Answer
\[y(x)=2\left[\tan (2x+C)-2x-1\right]\]
Work Step by Step
\[y'=(4x+y+2)^2\;\;\;\ldots (1)\]
Substitute $\: V=4x+y+2$ ____(2)
Differentiate (2) with respect to $x$
\[\frac{dV}{dx}=4+\frac{dy}{dx}\]
From (1) and (2)
\[\frac{dV}{dx}=4+V^2\]
Separating variables
\[\frac{dV}{4+V^2}=dx\]
Integrating,
\[\int\frac{dV}{2^2+V^2}=\int dx+C_1\]
Where $C_1$ is constant of integration
\[\frac{1}{2}\tan^{-1}\frac{V}{2}=x+C_1\]
\[\tan^{-1}\frac{V}{2}=2x+2C_1\]
\[V=2\tan(2x+2C_1)\]
From (2)
\[y=2\tan(2x+2C_1)-4x-2\]
\[y=2[\tan(2x+C)-2x-1]\]
Where $\:C=2C_1$
Hence \[y(x)=2\left[\tan (2x+C)-2x-1\right].\]
