Integral of (6t^2-1) 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 is the constant times the variable of integration:

      $$\int \left(-1\right)\, dt = - t$$

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

2. Add the constant of integration:

   $$2 t^{3} - t+ \mathrm{constant}$$

The answer is:

$$2 t^{3} - t+ \mathrm{constant}$$