How can I find the integral? I want to find the integral $$\int_{0}^{1}{\sqrt[3]{2x^3-3x^2-x+1}}\,\mathrm{d}x.$$
I tried, $$\int_{0}^{1}{\sqrt[3]{(-1 + 2 x) (-1 - x + x^2)}}\,\mathrm{d}x.$$
Put $t =-1 - x + x^2$, then $\mathrm{d} t = (2x + 1)\mathrm{d}x.$ And now, I can not find the integral. 

Note that $2x^3-3x^2-x+1=2(x-\frac{1}{2})^3-\frac{5}{2}(x-\frac{1}{2})$. Hence the integrand is symmetric about $x=\frac{1}{2}$, and since you are integrating with limits also symmetric about $\frac{1}{2}$, the value of the integral is just $0$.
$$\int_{0}^{1}{\sqrt[3]{2x^3-3x^2-x+1}}\,\mathrm{d}x=0$$
