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.