it just means that the prefactor is 'one'. But rather being the precise 1, it can also be an approximate (machine-precision) one. This is the case here. it is not exactly integer 1, but a floating point number 1. If you do:
D[(1/2)*x1^2 , x1]
you will see x1. no prefactor. The entire idea (apart from a few exception) of Wolfram Language is exact input = exact output. Machine-precision input = Machine-precision output. etc
