jmc

Cross-multiplying (which is the same as multiplying both sides by $x-1$ and by $x+2$) gives $$(x+1)(x+2)=(x-2)(x-1).$$ Expanding the products on both sides gives $$x^2+3x+2=x^2-3x+2.$$ Subtracting $x^2$ and 2 from both sides gives $3x=-3x$, so $6x=0$ and $x=\boxed{0}$.