jmc

Multiplying both sides by $(4-7x)(x^3+2),$ we get $$x(x^2-56)(x^3+2)-(4-7x)(21x+22)=4(4-7x)(x^3+2).$$This simplifies to $$x^6-28x^4-14x^3+147x^2+14x-120=0.$$First, we try looking for nice roots. Noting that both $x=1$ and $x=-1$ are roots, we can factor out $x-1$ and $x+1,$ to get $$(x-1)(x+1)(x^4-27x^2-14x+120)=0.$$By the Integer Root Theorem, any integer roots must be factors of 120 (including negative factors). Trying several factors, we notice that 2, 5, $-3,$ and $-4$ work, so the solutions are $\boxed{-4,-3,-1,1,2,5}.$