Zion Wheeler

2022-06-21

Explanation how it can be that ${f}^{\prime}(x)=g(x)$ but wolfram alpha says $\int g(x)\ne f(x)$?

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

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

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

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

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

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

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

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

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

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

scoseBexgofvc

Beginner2022-06-22Added 20 answers

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

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

Fletcher Hays

Beginner2022-06-23Added 6 answers

Explanation:

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

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