Integral of x(2x²-5)dx dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = x^{2}$.

      Then let $du = 2 x dx$ and substitute $du$:

      $$\int \left(u - \frac{5}{2}\right)\, du$$

      1. Integrate term-by-term:

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

            $$\int u\, du = \frac{u^{2}}{2}$$

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

            $$\int \left(- \frac{5}{2}\right)\, du = - \frac{5 u}{2}$$

         The result is: $\frac{u^{2}}{2} - \frac{5 u}{2}$

      Now substitute $u$ back in:

      $$\frac{x^{4}}{2} - \frac{5 x^{2}}{2}$$

   Method #2

   1. Rewrite the integrand:

      $$x \left(2 x^{2} - 5\right) 1 = 2 x^{3} - 5 x$$

   2. Integrate term-by-term:

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

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

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

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

         So, the result is: $\frac{x^{4}}{2}$

      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: $\frac{x^{4}}{2} - \frac{5 x^{2}}{2}$

2. Now simplify:

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

3. Add the constant of integration:

   $$\frac{x^{2} \left(x^{2} - 5\right)}{2}+ \mathrm{constant}$$

The answer is: