How to explain two answers for of $ \int dx \frac{1}{x-1}$?

Question

I noticed this issue was throwing off a more sophisticated problem I'm working on. When computing the indefinite integral

$$ I(x) = \int \frac{dx}{1-x} = \log | 1-x | + C,$$

I realized I could equivalently write

$$ I(x) = - \int \frac{dx}{x-1} = -\log|x-1| +C = \log \frac{1}{|1-x|} + C.$$

How are these two answers compatible? What am I missing here?

Answer 1

The first answer is wrong. Note that$$\int\frac{\mathrm dx}{a+bx}=\frac1b\log\lvert a+bx\rvert+C.$$In particular$$\int\frac{\mathrm dx}{1-x}=-\log\lvert1-x\rvert+C.$$

Answer 2

The first one is wrong . You have missed a minus sign.

Also $\log (\frac 1 c)=\log 1 -\log c=-\log c$.

Answer 3

You forgot to use the chain rule when doing the first integral. $$\int\frac{dx}{1-x}=-\log|1-x|+C,$$ which is the same as the second one you gave.

Answer 4

$$I=\int \frac{dx}{1-x}=-\log (x-1), ~if~ x>1~~~(1)$$ and $$I=\int \frac{dx}{1-x}=-\log (1-x), ~if~ x<1~~~(2)$$

By the way $$I=\int \frac{dx}{1-x} = - \log |1-x|~~~~(3)$$ will perform like (1) or (2) for the integrals $$\int_{2}^{5} \frac{dx}{1-x}= -\log 4 ~and ~\int_{-3}^{-1} \frac{dx}{1-x}=\log 2$$

But when one limit is negative and the other one is positive (3) will yield $$I=\int_{-2}^{3} \frac{dx}{1-x}= \log (3/2),$$ which is only the principal value of the integral.