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

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

   $$\int f \left(2 x - 1\right) \left(3 x + 4\right) 1\, dx = f \int \left(2 x - 1\right) \left(3 x + 4\right)\, dx$$

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

   So, the result is: $f \left(2 x^{3} + \frac{5 x^{2}}{2} - 4 x\right)$

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is: