Stupid question (*don't laugh), but can you move an existing dt to the other side before integration?

The derivative of a function of one variable does behave a lot like a quotient of differentials; in particular, df=(df/dt)dt.

Then what you appear to have is int(1,A,A_init,A_fin)=int(K,t,t_init,t_fin), which can be simplified if K is constant with respect to t.

Yes this is allowed in certain situations. Lots of Engineering Math is based on this technique.

You can do this as a shortcut, but it's not because you're allowed to just multiply and divide by differentials.

The reason is that if two indefinite integrals are equal then their integrands are equal. Note that this is not true for definite integrals. The integral from -1 to 1 of sin(x) is equal to the integral from -1 to 1 of x, for instance.

So let's say we've got df/dt = X(t) where X is some arbitrary function of time. If we integrate both sides with respect to t, then we get int[(df/dt)dt] = int[X(t)dt] and because the integral of the left hand side int[(df/dt)dt] = f(t) = int[df], we get int[df] = int[X(t)dt] and therefore df = X(t)dt.

So it happens because of the rules for integration and not because we're multiplying and cancelling out differentials.

If you want to see an example of how this reasoning can break down, consider what might happen if you want to get the derivative of the scalar-valued function f:R^2 -> R with respect to its input vector, that is, a function f(r) that takes a vector r = [x, y] as its input and gives a scalar (a single number) as its output. The differential of r is itself a vector, dr, and the differential of f is a scalar, df. Division of a scalar by a vector isn't defined, although there's still nothing strictly wrong with writing the derivative of f with respect to r as df/dr.

EDIT: I think the reason that you got some downvotes might be because this would be better suited for r/mathhelp, and also because the only "stupid" question in math or science is one you didn't bother to ask =)

That's fine but don't then arbitrarily put another dt on the LHS.

Notationally they're equivalent, and both notations are certainly used. Neither one is a proof, however.