Integral of x(1/2x-1) dx

1. Rewrite the integrand:

   $$x \left(\frac{x}{2} - 1\right) = \frac{x^{2}}{2} - x$$

2. Integrate term-by-term:

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \frac{x^{2}}{2}\, dx = \frac{\int x^{2}\, dx}{2}$$

      1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$:

         $$\int x^{2}\, dx = \frac{x^{3}}{3}$$

      So, the result is: $\frac{x^{3}}{6}$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \left(- x\right)\, dx = - \int x\, dx$$

      1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$:

         $$\int x\, dx = \frac{x^{2}}{2}$$

      So, the result is: $- \frac{x^{2}}{2}$

   The result is: $\frac{x^{3}}{6} - \frac{x^{2}}{2}$

3. Now simplify:

   $$\frac{x^{2} \left(x - 3\right)}{6}$$

4. Add the constant of integration:

   $$\frac{x^{2} \left(x - 3\right)}{6}+ \mathrm{constant}$$

The answer is: