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 $\frac{1}{2}ln(x+1)$
 A: 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 $\int 2x\,dx=x^2+47+C$.
A: $\frac{1}{2}\ln(2(x+1))=\ln\sqrt{2}+\frac{1}{2}\ln(x+1)\neq \frac{1}{2}\ln(x+1)$
But as far as $$\int\frac{1}{(2x+2)}dx$$ is concerned , it comes out to be $$\frac{1}{2}\ln(2x+2)+c$$ or you can write it as $$\frac{1}{2}\int\frac{1}{(x+1)}dx$$ which comes out to be $$\frac{1}{2}\ln(x+1)+k$$
Here , the extra $\ln\sqrt{2}$ has been taken care of in constant terms of integration.so, as far as indefinite integration is concerned, it gives family of curves as its solution, and they both represent same family of curves and differ by a constant only.
