Integral of 24*x*y+18*x^3*y^2 dy

1. Integrate term-by-term:

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

      $$\int 18 x^{3} y^{2}\, dy = 18 x^{3} \int y^{2}\, dy$$

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

         $$\int y^{2}\, dy = \frac{y^{3}}{3}$$

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

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

      $$\int 24 x y\, dy = 24 x \int y\, dy$$

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

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

      So, the result is: $12 x y^{2}$

   The result is: $6 x^{3} y^{3} + 12 x y^{2}$

2. Now simplify:

   $$6 x y^{2} \left(x^{2} y + 2\right)$$

3. Add the constant of integration:

   $$6 x y^{2} \left(x^{2} y + 2\right)+ \mathrm{constant}$$

The answer is:

$$6 x y^{2} \left(x^{2} y + 2\right)+ \mathrm{constant}$$