Explanation how it can be that $f'(x) = g(x)$ but wolfram alpha says $\int g(x) \neq f(x)$?

Question

I've just started learning about antiderivatives/primitive functions/indefinite integrals, and I have the functions

$f(x) = 3 \ln((\frac{x+2}{3})^2 + 1)$

$g(x) = \dfrac{2\frac{x+2}{3}}{(\frac{x+2}{3})^2 + 1}$

I came to the conclusion that $f'(x) = g(x)$ so that $\int g(x) = f(x)$ and I wanted to check with wolfram alpha, but wolfram says that $\int g(x) \neq f(x)$ even though it says $f'(x) = g(x)$.

It seems to me that this violates the definition of antiderivative?

Answer 1

The symbol $\int g(x) dx$ means the set of all antiderivatives of $g$. Since $f$ is an antiderivative of $g$ , we have

$\int g(x) dx=f(x)+C.$

Answer 2

The function $f(x)$ is one of the infinite number of antiderivatives of the function $g(x)$ which differ only by a constant. In the case of $f(x)$, that constant, typically denoted as $C$, is 0. When you integrate $g(x)$, you get a whole set of functions. Not just one function. That set is usually denoted $f(x)+C$ where $C\in\mathbb{R}$. Therefore, this means that there are going to be as many functions in that set as there as real numbers—that is, an infinite number. That's why, technically speaking, $\int g(x)\,dx\neq f(x)$. The result of the integration process is not a function, but a set of functions. Those are slightly different concepts.