Using the substitution $p=x+y$, find the general solution of $$\frac{dy}{dx}=(3x+3y+4)/(x+y+1)$$ Here are my steps:
Since $p=x+y$, $$\frac{3x+3y+4}{x+y+1}=\frac{3p+4}{p+1}=\frac{1}{p+1}+3$$
Therefore, integrate both sides $$y=\ln(p+1)+3p+c$$
$$y=\ln(x+y+1)+3(x+y)+c$$
But the answer in my book is $$x+y-\frac{1}{4}\ln(4x+4y+5)=4x+c$$ Is that correct?