Your calculus prof lied to you (probably)

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Recommends user905694 · 2022-06 · Mathematics Stack Exchange · on “Integrating $\frac{1}{x}$”

There is a general (mis)conception that $$\int \frac{1}{x} \, \mathrm{d}x = \ln|x| + C \label{1} \tag{1}$$ However, taking the indefinite integral as the set of all functions whose derivative is $\frac{1}{x}$ , the technically correct answer should be $$\int \frac{1}{x} \, \mathrm{d}x = \begin{cases} \ln(x) + C & \text{if $x > 0$} \\ \ln(-x) + D & \text{if $x < 0$} \end{cases} \label{2} \tag{2}$$ See Your calculus prof lied to you (probably) [this video] for more details. Is $\eqref{2}$ correct and is $\eqref{1}$ incorrect? Or are they both acceptable? I suppose $\eqref{2}$ is more correct but it perhaps doesn’t have any additional value compared to $\eqref{1}$ as the definite integral will give

Cites ProfInsall · 2022-06 · Mathematics Stack Exchange · on “Integrating $\frac{1}{x}$”

Use the notion of pseudo-constants, differentiable functions whose derivative is zero on their domain. My replies to comments on the video at [this video] give definitions and results and examples, based upon published work with Charatonik (r.i.p.) on absolute differentiation in metric spaces and unpublished work with him on differentiation in topological fields. In this special case, formulas (1) and (2) are unified in a formula that replaces C with a component-wise constant function and while I think that you're probably deluding yourself about your understanding if you use the term "arbitrary", feel free to call it "an arbitrary pseudo-constant".

Cites Accelerator · 2022-08 · Mathematics Stack Exchange · on “Anti-Derivative Question”

The textbook says the piecewise function is the correct answer because the domain of $F(x)$ is a disjoint union of two intervals, namely $(-\infty,0)$ and $(0,\infty)$ , so a different constant of integration (like $C_1$ and $C_2$ ) may be chosen for each of the intervals. The textbook is just showing the "most general" antiderivative there is. Whenever you have an antiderivative $F(x)$ , it's always implied there is a domain to it because functions always need a domain. In this case, the domain of $F(x)$ is all real numbers except $0$ since $x=0$ makes $F(x)$ undefined. The reason why people just casually take the indefinite integral and add that $+C$ at the end is because of a corollary of

Cites Torsten Schoeneberg · 2024-01 · Mathematics Stack Exchange · on “The $\int \frac{x^2}{x^2+x-2}$ only seems to exist for parts of its domain. But it must have an integral for its entire domain. What's happening here?”

First of all, as you correctly say, the integrand $f(x) =\frac{x^2}{x^2+x-2}$ has vertical asymptotes at $x=-2$ and at $x=1$ . This means that its domain is not, as you seem to think, the entire number line, but rather the number line with $-2$ and $1$ removed, i.e. $(-\infty,-2)\cup(-2,1)\cup(1,\infty)$ . Then, you do not distinguish clearly enough between the indefinite integral $$\int \frac{x^2}{x^2+x-2} dx$$ and a definite integral (with one fixed bound $a$ and one moving boundary $x$ ) of the form $$F_a(x) = \int_a^x \frac{t^2}{t^2+t-2} dt$$ The indefinite integral $$\int \frac{x^2}{x^2+x-2} dx$$ is a purely symbolic notation which asks for the most general form of all antiderivatives o