The given differential equation is
\(\frac {dy}{dx} = (x + y + 1)^2 \quad ....(i)\)
Let \(x + y + 1 = v\)
⇒ \(\frac{dy}{dx} = \frac {dv}{dx} - 1\)
\(\frac {dv}{dx} - 1 = v^2\)
⇒ \(\frac{dv}{v^2 + 1} = dx\)
Integrating, we get
\(\int \frac{dv}{v^2 + 1} = \int dx\)
\(\tan^{–1} (v) = x + c\)
\( \tan^{–1} (x + y + 1) = x + c\)
which is the required solution.