jmc

We can start by multiplying both sides of the equation by $(x+1{)}^3$. This gives $$5x-7=A(x-1{)}^2+B(x-1)+C.$$Expanding and rearranging the right hand side gives $$5x-7=A{x}^2+(B-2A)x-A-B+C.$$By comparing coefficients we know that $A=0$, $B-2A=5$, and $-A-B+C=-7.$ So $B=5$ and $C=-7+5=-2$. Therefore $A+B+C=\boxed{3}.$

Alternatively, we can substitute $x=2$ into the given equation, which gives us $A+B+C=3$ right away.