Integral of 3xy^2+7xy dy

1. Integrate term-by-term:

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

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

      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: $x y^{3}$

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

      $$\int 7 x y\, dy = 7 x \int y\, dy$$

      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: $\frac{7 x y^{2}}{2}$

   The result is: $x y^{3} + \frac{7 x y^{2}}{2}$

2. Now simplify:

   $$\frac{x y^{2} \cdot \left(2 y + 7\right)}{2}$$

3. Add the constant of integration:

   $$\frac{x y^{2} \cdot \left(2 y + 7\right)}{2}+ \mathrm{constant}$$

The answer is:

$$\frac{x y^{2} \cdot \left(2 y + 7\right)}{2}+ \mathrm{constant}$$