1
$\begingroup$

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?

$\endgroup$
4
  • 1
    $\begingroup$ $\log(2x-2)=\log 2+\log(x-1)$ And then combine the $\log 2$ term with your added constant $+C$ $\endgroup$
    – Frank W
    Sep 3, 2022 at 22:35
  • $\begingroup$ @FrankW Yeah, didn't think about that! Thanks! $\endgroup$
    – Seeker
    Sep 3, 2022 at 22:38
  • $\begingroup$ 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 $\sec^2(x)+C$. $\endgroup$
    – 2'5 9'2
    Sep 4, 2022 at 3:07
  • $\begingroup$ Check all answers will differ by a constant. $\endgroup$
    – Z Ahmed
    Sep 4, 2022 at 3:53

1 Answer 1

Reset to default
4
$\begingroup$

Remember that an antiderivative (= indefinite integral) is defined only up to the sum of a constant, thus:

$$\frac12\ln|2x-2|=\frac12\left(\ln2+\ln|x-2|\right)+c=\frac12\ln|x-2|+C$$

with $\;c,\,C\;$ arbitrary constants.

$\endgroup$
1
  • 1
    $\begingroup$ Didn't think about combining them, Thanks! $\endgroup$
    – Seeker
    Sep 3, 2022 at 22:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .