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

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We start with the differential equation

$$
y' = (x + y + 1)^2.
$$

**Step 1. Change of Variable**

Introduce a new variable

$$
w = x + y + 1.
$$

Differentiate with respect to $$ x $$:

$$
\frac{dw}{dx} = 1 + y'.
$$

**Step 2. Substitute the Original Equation**

Given $$ y' = w^2 $$, substitute into the derivative of $$ w $$:

$$
\frac{dw}{dx} = 1 + w^2.
$$

**Step 3. Separate Variables**

Separate the variables so that all $$ w $$-terms are on one side:

$$
\frac{dw}{1 + w^2} = dx.
$$

**Step 4. Integrate Both Sides**

Integrate the left-hand side with respect to $$ w $$ and the right-hand side with respect to $$ x $$:

$$
\int \frac{dw}{1 + w^2} = \int dx.
$$

The integral of the left-hand side is

$$
\arctan(w) + C_1,
$$

and the right-hand side integrates to

$$
x + C_2.
$$

We can combine the constants $$ C_1 $$ and $$ C_2 $$ into a single constant $$ C $$:

$$
\arctan(w) = x + C.
$$

**Step 5. Solve for $$ w $$**

Solve for $$ w $$ by taking the tangent of both sides:

$$
w = \tan(x + C).
$$

**Step 6. Substitute Back for $$ y $$**

Recall that $$ w = x + y + 1 $$, so substitute back:

$$
x + y + 1 = \tan(x + C).
$$

Finally, solve for $$ y $$:

$$
y = \tan(x + C) - x - 1.
$$

This is the general solution of the differential equation.
The general solution to the differential equation $$ y' = (x + y + 1)^2 $$ is: $$ y = \tan(x + C) - x - 1 $$ where $$ C $$ is an arbitrary constant.

\( y ^ { \prime } = ( x + y + 1 ) ^ { 2 } \)

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