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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 $\frac{1}{2}ln(x+1)$

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It isn't. But if this comes from an integration, note that the two functions differ by a constant, so the answer to an integration problem can be put as either one of them plus $C$. – André Nicolas Jan 24 '13 at 18:23
Yes this comes from integrating $ \int \frac{1}{2x+2} dx $ – neverloggedin Jan 24 '13 at 18:28

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 $\int 2x\,dx=x^2+47+C$.

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$\frac{1}{2}\ln(2(x+1))=\ln\sqrt{2}+\frac{1}{2}\ln(x+1)\neq \frac{1}{2}\ln(x+1)$

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, 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.

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