∫ (x)* (x-1) 2dx ??

2*[(X^3/3)-(x^2/2)]+C

∫(x²-x) 2x dx

u = x² - x

du = 2x dx

= 2x³/3 - x² + c 

Para resolver a integral dada devemos realizar os seguintes cálculos abaixo:

\(\begin{align} & f=x{{(x-1)}^{2}} \\ & f=x({{x}^{2}}-2x+1) \\ & \int_{{}}^{{}}{f(x)=\int_{{}}^{{}}{{{x}^{3}}-2{{x}^{2}}+x}} \\ & \int_{{}}^{{}}{f{{(x)}^{n}}=\frac{{{x}^{n+1}}}{n+1}} \\ & \int_{{}}^{{}}{f(x)=\frac{{{x}^{3+1}}}{3+1}-2\frac{{{x}^{2+1}}}{2+1}+\frac{{{x}^{1+1}}}{1+1}} \\ & \int_{{}}^{{}}{f(x)=\frac{{{x}^{4}}}{4}-2\frac{{{x}^{3}}}{3}+\frac{{{x}^{2}}}{2}} \\ & \int_{{}}^{{}}{f(x)=\frac{{{x}^{4}}}{4}-2\frac{{{x}^{3}}}{3}+\frac{{{x}^{2}}}{2}}+C \\ \end{align} \)

Portanto, o valor da integral dada será \(\boxed{\int_{}^{} {f(x) = \frac{{{x^4}}}{4} - 2\frac{{{x^3}}}{3} + \frac{{{x^2}}}{2}} + C}\).