jmc

Multiplying both sides by $(x+1{)}^2,$ we get $$(13x-x^2)(x(x+1)+(13-x))=42(x+1{)}^2.$$This expands to $x^4-13x^3+55x^2-85x+42=0,$ which factors as $(x-1)(x-6)(x^2-6x+7)=0.$ By the quadratic formula, the roots of $x^2-6x+7=0$ are $3\pm \sqrt{2}.$ Therefore, the solutions are $\boxed{1,6,3+\sqrt{2},3-\sqrt{2}}.$