jmc

By partial fractions, $$\frac{-x^2+3x-4}{x^3+x}=\frac{-x^2+3x-4}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}$$Multiplying by $x(x^2+1)$ gives $$-x^2+3x-4=(A+B)x^2+Cx+A.$$By comparing coefficients, we can see that $A=-4$ and $C=3.$ Then, $-4+B=-1$ which means $B=3$. Thus, $$\frac{-x^2+3x-4}{x^3+x}=\frac{-4}{x}+\frac{3x+3}{x^2+1}.$$and $(A,B,C)=\boxed{(-4,3,3)}.$