Integral of 6t^2-4t dt

1. Integrate term-by-term:

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

      $$\int 6 t^{2}\, dt = 6 \int t^{2}\, dt$$

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

         $$\int t^{2}\, dt = \frac{t^{3}}{3}$$

      So, the result is: $2 t^{3}$

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

      $$\int \left(- 4 t\right)\, dt = - \int 4 t\, dt$$

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

         $$\int 4 t\, dt = 4 \int t\, dt$$

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

            $$\int t\, dt = \frac{t^{2}}{2}$$

         So, the result is: $2 t^{2}$

      So, the result is: $- 2 t^{2}$

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

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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