Why does d/dt transform to a partial derivative in the integral?

[Lasha]

In a normalization chapter there's an equation(1.21) which says: d/dt ∫|ψ(x,t)|$^{2}$dx=∫∂/∂t |ψ(x,t)|$^{2}$dx
there was a description:(Note that integral is a function only of t,so I use a total derivative (d/dt) in the first expression,but the integrand is a function of x as well as t , so it's a partial derivative in the second one (∂/∂t) )
so this textbook started very simple and intuitive, but now I'm really confused.First of all why did d/dt appear and why did it "transform" to a partial derivative as it "entered" the integral?

 

[vanhees71]

The integral
$$N(t)=\int_{\mathbb{R}} \mathrm{d} x \; |\psi(x,t)|^2$$
is a function of $t$ only, because you integrate over $x$. Thus you can take only the derivative wrt. $t$.

Now, because the integration range $\mathbb{R}$ doesn't change with time, you can as well take first the time derivative of the integrand and then integrate wrt. $x$. Now, since you have a function of $x$ and $t$ you must indicate that you take the time derivative at fixed $x$. That's why you have to use a partial derivative:
$$\frac{\mathrm{d}N(t)}{\mathrm{d} t} = \int_{\mathbb{R}} \mathrm{d} x \frac{\partial}{\partial t} |\psi(x,t)|^2.$$