Integral of (6x²-2x+1) dx

1. Integrate term-by-term:

   1. Integrate term-by-term:

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

         $$\int 6 x^{2}\, dx = 6 \int x^{2}\, dx$$

         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: $2 x^{3}$

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

         $$\int \left(- 2 x\right)\, dx = - 2 \int x\, dx$$

         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: $- x^{2}$

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

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

      $$\int 1\, dx = x$$

   The result is: $2 x^{3} - x^{2} + x$

2. Now simplify:

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

3. Add the constant of integration:

   $$x \left(2 x^{2} - x + 1\right)+ \mathrm{constant}$$

The answer is:

$$x \left(2 x^{2} - x + 1\right)+ \mathrm{constant}$$