Integral of x(x^2-4x+5)dx dx

1. Rewrite the integrand:

   $$x \left(\left(x^{2} - 4 x\right) + 5\right) = x^{3} - 4 x^{2} + 5 x$$

2. Integrate term-by-term:

   1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$:

      $$\int x^{3}\, dx = \frac{x^{4}}{4}$$

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

      $$\int \left(- 4 x^{2}\right)\, dx = - 4 \int x^{2}\, dx$$

      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: $- \frac{4 x^{3}}{3}$

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

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

3. Now simplify:

   $$\frac{x^{2} \left(3 x^{2} - 16 x + 30\right)}{12}$$

4. Add the constant of integration:

   $$\frac{x^{2} \left(3 x^{2} - 16 x + 30\right)}{12}+ \mathrm{constant}$$

The answer is:

$$\frac{x^{2} \left(3 x^{2} - 16 x + 30\right)}{12}+ \mathrm{constant}$$