(2+x)y' = 3y Find the general solution of the differential equation

Recall that y' is the same as (dy)/(dx) . Then in the given problem: (2+x)y'=3y , we may write it as:
(2+x) (dy)/(dx) = 3y.
 This will help to follow the variable separable differential equation in a form of N(y) dy = M(x) dx.
To rearrange (2+x) (dy)/(dx) = 3y ,cross-multiply (dx) to the other side:
(2+x)dy =3y dx
Divide both sides by (2+x) :
((2+x)dy)/(2+x) =(3y dx)/(2+x)
dy =(3y dx)/(2+x)
Divide both sides by y :
(dy )/y=(3y dx)/((2+x)y)
(dy)/y=(3dx)/(2+x)
 To solve for the general solution of the differential equation, apply direct integration on both sides:
int (dy)/y=int (3dx)/(2+x)
For the left side, apply the basic integration formula for logarithm: 
int (dy)/y= ln|y|
For the right side, we may apply the basic integration property: int c*f(x) dx = c int f(x)dx .
int (3dx)/(2+x)= 3 int (dx)/(2+x)
 Let u =2+x then du= dx
The integral becomes:
3 int (dx)/(2+x) = 3 int (du)/u
We can now apply the  basic integration formula for logarithm on the integral part:
3 int (du)/u= 3ln|u| +C
Recall u =(2+x) then 3 int (dx)/(2+x) =3ln|2+x| +C
Combining the results from both sides, we get:
ln|y|=3ln|2+x| +C
y=e^(3ln|x+2|+C)
y= e^(ln(x+2)^3+C) 
Law of Exponents: x^(n+m)= x^n*x^m
y= e^(ln(x+2)^3)*e^C
e^C =C is an arbitrary constant, so
y= Ce^(ln(x+2)^3)
y = C(x+2)^3