Integral of (2x^2+x-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:

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

         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: $\frac{2 x^{3}}{3}$

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

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

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

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

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

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

2. Now simplify:

   $$\frac{x \left(4 x^{2} + 3 x - 6\right)}{6}$$

3. Add the constant of integration:

   $$\frac{x \left(4 x^{2} + 3 x - 6\right)}{6}+ \mathrm{constant}$$

The answer is:

$$\frac{x \left(4 x^{2} + 3 x - 6\right)}{6}+ \mathrm{constant}$$