How do I solve for this?
dy/dx =yx(x + 3)

Answer:
[tex]{ \tt{ \frac{dy}{dx} = yx( x + 3)}} \\ \\ { \tt{ \frac{dy}{y} = \frac{x(x + 3) \: dx}{1} }} \\ \\ { \tt{ \int \frac{dy}{y} = \int( {x}^{2} + 3x) \: dx }} \\ \\ { \tt{ ln(y) = \frac{ {x}^{3} }{3} + \frac{ {3x}^{2} }{2} + c }} \\ \\ [/tex]