# Integrating $1/{(1-x)}$

Given $y = 1 - x = -1 \cdot (x-1)$ with $x \in (0,1)$ then

$$\int \frac{1}{y} \: \mathrm{d}x = \int \frac{1}{1-x} \: \mathrm{d}x = \ln(1-x) + C$$ but also $$\int \frac{1}{y} \: \mathrm{d}x = - \int \frac{1}{x-1} \: \mathrm{d}x = -\ln(x-1) + C$$ which are not equal. What am I missing?

• Differentiate the first one, don't forget the chain rule. Feb 27 '15 at 15:21
• only the second one is right. but you still need an absolute value sign for the argument of $\ln$
– abel
Feb 27 '15 at 15:21

Your first attempt is not correct: Given the integral $$\int \frac{1}{1-x} \,dx$$ let's put $1-x = u,$ and so $\,du = (1-x)'\,dx = -dx \iff dx = -du.$
Upon substitution, $$\int \frac{1}{1-x} \,dx = -\int \frac{1}{u} \,du = -\ln|u| + C = -\ln|1-x| + C = -\ln|x-1| + C$$
Since $x \in (0, 1)$, we can write the answer as $$- \ln(1-x)+ C\tag{\dagger}$$ $\dagger:\,$ We can drop the absolute value sign because $(1-x)>0$.
$\int\frac{1}{u}du=\ln(u) +C$.