General property of an eigenfunction of a linear ODE Consider the second order ODE (actually it is time-independent Schrödinger equation)
$$\frac{d^2\phi}{dx^2} + V(x)\phi = E\phi$$
We know that $\phi$ is a bound solution. This means the following two things:
(i)$$\lim_{x\to \pm\infty}\phi(x)=0 .$$
(ii) $0<E<\max[V(x)]$.
Using this information, can we prove
$$\int_{-\infty}^{\infty}\vert\phi\vert^2\frac{dV}{dx} dx=0\,?$$ 
You are free to make any reasonable assumptions about the nature of $V(x)$ and $\phi(x)$ along your way.
 A: If you assume that the potential $V(x)$ is bounded, then we can prove it by integration by parts as suggested by anon:
$$
\int^\infty_{-\infty}\phi^2\frac{dV}{dx}{dx}=\int^\infty_{-\infty}\phi^2dV
=\phi^2V\,\big|^\infty_{-\infty}-2\int^\infty_{-\infty}V\phi\frac{d\phi}{dx}dx
$$
where the last equality follows from integration by parts. Since $\lim_{x\to \pm\infty}\phi(x)=0$ by your assumption and $V(x)$ is bounded by my assumption, $\phi^2V\,\big|^\infty_{-\infty}=0$. Now using the equation $\frac{d^2\phi}{dx^2} + V(x)\phi = E\phi$, we can rewrite the above equation as
$$
\int^\infty_{-\infty}\phi^2\frac{dV}{dx}{dx}=-2\int^\infty_{-\infty}V\phi\frac{d\phi}{dx}dx=2\int^\infty_{-\infty}\frac{d^2\phi}{dx^2}\frac{d\phi}{dx}dx-2E\int^\infty_{-\infty}\frac{d\phi}{dx}dx:=I+II.
$$
Note that 
$$I=2\int^\infty_{-\infty}\frac{d^2\phi}{dx^2}\frac{d\phi}{dx}dx=2\int^\infty_{-\infty}\frac{d\phi}{dx}d\big(\frac{d\phi}{dx}\big)=\big(\frac{d\phi}{dx}\big)^2\Big|^\infty_{-\infty}=0$$
since $\lim_{x\to \pm\infty}\phi(x)=0$ implies that $\lim_{x\to \pm\infty}\frac{d\phi}{dx}(x)=0$. Note also that 
$$II=-2E\int^\infty_{-\infty}\frac{d\phi}{dx}dx=-2E\phi\,\big|^\infty_{-\infty}=0.$$
Now the result follows. 
Note added: To see that $\lim_{x\to \pm\infty}\frac{d\phi}{dx}(x)=0$, first note that $\frac{d\phi}{dx}$ is bounded. To see this, since $\frac{d^2\phi}{dx^2} + V(x)\phi = E\phi$, $\lim_{x\to \pm\infty}\phi(x)=0$ and $V(x)$ is bounded, we have $\lim_{x\to \pm\infty}\frac{d^2\phi}{dx^2}=0$. This implies that $\frac{d\phi}{dx}$ is bounded. 
Next, if we multiply $\phi$ to $\frac{d^2\phi}{dx^2} + V(x)\phi = E\phi$, we get
$$\phi\frac{d^2\phi}{dx^2}=(E-V(x))\phi^2$$
Now integrate it from $M$ to $\infty$ and we get
$$\int_M^\infty\phi\frac{d^2\phi}{dx^2}dx=\int_M^\infty(E-V(x))\phi^2dx.$$
The right hand side is zero when $M$ tends to infinity since $V(x)$ is bounded and the eigenfunction $\phi$ is $L^2$. On the other hand, by integration by parts, the left hand side is 
$$\int_M^\infty\phi\frac{d^2\phi}{dx^2}dx=(\phi\frac{d\phi}{dx})\Big|^\infty_M-\int_M^\infty(\frac{d\phi}{dx})^2dx=-\phi(M)\frac{d\phi}{dx}(M)-\int_M^\infty(\frac{d\phi}{dx})^2dx.$$
The first term goes to zero since $\lim_{x\to \pm\infty}\phi(x)=0$ and $\frac{d\phi}{dx}$ is bounded. Combining all these, we have $\lim_{x\to \infty}\frac{d\phi}{dx}(x)=0$. Similarily we can prove that  $\lim_{x\to -\infty}\frac{d\phi}{dx}(x)=0$.
