Is indefinite integration suspect?

Basically, there's a type error: "$\int f(x)\,dx$" is a perfectly meaningful thing, but that thing is not a single function - rather, it's a set of functions.

The point is that a function doesn't have a unique antiderivative. For example, ${x^2\over 2}$ is an antiderivative of $x$ (with respect to $x$ of course), but so is ${x^2\over 2}-4217$. It's not the indefinite integral which is suspect, but rather the notation we use around it - specifically, the way we use "$=$." Properly speaking, $\int f(x)dx$ refers to a set of functions.

This is generally addressed by including a constant of integration, so that we write $$\int x\,dx={x^2\over 2}+C$$ to mean "The set of antiderivatives of $x$ is the set of functions of the form ${x^2\over 2} + C$ for $C\in\mathbb{R}$."

• That said, blindly adding a constant of integration still doesn't always fix the problem: let $f(x)=-{1\over x^2}+1$ if $x>0$ and $-{1\over x^2}-1$ if $x<0$; what's the derivative of $f$, and does $f$ have the form $-{1\over x^2}+C$ for some fixed real number $C$?

The other answers have made good points about constants of integration but this is not actually what I meant, although it is related. What I meant is what lulu says in the comments: writing antiderivatives this way misleads you about the relationship between the $x$ on the LHS (which is a dummy variable) and the $x$ on the RHS (which is not). The "real" $x$ on the LHS is one of the bounds of integration, which is being suppressed in the notation.

The sense in which this is misleading becomes clearer once you start considering double integrals, which is the context of the question you link to. If it makes sense to write $\int f(x) \, dx = g(x)$, then surely it also makes sense to write $\int g(x) \, dx = h(x)$, right? Then does it make sense to write

$$\iint f(x) \, dx \, dx = h(x)$$

or not? What do you think?

For one example, the familiar old "formula" $$ \int \frac{1}{x} \ dx = \ln|x| + C $$ is false (unless you define the indefinite integral VERY carefully). This purports to say that any antiderivative of $f(x) = \frac{1}{x}$ must take the form $F(x) = \ln|x|+C$ for some fixed constant $C$. But this is only true over a connected interval. For example, the function $$ G(x) = \begin{cases} \ln|x| +1, & x < 0\\ \ln|x|-1, & x > 0\end{cases} $$ satisfies $G'=f$, even though it is not expressible in the form $\ln|x|+C$. Done right, we should only define indefinite integrals over intervals (this is due to the Mean Value Theorem).