<s>[INST] Solving $(y'+1)\ln{\frac{x+y}{x+3}} = \frac{x+y}{x+3}$ Hello everyone :) This is another task I'm trying to solve and can't seem to get the same result as Wolfram Alpha. 

Solve the following differential equation: $$(y'+1)\ln{\frac{x+y}{x+3}} = \frac{x+y}{x+3}$$

My attempt:
Use substitution $$e^z = \frac{x+y}{x+3}$$
and 
$$y=e^z(x+3)-x$$
$$y=xe^z+3e^z-x$$ 
and
$$y'=e^z+z'xe^z+3z'e^z-1$$
Hence:
$$
(e^z+z'xe^z+3z'e^z)z = e^z
$$
Deviding by $e^z$
$$
z+zz'x+3zz' = 1$$
$$
1+z'x+3z' = \frac{1}{z}
$$
$$
z'(x+3)= \frac{1-z}{z}
$$
$$
\int \frac{z}{1-z}dz = \int \frac{1}{x+3} dx
$$
$$
-\ln{\frac{x+y}{x+3}} - \ln{\bigg|1-\ln{\frac{x+y}{x+3}} \bigg|=\ln{(x+3)}}
$$
EDIT:
After fixing the solution, I got the above solution. But still, this solution does not correspond to that on Wolfram Alpha nor can i  find a way to transform them to be equal. 
 [/INST] I managed to solve this differential equation. Following from the last part of the question:
$$
-\ln{\frac{x+y}{x+3}} - \ln{\bigg|1-\ln{\frac{x+y}{x+3}} \bigg|=\ln{(x+3)}} +{C_1}
$$
Next is:
$$
-\bigg(\ln{\frac{x+y}{x+3}}+\ln{\bigg| 1- \ln{\frac{x+y}{x+3}}\bigg|}\bigg) = \ln{{(x+3)}}+\ln{C_2}
$$
$$
-\ln{\frac{x+y}{x+3}\bigg| 1-\ln{\frac{x+y}{x+3}}  \bigg| } = \ln{(x+3)} + \ln{C_2}= \ln{(x+3)C_2}
$$
$$
\ln{\frac{x+y}{x+3}\bigg| 1-\ln{\frac{x+y}{x+3}}  \bigg| } = \ln{\frac{1}{x+3}}
$$
$$
\frac{x+y}{x+3}\bigg( 1-\ln{\frac{x+y}{x+3}}  \bigg)  = \frac{C_3}{x+3}
$$
$$
{(x+y)}\bigg( 1-\ln{\frac{x+y}{x+3}}  \bigg)  = {C_3}
$$
$$
{(x+y)}\bigg( 1-\ln{\frac{x+y}{x+3}}  \bigg)  = {C_3}
$$
$$
x-x\ln{\frac{x+y}{x+3}}+y-y\ln{\frac{x+y}{x+3}} = -C
$$
$$
x\ln{\frac{x+y}{x+3}}+y \bigg( \ln{\frac{x+y}{x+3}} - 1 \bigg) - x = C
$$
</s>