According to the special relativity theory, moving object have kinetic energy given by formula $KE = (\gamma - 1) m c^2 $. So this is equivalent with classical formulation $KE = (1/2) m v^2$ for the case $v^2 / c^2 \ll 1 $ (using binomial expansion). But in classical formulation, moving object just have kinetic energy and there are no form of energy other than that. For special relativity say it statement for KE is equivalent with classical formulation, then we can say there is no component of $m c^2$ for moving object too, because total energy for moving object in classical formulation is just their kinetik energy.
If not (there is component $m c^2$ in moving object), then what is equivalent statement for total energy $\gamma m c^2$ in classical formulation? I mean is there any procedur to exploit total energy other than making a bomb (complex nuclear reaction)?
Einstein derive his statement for energy just using integration to $F ds$, a classical defenition of work, but like a magic trick there is appear $m c^2 $ term, where $m c^2 \gg (1/2) m v^2 $ in resultant equation. So it is safe for me to say that when an object moving on $ds$ path it posses an intrinsic $m c^2$ property? Remember $F ds $ is statement for the work done by that object, so this defenition coupled with the path $ds $ and also movement of that object. If that object at rest, then $ds = 0 $, there is no work done by that object, and maybe there is nothing $m c^2$ too.
Precisely, is there any way to derive $m c^2$ term without using integration to $ds$ path or wihout assumption that the object is moving?