Should the differential of a wavefunction have a time partial derivative?

Question

In chapter 1 of Griffths' QM text, he shows that $\frac{\mathrm{d}}{\mathrm{d}t}\int_{-\infty}^{\infty}|\Psi|^2\,\mathrm{d}x=0$ by noting

$$\begin{align} \frac{\mathrm{d}}{\mathrm{d}t}\int_{-\infty}^{\infty}|\Psi|^2\,\mathrm{d}x &= \int_{-\infty}^{\infty}\frac{\partial}{\partial t}|\Psi|^2\,\mathrm{d}x \\ &= \int_{-\infty}^{\infty}\frac{i\hbar}{2m}\biggl(\bar{\Psi}\frac{\partial^2}{\partial x^2}\Psi-\Psi\frac{\partial^2}{\partial x^2}\bar{\Psi}\biggr)\mathrm{d}x \\ &= \int_{-\infty}^{\infty}\frac{\partial}{\partial x}\biggl[\frac{i\hbar}{2m}\biggl(\bar{\Psi}\frac{\partial}{\partial x}\Psi-\Psi\frac{\partial}{\partial x}\bar{\Psi}\biggr)\biggr]\,\mathrm{d}x \\ &= \biggl[\frac{i\hbar}{2m}\biggl(\bar{\Psi}\frac{\partial}{\partial x}\Psi-\Psi\frac{\partial}{\partial x}\bar{\Psi}\biggr)\biggr]_{-\infty}^{\infty} \end{align}$$

My confusion is in the last step. I think what he's doing is expressing the integrand as the differential $\mathrm{d}f$, where

$$f=\frac{i\hbar}{2m}\biggl(\bar{\Psi}\frac{\partial}{\partial x}\Psi-\Psi\frac{\partial}{\partial x}\bar{\Psi}\biggr).$$

But isn't $f$ a function of $x$ and $t$, making the differential $\mathrm{d}f=\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial t}\mathrm{d}t$ instead of just $\mathrm{d}f=\frac{\partial f}{\partial x}\mathrm{d}x$ as Griffths has?

Answer 1

The integrals with $dx$ are partial integrals; $t$ is assumed constant inside them. You could argue that by analogy with partial differentiation they should be written with $\partial x$ instead of $dx$ but historically that's not how they're written. So basically the antidifferentiation performed by the integral consists of inverting a $\partial/\partial x$ operation, not a $d/dx$ operation, and you don't need to take $t$ into account.