Integral of -x^2+6x-5 dx

1. Integrate term-by-term:

   1. Integrate term-by-term:

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

         $$\int \left(- x^{2}\right)\, dx = - \int x^{2}\, dx$$

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

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

         $$\int 6 x\, dx = 6 \int x\, dx$$

         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: $3 x^{2}$

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

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

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

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

2. Now simplify:

   $$\frac{x \left(- x^{2} + 9 x - 15\right)}{3}$$

3. Add the constant of integration:

   $$\frac{x \left(- x^{2} + 9 x - 15\right)}{3}+ \mathrm{constant}$$

The answer is:

$$\frac{x \left(- x^{2} + 9 x - 15\right)}{3}+ \mathrm{constant}$$