Solving the differential equation $dy/dx=\frac{6}{x+y}$ 
Let $ \text{  }\dfrac{dy}{dx}=\frac{6}{x+y}$ where $y(0)=0$. Find the value of $y$ when $x+y=6$.

Let $x+y=v$. Thus $$ \text{  }\dfrac{dy}{dx}=\dfrac{dv}{dx}-1$$
Therefore $$ \dfrac{dv}{dx}=\dfrac{6+v}{v}$$
On separating the variables and integrating, I get
$$y=6\ln(x+y+6
)+C$$
Could somebody please show me where I've gone wrong?

$u = x + y \implies u' = 1 + y' $. Thus, 
$$ u' - 1 = \frac{6}{u} \implies u' = \frac{6+u}{u} \implies \int \frac{ u du }{6 + u} = x +C \implies u - 6 \ln|u+6| = x + C $$
Since $u = x+y$, then 
$$ x + y - 6 \ln | x + y + 6 | = x + C $$
Since $y(0) = 0$, then 
$$ 0 + 0 - 6 \ln 6 = C \implies C = - 6 \ln 6 $$
