jmc

By partial fractions, $$\begin{array}{rl}\frac{1}{(x-1)(x-2)}& =\frac{1}{x-2}-\frac{1}{x-1},\\ \frac{1}{(x-2)(x-3)}& =\frac{1}{x-3}-\frac{1}{x-2},\\ \frac{1}{(x-3)(x-4)}& =\frac{1}{x-4}-\frac{1}{x-3},\end{array}$$so the given equation reduces to $$\frac{1}{x-4}-\frac{1}{x-1}=\frac{1}{6}.$$Multiplying both sides by $6(x-4)(x-1),$ we get $$6(x-1)-6(x-4)=(x-4)(x-1),$$which simplifies to $x^2-5x-14=0.$ This factors as $(x-7)(x+2)=0,$ so the solutions are $\boxed{7,-2}.$