By the Integer Root Theorem, any integer root must be a divisor of the constant term -- thus, in this case, a (positive or negative) divisor of $ 72 $. However, this leaves quite a lot of candidates: $$\pm 1,\ \pm 2,\ \pm 3,\ \pm 4,\ \pm 6,\ \pm 8,\ \pm 9,\ \pm 12,\ \pm 18,\ \pm 24,\ \pm 36,\ \pm 72.$$To narrow our choices, we define another polynomial. Note that $ g(1) = 77 $. Then by the Factor Theorem, $ g(x) - 77 $ is divisible by $ x - 1 $. In other words, $$g(x) = (x-1)q(x) + 77$$for some polynomial $ q(x) $. Thus if we define $ h(x) = g(x+1) $, then we have $$h(x) = xq(x+1) + 77,$$so $ h(x) $ has a constant term of $ 77 $. Thus any integer root of $ h(x) $ is a divisor of $ 77 $; the possibilities are $$-77,\ -11,\ -7,\ -1,\ 1,\ 7,\ 11,\ 77.$$This is useful because, if $ x $ is a root of $ g(x) $, then $ h(x-1)=g(x)=0 $, so $ x-1 $ must appear in the list of roots of $ h(x) $. In particular, $ x $ must be $ 1 $ more than a root of $ h(x) $, which gives the possibilities $$-76,\ -10,\ -6,\ 0,\ 2,\ 8,\ 12,\ 78.$$Of these, only $ -6 $, $ 2 $, $ 8 $, and $ 12 $ were candidates in our original list. Testing them one by one, we find that $ x=\boxed{12} $ is the only integer root of $ g(x) $.