Psi^*(V_{,x}\Psi + V\Psi_{,x}) = - \Psi^*V_{,x}\Psi.$$
The global conjugation flip over the entire integral is itself valid if that integral is real (since $\langle a|b \rangle = \langle b|a \rangle^*$), and we should expect it to be real on physical grounds, being $d\langle p\rangle/dt$. Though why would one want to do anything of the sort instead of directly proceeding to the goal, I really have no idea. In any case, this makes it clear that what's really missing is a proof of
$$\int\left(\frac{\partial^3\Psi^*}{\partial x^3}\Psi -\frac{\partial\Psi^*}{\partial x} \frac{\partial^2\Psi}{\partial x^2}\right)~\mathrm{d}x = 0 $$
or
$$\int \left( \Psi^* \frac{\partial^3 \Psi}{\partial x^3} -\frac {\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x} \right)~\mathrm{d}x = 0$$
These are also correct for nice enough boundary conditions, and can be shown using integration by parts twice on the first term.
Another method: by differentiation the definition of expectation value and substituting the Schrödinger equation for the time derivative of the state,
$$\begin{eqnarray*}
\frac{d\langle A\rangle}{dt} &=& \langle\dot{\psi}|A|\psi\rangle + \langle\psi|\dot{A}|\psi\rangle + \langle\psi|A|\dot{\psi}\rangle\\
&=& -\frac{1}{i\hbar}\langle\psi|H^\dagger A|\psi\rangle + \left\langle\frac{\partial A}{\partial t}\right\rangle + \frac{1}{i\hbar}\langle\psi|AH|\psi\rangle\\
&=& \left\langle\frac{\partial A}{\partial t}\right\rangle + \frac{1}{i\hbar}\left\langle[A,H]\right\rangle
\end{eqnarray*}$$
Since $[p,H] = [p,V]$ is obvious, it is then easy to verify that $[p,V] = -i\hbar(\nabla V)$, and $p$ is not explicitly dependent on time, so that term doesn't even appear in that case. This scheme can be done without explicit bra-ket notation, but even the position-space integral representation is simpler if one leaves the Hamiltonian as an abstract operator for a while.
A: $$
\newcommand \ps   [1] {\left(#1\right)}                               %  parentheses (auto)
\newcommand \bs   [1] {\left[#1\right]}                               %  brackets (auto)
\newcommand \expt [1] {\left\langle#1\right\rangle}                   %  expectation value
\newcommand \de   [2] {\frac{\mathrm d #1}{\mathrm d#2}}              %  first order derivative
\newcommand \den  [3] {\frac{\mathrm d^#3 #1}{\mathrm d#2^#3}}        %  n-th derivative
\newcommand \pa   [2] {\frac{\partial #1}{\partial #2}}               %  partial derivative
\newcommand \pan  [3] {\frac{\partial^#3 #1}{\partial#2^#3}}          %  n-th partial derivative
\newcommand \inti [2]{\int\limits_{-\infty}^\infty{#1}\,\mathrm d #2} %  integral -oo -> oo
$$
\begin{align*}
\de{\expt p}t 
    &= \de{}t\inti{\Psi^*\ps{\frac\hbar i\pa{}x}\Psi}x\\
    &= \frac\hbar i\inti{\pa{}t\ps{\Psi^*\pa\Psi x}}x\\
    &= \frac\hbar i\inti{\left(\pa{\Psi^*}t\pa\Psi x+\Psi^*\pa{}x\pa\Psi t\right)}x
%   &\hspace{3cm} \text{from the Schr\"odinger Equation}\\
  &i\hbar\pa\Psi t &= -\frac{\hbar^2}{2m}\pan\Psi x2+V\Psi\\
 && \pa\Psi t &= \frac{i\hbar}{2m}\pan\Psi x2-\frac i\hbar V\Psi\\
 && \pa{\Psi^*}t &= -\frac{i\hbar}{2m}\pan{\Psi^*}x2+\frac i\hbar V\Psi^*\\[-4ex]
    &= \frac\hbar i\inti{\ps{\bs{-\frac{i\hbar}{2m}\pan{\Psi^*}x2+\frac i\hbar V\Psi^*}\pa\Psi x
     +\Psi^*\pa{}x \bs{\frac{i\hbar}{2m}\pan{\Psi}x2-\frac i\hbar V\Psi}}}x\\
% &\hspace{3cm} \text{rearranging}\\
    &= \frac\hbar i\inti{\frac{i\hbar}{2m}\ps{\Psi^*\pan\Psi x3-\pan{\Psi^*}x2\pa\Psi x}           
      + \frac i\hbar\ps{V\Psi^*\pa\Psi x-\Psi^*\pa{}x(V\Psi)}}x\\
 &= \frac{\hbar^2}{2m}\inti{\pa{}x\ps{\Psi^*\pan\Psi x2-\pan{\Psi^*}x2\Psi}}x       
      + \inti{\ps{V\Psi^*\pa\Psi x-\Psi^*\pa Vx\Psi-\Psi^*V\pa\Psi x}}x\hspace{-4em}\\
 &= \frac{\hbar^2}{2m}\ps{\Psi^*\pan\Psi x2-\pan{\Psi^*}x2\Psi}\Big|_{-\infty}^\infty           
      - \inti{\Psi^*\pa Vx\Psi}x
    &\hspace{-5em} &\Psi(\pm\infty),\Psi^*(\pm\infty)=0\\[-2ex]
    &= 0-\expt{\pa Vx}\\
    &= \expt{-\pa Vx} 
\end{align*}
