Why are there apparently multiple general anti-derivatives for $f(x) = 1/x$?

Question

So I've learned that the general anti-derivative of $f(x)=1/x$ is:

$\ln|x|+C$

But it is also true, according to Wolfram Alpha, that the derivative of $\ln(2x)$ is also $1/x$.

So what gives? How can these facts both be true? Why isn't $\ln(2x) + C$ also an anti-derivative of $1/x$?

Is there something about the definition of anti-differentiation that I'm missing?

NOTE: I am in high school Calculus, and we have not yet covered integration. That being said, I have some limited familiarity with the notation for indefinite integrals (which I understand to be the same as anti-derivatives?), so as long as you're careful to explain things and use precise language, I suppose I'm fine with using integrals in any answers.

Answer 1

Remember that functions that differ by a constant have the same derivative (the converse is also true on intervals: if $f'(x) = g'(x)$ for all $x$ on an interval, then $g(x) = f(x)+C$ for some constant $C$).

That means that given any function $f(x)$, if there is one antiderivative then there are infinitely many: because if $F(x)$ is an antiderivative, then so is $F(x)+C$ for any constant $C$, and there are infinitely many choices of $C$.

Now notice that $\ln(x)$ and $\ln(kx)$ actually differ by a constant: the properties of the logarithm give you that $\ln(kx) = \ln(k)+\ln(x) =\ln(x)+C$, with $C=\ln(k)$. That's why $\ln(2x)$ is also an antiderivative for $\frac{1}{x}$ on $(0,\infty)$.

Answer 2

Every function that has an anti derivative has infinitely many antiderivatives. Let $f(x)$ be a function with an antiderivative $F(x)$. Then every antiderivative of $f(x)$ is of the form $F(x)+C$ for some constant $C$.

But wait, $\ln|2x|$ doesn't appear to be of the form $\ln|x|+C$! But in fact it is of this form. Using properties of the logarithm we see: \begin{align*} \ln|2x|=\ln(2|x|) = \ln(2)+\ln|x| \end{align*} so this is of the form $\ln|x|+C$ with $C=\ln(2)$.

It follows that $\ln|2x|+C$ is also an antiderivative of $\frac{1}{x}$. It can be re-written as: $$\ln|2x|+C=\ln|x|+(\ln 2+ C)$$ or as: $$\ln|2x|+\ln(e^C) = \ln|2x\cdot e^C| = \ln|(2e^C)x|.$$

The logarithm is a bit of a funny function in that multiplying the argument $x$ by a constant (i.e., squishing or stretching the function horizontally) is the same as translating it vertically by a constant. It seems very counterintuitive that these two actions could result in the same function, but it all boils down to the fact that $\ln(ab)=\ln(a)+\ln(b)$.