Given, $\frac{d y}{d x}=\frac{1}{x+y+1}$
Let $\quad x+y+1=v$
$$
\begin{aligned}
& \Rightarrow \quad 1+\frac{d y}{d x}=\frac{d v}{d x} \Rightarrow \frac{d y}{d x}=\frac{d v}{d x}-1 \\
& \therefore \quad \frac{d v}{d x}-1=\frac{1}{v} \\
& \Rightarrow \quad \frac{d v}{d x}=\frac{1}{v}+1 \Rightarrow \frac{d v}{d x}=\frac{1+v}{v} \\
& \Rightarrow \quad \frac{v}{1+v} d v=\mathrm{l} d x \Rightarrow \frac{\mathrm{l}+v-\mathrm{l}}{\mathrm{l}+v} d v=\mathrm{l} d x \\
& \Rightarrow \quad\left[\mathrm{l}-\frac{\mathrm{l}}{\mathrm{l}+V}\right] d v=\mathrm{l} d x \\
&
\end{aligned}
$$
On integrating both sides, we get
$$
\begin{array}{ccc}
& & \int 1 d v-\int \frac{1}{1+v} d x=\int 1 d x \\
\Rightarrow & v-\log (1+v)=x+c \\
\Rightarrow & x+y+1-\log (x+y+2=x+c \\
\Rightarrow & y+1-\log (2+x+y)=c \\
\Rightarrow & y & =\log (x+y+2)+c-1 \\
\Rightarrow & & y=\log (x+y+4-\log k \\
& & {[\because \log k=c-1]} \\
\Rightarrow & & y=\log \left(\frac{x+y+2}{k}\right)
\end{array}
$$