do you need to include the wiggle infront of the first bracket? if not;
(x+1) ÷ [(x^2+2) x (2x-3dx)]
x^2 x 2x = 2x^3
x^2 x -3dx = -3dx^3
2 x 2x = 4x
2 x -3dx = - 6dx
i cant find a way to make it equal 0 so i think the answer is just
x+1 over 2x^3 - 3dx^3 + 4x - 6dx as a fraction
