jmc

To eliminate the fractions, we multiply by $(x^2+4x+1)(x^2-10x)$ on both sides, giving $$(x-2)(x^2-10x)=(x-5)(x^2+4x+1).$$Expanding both sides yields $$x^3-12x^2+20x=x^3-x^2-19x-5,$$and so $$0=11x^2-39x-5.$$By Vieta's formulas, the sum of the roots of this equation is $\boxed{\frac{39}{11}}.$ (One can compute the roots explicitly and check that they do not make any of the denominators of the original equation equal to zero.)