Q:

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. 

A:

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 $0$.
$$\int_{0}^{1}{\sqrt[3]{2x^3-3x^2-x+1}}\,\mathrm{d}x=0$$

A:

I have just found one solution.
Write the given integral has the form
$$I=\int_0^1{\sqrt[3]{(-2(1-x)^3+3(1-x)^2+(1-x)-1)}\,\mathrm{d}x}.$$
And then, put $t=1-x$, we have
$$I=-\int_0^1{\sqrt[3]{(2t^3-3t^2-t+1)}\,\mathrm{d}t}=-I.$$
Therefore $I=0.$

