jmc

Multiplying both sides by $x(x+1)(x+3),$ we get $$x+2=A(x+1)(x+3)+Bx(x+3)+Cx(x+1).$$Setting $x=0,$ we get $3A=2,$ so $A=\frac{2}{3}.$

Setting $x=-1,$ we get $-2B=1,$ so $B=-\frac{1}{2}.$

Setting $x=-3,$ we get $6C=-1,$ so $C=-\frac{1}{6}.$ Hence, $$ABC=\frac{2}{3}\cdot \left(-\frac{1}{2}\right)\cdot \left(-\frac{1}{6}\right)=\boxed{\frac{1}{18}}.$$