Integral of 8*(x-1) dx

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

   $$\int 8 \left(x - 1\right)\, dx = 8 \int \left(x - 1\right)\, dx$$

   1. Integrate term-by-term:

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

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

      1. The integral of a constant is the constant times the variable of integration:

         $$\int \left(-1\right)\, dx = - x$$

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

   So, the result is: $4 x^{2} - 8 x$

2. Now simplify:

   $$4 x \left(x - 2\right)$$

3. Add the constant of integration:

   $$4 x \left(x - 2\right)+ \mathrm{constant}$$

The answer is:

$$4 x \left(x - 2\right)+ \mathrm{constant}$$