Differential equation: $\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}$ 
The solution of $\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}$ is given by: 
a) $(x+2)^4 (1+\frac{2y}{x})= ke^{\frac{2y}{x}}$
b) $(x+2)^4 (1+ 2\frac{(y-2)}{x+2})= ke^{\frac{2(y-2)}{x+2}}$
c) $(x+2)^3 (1+ 2\frac{(y-2)}{x+2})= ke^{\frac{2(y-2)}{x+2}}$
d) None of these

Attempt: 
I have expanded and checked but couldn't spot any exact differentials. 
Secondly, it's not a homogeneous equation, so couldn't use $y = vx$. 
How do I go about solving this problem? 

$$\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}=\left(\dfrac{x+y}{x+2}\right)^2\dfrac{x+2}{y-2}$$
let $w=\dfrac{x+y}{x+2}$ then $y=w(x+2)-x$ and
$$w'(x+2)+w-1=y'=w^2\dfrac{1}{w-1}$$
which is separable
$$\dfrac{w-1}{2w-1}dw=\dfrac{dx}{x+2}$$
with integration the solution will be obtained.
