y' = x(1+y) Solve the differential equation

 An ordinary differential equation (ODE) has differential equation for a function with single variable. A first order ODE follows (dy)/(dx)= f(x,y) .
It can also be in a form of N(y) dy= M(x) dx as variable separable differential equation.
 To be able to set-up the problem as N(y) dy= M(x) dx , we let y' = (dy)/(dx) .
 The problem: y'=x(1+y) becomes:
(dy)/(dx)=x(1+y)
Rearrange by cross-multiplication, we get:
(dy)/(1+y)=xdx
Apply direct integration on both sides: int (dy)/(1+y)= int xdx to solve for the general solution of a differential equation.
For the left side, we consider u-substitution by letting:
u= 1+y then du = dy
The integral becomes:  int(dy)/(1+y)=int(du)/(u)
 Applying basic integration formula for logarithm:
int(du)/(u)=ln|u|
Plug-in u = 1+y on ln|u| , we get:
int(dy)/(1+y)=ln|1+y|
For the right side, we apply the Power Rule of integration: int x^n dx = x^(n+1)/(n+1)+C
int x* dx= x^(1+1)/(1+1)+C
              = x^2/2+C
Combining the results from both sides, we get the general solution of the differential equation as:
ln|1+y|= x^2/2+C
or
y =e^((x^2/2+C))-1