Answer
$y=Ke^{x}-x-1,K \in \mathbb R $
Work Step by Step
$$y'=x+y, u=x+y$$
$\to$
$$y'=u$$
Using the implicit differentiating it follows:
$$u=x+y \to u'=1+y'$$
$$u'=1+u$$
$$\frac{du}{dx}=1+u$$
$$\frac{du}{1+u}=dx$$
$$\int \frac{du}{1+u}=\int dx$$
$$\ln(|1+u|)=x+c$$
$$|1+u|=e^{x+c}$$
$$1+u=e^{x+c}~~\text{or}~~-(1+u)=e^{x+c}$$
$$1+u=e^{x+c}~~\text{or}~~-1-u=e^{x+c}$$
$$u=e^{x+c}-1~~\text{or}~~u=-e^{x+c}-1$$
Since $u=x+y \to e^{x+c}-1=x+y \to y=-x+e^{x+c}-1 $
Since $u=x+y \to -e^{x+c}-1=x+y \to y=-x-e^{x+c}-1 $
Therefore, the solution is:
$$y=e^ce^{x}-x-1,~~c \in \mathbb R $$
$\text{or}$
$$y=-e^ce^{x}-x-1,~~c \in \mathbb R $$
so $y=Ke^x-x-1,K\in\mathbb{R}$
