Given differential equation, $\frac{d y}{d x}=\frac{x+y+1}{2 x+2 y+1}$
Put $x+y=v$ $\Rightarrow \quad 1+\frac{d y}{d x}=\frac{d v}{d x} \Rightarrow \frac{d y}{d x}=\frac{d v}{d x}-1$
$\therefore$ Eq.
(i) becomes, $\frac{d y}{d x}=\frac{v+1}{2 v+1}$
$\Rightarrow \quad \frac{d v}{d x}-1=\frac{v+1}{2 v+1} \Rightarrow \frac{d v}{d x}=\frac{v+1}{2 v+1}+1$
$\Rightarrow \quad \frac{d v}{d x}=\frac{v+1+2 v+1}{2 v+1} \Rightarrow \frac{d v}{d x}=\frac{3 v+2}{2 v+1}$
$\Rightarrow \frac{2 v+1}{3 v+2} d v=d x \Rightarrow \frac{\left[\frac{2}{3}(3 v+2)-\frac{1}{3}\right]}{3 v+2} d v=d x$
$\Rightarrow \quad\left[\frac{2}{3}-\frac{1}{3} \times \frac{1}{(3 v+2)}\right] d v=d x$
Integrating both sides, we get $\Rightarrow \quad \int\left(\frac{2}{3}-\frac{1}{3} \times \frac{1}{(3 v+2)}\right) d v=\int d x+C^{\prime}$
Where $c$ is a constant of integration.
$\Rightarrow \quad \frac{2}{3} v-\frac{1}{3} \frac{\log (3 v+2)}{3}=x+C^{\prime}$
$\Rightarrow \quad \frac{2}{3}(x+y)-\frac{1}{9} \log (3 x+3 y+2)=x+C$
$\Rightarrow \quad 6 x+6 y-\log (3 x+3 y+2)=9 x+C$
$\Rightarrow \quad 3 x-6 y+\log (3 x+3 y+2)=C$