# Integral of $\frac{1}{1-x}$

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?

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

• $-\ln(1-x) + C$ 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.