Evaluate $ \int_{0}^{1} \sqrt{1+\sqrt[3] {x}}\,dx,$ Evaluate 
  $ \int_{0}^{1} \sqrt{1+\sqrt[3] {x}} \,dx,$ 


So,i've tried doing the substitution $ u=\sqrt[3] {x}$( this was a suggestion by wolfram), but frankly i am reaching a point where it doesn't work . Could help me out ?
If $u = \sqrt{1+\sqrt[3]{x}}$, then the integral becomes
$$\int_0^1 \sqrt{1+\sqrt[3]{x}} \ dx = 6 \int_1^\sqrt{2} u^2 \, (u^2-1)^2du.$$
Now, expand the integrand and integrate a polynomial.
Hint:

Let $\sqrt{1+\sqrt[3]x}= y\implies x=((y^2-1)^3, dx=?$

If $x=0,y=1; x=1, y=?$