Integral of f³¹(2x-5)(x+2)dx dx

1. Rewrite the integrand:

   $$f^{3} \left(2 x - 5\right) \left(x + 2\right) = 2 f^{3} x^{2} - f^{3} x - 10 f^{3}$$

2. Integrate term-by-term:

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

      $$\int 2 f^{3} x^{2}\, dx = 2 f^{3} \int x^{2}\, dx$$

      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: $\frac{2 f^{3} x^{3}}{3}$

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

      $$\int \left(- f^{3} x\right)\, dx = - f^{3} \int x\, dx$$

      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: $- \frac{f^{3} x^{2}}{2}$

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

      $$\int \left(- 10 f^{3}\right)\, dx = - 10 f^{3} x$$

   The result is: $\frac{2 f^{3} x^{3}}{3} - \frac{f^{3} x^{2}}{2} - 10 f^{3} x$

3. Now simplify:

   $$\frac{f^{3} x \left(4 x^{2} - 3 x - 60\right)}{6}$$

4. Add the constant of integration:

   $$\frac{f^{3} x \left(4 x^{2} - 3 x - 60\right)}{6}+ \mathrm{constant}$$

The answer is: