I am trying to solve this integral and after a substitution I come to this result

$$\int \frac{1}{1-x} \,dx=-\ln |1-x|$$

Now I have the two cases $-\ln (1-x)$ and $-\ln(x-1)$.

According to my lecture, however, there is only one result namely $-\ln (1-x)$. Why? What happens to the second case?

  • $\begingroup$ Are you just given this integral or is it a part of a bigger problem? $\endgroup$ – kingW3 Nov 22 '15 at 22:06
  • 1
    $\begingroup$ You forgot $+C$. $\endgroup$ – user236182 Nov 22 '15 at 22:08
  • $\begingroup$ What's the context of the lecture? What was the problem you were trying to solve? $\endgroup$ – Dylan Nov 23 '15 at 2:13

Actually, both are correct:

  • $-\ln(1-x) + C$ is the right answer on the domain $x < 1$.

  • $-\ln(x-1) + C$ is the right answer on the domain $x > 1$.

In many contexts, the domain will be a neighborhood of $0$, so the first will be the right answer. Maybe that is why your lecture only listed it. But both are really correct, depending on the domain.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.