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

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

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    $\begingroup$ Differentiate the first one, don't forget the chain rule. $\endgroup$
    – Git Gud
    Feb 27 '15 at 15:21
  • $\begingroup$ only the second one is right. but you still need an absolute value sign for the argument of $\ln$ $\endgroup$
    – abel
    Feb 27 '15 at 15:21
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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 $(1-x)>0$.

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I believe you may have misapplied the chain rule?

$\int\frac{1}{u}du=\ln(u) +C$.

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