Integral of (2x-1)(3x+4) dx

1. Rewrite the integrand:

   $$\left(2 x - 1\right) \left(3 x + 4\right) = 6 x^{2} + 5 x - 4$$

2. 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:

      $$\int 5 x\, dx = 5 \int x\, dx$$

      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: $\frac{5 x^{2}}{2}$

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

      $$\int \left(-4\right)\, dx = - 4 x$$

   The result is: $2 x^{3} + \frac{5 x^{2}}{2} - 4 x$

3. Now simplify:

   $$\frac{x \left(4 x^{2} + 5 x - 8\right)}{2}$$

4. Add the constant of integration:

   $$\frac{x \left(4 x^{2} + 5 x - 8\right)}{2}+ \mathrm{constant}$$

The answer is: