@Maqroll Consider 2 ways to define the function: h:=x->expr and h:=unapply(expr, x) , where expr is an expression. If the expression expr does not require any evaluation, then these 2 ways are equivalent, for example h:=x->x^2+1 and h:=unapply(x^2+1, x) .
If expr requires calculation, for example, it is given as an integral as in your example or in some other way, then at the moment of the definition of an arrow-function it is not calculated, but will be calculated only when you call this function with some argument value. In the case of unapply-function, the expression expr is evaluated immediately when this function is specified. Compare
f := x->a*x/(4*x^2+b):
h:=x->int(f(t), t=0..x, continuous);
h:=unapply(int(f(t), t=0..x, continuous), x);