Which one of them is correct: $F(x)= \int f(x)\mathrm dx + C$ or $\int f(x)\mathrm dx= F(x) + C$?

Question

I was doodling a bit at my leisure where I came across this:

$$\frac{\mathrm d}{\mathrm dx}F(x)= f(x) \\ \implies \mathrm dF(x)= f(x)\mathrm dx \;;$$ integrating $$\int \mathrm dF(x) = \int f(x)\mathrm dx \\ \implies F(x) - F(x_0) = \int f(x)\mathrm dx \\ \implies F(x)= \int f(x)\mathrm dx + F(x_0) = \int f(x)\mathrm dx + C\;.$$ where $F(x_0)$ is an arbitrary constant.

But I learned that

$$\int f(x)\mathrm dx = F(x) + C\;.$$

Both on differentiation give $$\frac{\mathrm d}{\mathrm dx} F(x) = f(x)\;.$$

But if both are correct, then that would mean $$\int f(x)\mathrm dx + C = \int f(x)\mathrm dx - C\;,$$ which wouldn't be true unless $C= 0.$ I don't know which one of them is correct. Can anyone tell me the correct piece? I've always learnt the later, after all:(

Answer 1

At this line $$F(x) - F(x_0) = \int f(x)dx$$ let $-F(x_0)=C$ and you're done.

Answer 2

You cannot take same $C$ on both expressions. You do not know that both of the constants will be same.

Answer 3

They both are the same. $$F(x)=\int f(x)dx+C\implies F(x)-C=\int f(x)dx\underset{D=-C}{\implies} F(x)+D=\int f(x)dx$$

Answer 4

You wouldn't have such a doubt if you had understood what $\int f(x)\, dx$ really means. It is the set of all functions whose derivative is $f$. Now, fix a single $F$ such that $F'=f$, and remark that $$ \left\{ F(\cdot )+C \mid C \in \mathbb{R} \right\} = \left\{ F(\cdot)-C \mid C \in \mathbb{R} \right\} $$ In any case, both identities in the title are essentially meaningless...