# Getting two different answers for $\int \frac 1 {2x-2}dx$

When I tried to compute the integral $$\int \frac 1 {2x-2}dx$$, the following was my attempt. $$\int \frac 1 {2x-2}dx=\frac 1 2 \int \frac 1 {x-1}dx=\frac 1 2\ln\left(|x-1|\right)+c$$

But then I saw someone on youtube doing the same integral and getting the answer $$\int \frac 1 {2x-2}dx=\frac 1 2 \ln\left(|2x-2|\right)+c$$

Now I immediately thought that something was wrong, so I tried to put the integral into Symbolab. To my surprise, Symbolab seemed to agree with the second answer. Then I put the integral into Wolfram and I got the answer that I had at first.

After that I decided to input the following into Symbolab. $$\int \frac 1 2 \cdot\frac 1 {x-1}dx$$ and Symbolab seemed to agree with my first answer.

Now, it could be possible that I am missing something very obvious, but I don't know what is it. So, am I wrong or was Symbolab wrong?

• $\log(2x-2)=\log 2+\log(x-1)$ And then combine the $\log 2$ term with your added constant $+C$ Sep 3, 2022 at 22:35
• @FrankW Yeah, didn't think about that! Thanks! Sep 3, 2022 at 22:38
• Reminds me of $\int2\tan(x)\sec(x)\sec(x)\,dx$. Use $u=\tan(x)$ and you get $\tan^2(x)+C$. Use $u=\sec(x)$ and you get $\sec^2(x)+C$. Sep 4, 2022 at 3:07
• Check all answers will differ by a constant. Sep 4, 2022 at 3:53

$$\frac12\ln|2x-2|=\frac12\left(\ln2+\ln|x-2|\right)+c=\frac12\ln|x-2|+C$$
with $$\;c,\,C\;$$ arbitrary constants.