Differentiating $\ln(x-1)$ gives $1/(x-1)$, but integrating $1/(x-1)$ gives $\ln|x-1|$. Clearly there is a discrepancy. Why is the integral of $1/(x-1)$ given as $\ln|x-1|$? I have not been able to find a satisfactory explanation.
Isn't that technically expanding the scope of the domain to more than it should actually be?
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Hint:$$\int_{}^{}\dfrac 1x dx=ln|x|$$ because $$I_1=\int_{5}^{7}\dfrac 1x dx\\I_2=\int_{-3}^{-1}\dfrac 1x dx$$ both of them $I_1,I_2 $ have meaning (area under function)
but $x\in (-3,-1) $ $\ln x$ does not mean any thing...
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I wanted to explain a bit about the domains of these functions, rather than why they have the values they do, since that seems to be the confusion here.
Few instructors emphasize this until you get to a higher level of math, but it is important to know that a function is two things: a domain, and the "rule" which operates on that domain. In this case, when we say $f(x)=\ln(x-1)$, it is implied that the domain of $f$ is wherever the function makes sense. In this case, the domain is $(1,\infty)$. Thus, when we differentiate $f$, we want a function who gives the correct derivative at every point on the domain. Since $\frac{1}{x-1}$ works for every value in $(1,\infty)$, we are good to go. However, in this case, the function $f'(x)=\frac{1}{x-1}$ actually still has domain $(1,\infty)$, even though this function could make sense elsewhere. The reason the domain is only $(1,\infty)$ is because the purpose of $f'$ is to give the derivative of $f$, and $f$ doesn't even exist on $(-\infty ,1]$.
Now suppose we start with the function $f(x)=\frac{1}{x-1}$. Since it isn't otherwise specified, we assume that the domain is the largest possible one. In this case, the domain of $f$ is $(-\infty,1)\cup(1,\infty)$, otherwise written simply as $x\neq 1$. If we want to find an antiderivative of $f$, we need to find one that makes sense on the whole domain. If we say that the antiderivative is $F(x)=\ln(x-1)$, then this function only exists on part of the domain we want. Thus, by using $F(x)=\ln|x-1|$, we have an antiderivative on the entire domain of $f$.
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The function $$f(x)=\frac{1}{x-1}$$ exists at all $x$ other than $x=1$, so its antiderivative $F(x)$ should also exist and be differentiable at all $x$ other than $x=1$. Because $x-1$ is negative for $x \lt 1$, $\ln(x-1)$ is undefined for those values. However, because $f(x)$ is defined at those values, its antiderivative should be as well, so we let (almost just by convention) $$\int \frac{1}{x-1}=\ln|x-1|$$ so that it will be defined for the same values that $f(x)$ is.
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On $(1,\infty)$, $$ \int\frac1{x-1}\,\mathrm{d}x=\log(x-1)+C\tag{1} $$ and on $(-\infty,1)$, $$ \int\frac1{x-1}\,\mathrm{d}x=\log(1-x)+C\tag{2} $$ A shorthand for $(1)$ and $(2)$ is $$ \int\frac1{x-1}\,\mathrm{d}x=\log|x-1|+C\tag{3} $$ However, one must be careful about using $(3)$. It seems to indicate that $$ \int_0^2\frac1{x-1}\,\mathrm{d}x=0\tag{4} $$ but the integral in $(4)$ does not converge, so $(4)$ is not valid without the assumptions of the Cauchy Principal Value.
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Another way to prove that the derivative of $f(x)=\ln|x-1|$ is $f'(x)=\frac{1}{x-1}$ is to use the identity $$|x-1|=\sqrt{(x-1)^2}$$ and rewrite $f(x)$ as follow $$f(x)=\ln\left(\sqrt{(x-1)^2}\right) \ \ \ \forall x\in\mathbb{R}-\{1\}$$ Accordi with the log properties $f(x)=\ln(\sqrt{(x-1)^2})=\frac{1}{2}\ln((x-1)^2) \ \ \ \forall x\in\mathbb{R}-\{1\}$.
Now we use the chain rule: $$\begin{align}f'(x)&=\frac{1}{2}\cdot\frac{1}{(x-1)^2}\cdot\frac{d}{dx}[(x-1)^2]\\ &=\frac{1}{2(x-1)^2}\cdot 2(x-1)= \\ &=\frac{1}{x-1} \ \ \ \forall x\in\mathbb{R}-\{1\}\end{align}$$
Actually, we have
$$\frac{d\ln|x-1|}{dx}=\frac{1}{x-1}$$
for $x\ne1$.
When $x>1$, $\ln|x-1|=\ln(x-1)$ and so
$$\frac{d\ln|x-1|}{dx}=\frac{d\ln(x-1)}{dx}=\frac{1}{x-1}$$
When $x<1$, $\ln|x-1|=\ln(1-x)$ and so
$$\frac{d\ln|x-1|}{dx}=\frac{d\ln(1-x)}{d(1-x)}\cdot\frac{d(1-x)}{dx}=\frac{-1}{1-x}=\frac{1}{x-1}$$