Zion Wheeler

2022-06-21

Explanation how it can be that ${f}^{\prime }\left(x\right)=g\left(x\right)$ but wolfram alpha says $\int g\left(x\right)\ne f\left(x\right)$?
I've just started learning about antiderivatives/primitive functions/indefinite integrals, and I have the functions
$f\left(x\right)=3\mathrm{ln}\left(\left(\frac{x+2}{3}{\right)}^{2}+1\right)$
$g\left(x\right)=\frac{2\frac{x+2}{3}}{\left(\frac{x+2}{3}{\right)}^{2}+1}$
I came to the conclusion that ${f}^{\prime }\left(x\right)=g\left(x\right)$ so that $\int g\left(x\right)=f\left(x\right)$ and I wanted to check with wolfram alpha, but wolfram says that $\int g\left(x\right)\ne f\left(x\right)$ even though it says ${f}^{\prime }\left(x\right)=g\left(x\right)$.
It seems to me that this violates the definition of antiderivative?

### Answer & Explanation

scoseBexgofvc

Step 1
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 $f\left(x\right)+C$ 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, $\int g\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx\ne f\left(x\right)$. The result of the integration process is not a function, but a set of functions. Those are slightly different concepts.

Fletcher Hays

Explanation:
The symbol $\int g\left(x\right)dx$ means the set of all antiderivatives of g. Since f is an antiderivative of g , we have $\int g\left(x\right)dx=f\left(x\right)+C.$

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