Integral of (64x-5) dx

1. Integrate term-by-term:

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

      $$\int 64 x\, dx = 64 \int x\, dx$$

      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: $32 x^{2}$

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

      $$\int \left(-5\right)\, dx = - 5 x$$

   The result is: $32 x^{2} - 5 x$

2. Now simplify:

   $$x \left(32 x - 5\right)$$

3. Add the constant of integration:

   $$x \left(32 x - 5\right)+ \mathrm{constant}$$

The answer is:

$$x \left(32 x - 5\right)+ \mathrm{constant}$$