Integral of (7x-5)*x-2dx*(1/3) dx

1. Integrate term-by-term:

   1. Rewrite the integrand:

      $$x \left(7 x - 5\right) = 7 x^{2} - 5 x$$

   2. Integrate term-by-term:

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

         $$\int 7 x^{2}\, dx = 7 \int x^{2}\, dx$$

         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: $\frac{7 x^{3}}{3}$

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

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

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

      The result is: $\frac{7 x^{3}}{3} - \frac{5 x^{2}}{2}$

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

      $$\int \frac{\left(-1\right) 2}{3}\, dx = - \frac{2 x}{3}$$

   The result is: $\frac{7 x^{3}}{3} - \frac{5 x^{2}}{2} - \frac{2 x}{3}$

2. Now simplify:

   $$\frac{x \left(14 x^{2} - 15 x - 4\right)}{6}$$

3. Add the constant of integration:

   $$\frac{x \left(14 x^{2} - 15 x - 4\right)}{6}+ \mathrm{constant}$$

The answer is:

$$\frac{x \left(14 x^{2} - 15 x - 4\right)}{6}+ \mathrm{constant}$$