Let g(x) be the inverse of the function f(x), and f'(x)= 1/1+x3. then g(x) is equal to....

f(x) = ∫1/(1+x³)dx  1/(1+x³) = (x+1)/3 - (x/3)/(((x-0.5)^2) +0.75) +(2/3)/(((x-0.5)^2) +0.75)

= (x+1)/3 - (1/6)(2x-1)/(((x-0.5)^2) +0.75) - (1/2)(((x-0.5)^2) +0.75)

therefore, ∫1/(1+x³)dx  = f(x) = (1/3)ln(x+1) - (1/6)ln(x²-x+1) -(1/√3)arctan((2x-1)/√3)

g(x) = ƒ^-1(x) implies that f(g(x))= x

x = (1/3)ln(g(x)+1) - (1/6)ln((g(x))²-x+1) -(1/√3)arctan((2g(x)-1)/√3)

One has to solve this equation in g(x) in order to express g(x) explicitly. It is obvious that this equation cannot be solved.