The solution $y^{\mathrm{\prime }}=\frac{1-2y-4x}{1+y+2x}$ is given by 

Put $u=y+2x\Rightarrow \frac{dy}{dx}=\frac{du}{dx}-2$ 

so the given equation reduces to

 $\frac{du}{dx}-2=\frac{1-2u}{1+u}\Rightarrow \frac{du}{dx}=\frac{1-2u+2+2u}{1+u}$

$\Rightarrow \frac{1+u}{3}du=dx$

$\Rightarrow \left(u+\frac{u^2}{2}\right)=3x+C$

$\text{ i. e. }\left(y+2x+\frac{(y+2x{)}^2}{2}\right)=3x+C$

$y+\frac{1}{2}(y+2x{)}^2=x+C$