jmc

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 $g(x)$.