Integral of 3x³-4x² dx

1. Integrate term-by-term:

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

      $$\int 3 x^{3}\, dx = 3 \int x^{3}\, dx$$

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

         $$\int x^{3}\, dx = \frac{x^{4}}{4}$$

      So, the result is: $\frac{3 x^{4}}{4}$

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

      $$\int \left(- 4 x^{2}\right)\, dx = - 4 \int x^{2}\, dx$$

      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: $- \frac{4 x^{3}}{3}$

   The result is: $\frac{3 x^{4}}{4} - \frac{4 x^{3}}{3}$

2. Now simplify:

   $$\frac{x^{3} \left(9 x - 16\right)}{12}$$

3. Add the constant of integration:

   $$\frac{x^{3} \left(9 x - 16\right)}{12}+ \mathrm{constant}$$

The answer is: