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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?

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  • $\begingroup$ Are you just given this integral or is it a part of a bigger problem? $\endgroup$ – kingW3 Nov 22 '15 at 22:06
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    $\begingroup$ 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
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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.

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