Let f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R satisfy f(x3+1)=x6+4x3+8f(x^3+1) = x^6 + 4x^3 + 8f(x3+1)=x6+4x3+8 for all x∈Rx \in \mathbb{R}x∈R. Then ∫12f(x) dx\displaystyle \int_1^2 f(x)\,dx∫12f(x)dx equals
293\dfrac{29}{3}329
101010
111111
313\dfrac{31}{3}331