$\frac{d y}{d x} = (4 x + y + 1)^{2}$

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Solution

$\frac{d y}{d x} = (4 x + y + 1)^{2}$
Let, $(4 x + y + 1)^{2} = z$
So, $4 d x + d y = d z$
$⟹ d y = d z - 4 d x$
So, $\frac{d z - 4 d x}{d x} = x^{2} ⟹ \frac{d z}{d x} = z^{2} + 4$
So, $\frac{d z}{z^{2} + 4} = d x$
On integrating on both sides
$\frac{1}{2} {tan}^{- 1} (\frac{z}{2}) = x + c$
Now as $(4 x + y + 1)^{2} = z$
So, $\frac{1}{2} {tan}^{- 1} (\frac{(4 x + y + 1)}{2}) = x + c$

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