Simple question - Proof How is $\frac{1}{2}ln(2x+2) = \frac{1}{2}ln(x+1) $ ?
As $\frac{1}{2}ln(2x+2)$ = $\frac{1}{2}ln(2(x+1))$, how does this become$ \frac{1}{2}ln(x+1)$?
Initial question was $ \int \frac{1}{2x+2} $
What I done was $  \int \frac{1}{u}du $ with $u=2x+2$ which lead to $\frac{1}{2}ln(2x+2) $ but WolframAlpha stated it was $\frac{1}{2}ln(x+1)$

We have $\ln(2x+2)=\ln(2(x+1))=\ln 2+\ln(x+1)$. Thus
$$\frac{1}{2}\ln(2x+2)=\frac{1}{2}\ln 2+\frac{1}{2}\ln(x+1).$$
The two functions thus are definitely not equal. But they differ by a constant. 
So the answer to $\int \frac{dx}{2x+2}$ can be equally well put as $\frac{1}{2}\ln(|2x+2|)+C$ and $\frac{1}{2}\ln(|x+1|)+C$. (We can forget about the absolute value part if $x+1$ is positive in our application.) 
Remark: A simpler example: It is correct to say $\int 2x\,dx=x^2+C$. It is equally correct (but a little weird) to say $\int 2x\,dx=x^2+47+C$.
