Integral of y^3(2y^2-3)dy dx

1. There are multiple ways to do this integral.

   Method #1

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

      Then let $du = 2 y dy$ and substitute $du$:

      $$\int \left(u^{2} - \frac{3 u}{2}\right)\, du$$

      1. Integrate term-by-term:

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

            $$\int u^{2}\, du = \frac{u^{3}}{3}$$

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

            $$\int \left(- \frac{3 u}{2}\right)\, du = - \frac{3 \int u\, du}{2}$$

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

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

            So, the result is: $- \frac{3 u^{2}}{4}$

         The result is: $\frac{u^{3}}{3} - \frac{3 u^{2}}{4}$

      Now substitute $u$ back in:

      $$\frac{y^{6}}{3} - \frac{3 y^{4}}{4}$$

   Method #2

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

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

         $$\int 2 y^{5}\, dy = 2 \int y^{5}\, dy$$

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

            $$\int y^{5}\, dy = \frac{y^{6}}{6}$$

         So, the result is: $\frac{y^{6}}{3}$

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

         $$\int \left(- 3 y^{3}\right)\, dy = - 3 \int y^{3}\, dy$$

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

            $$\int y^{3}\, dy = \frac{y^{4}}{4}$$

         So, the result is: $- \frac{3 y^{4}}{4}$

      The result is: $\frac{y^{6}}{3} - \frac{3 y^{4}}{4}$

2. Now simplify:

   $$\frac{y^{4} \cdot \left(4 y^{2} - 9\right)}{12}$$

3. Add the constant of integration:

   $$\frac{y^{4} \cdot \left(4 y^{2} - 9\right)}{12}+ \mathrm{constant}$$

The answer is:

$$\frac{y^{4} \cdot \left(4 y^{2} - 9\right)}{12}+ \mathrm{constant}$$