Where is the mistake in this false proof using indefinite integral? Indefinite integral of a function $f$, i.e $\int f(x)dx$ gives as result a infinite amount of functions such that $F'(x) = f(x)$ that differs by a constant called integration constant($C$), that is $\int f(x)dx = F(x) +C$
Also integral satisfies the property $\int f(x)+g(x)dx=\int f(x)dx + \int g(x)dx$
Using the indefinite definition and that property i can do the following procedure:
$$\int f(x)dx = \int f(x)dx$$
$$\int f(x)dx = \int f(x)+0 dx$$
$$\int f(x)dx = \int f(x)dx + \int 0dx$$
$$\color{red}{\int f(x)dx } = \color{red}{\int f(x)dx} + C_1, \text{with } C_1 \in \mathbb{R}$$
$$\color{red}{F(x)+C}=\color{red}{F(x)+C} + C_1$$
$$0=C_1$$
And since $C_1$ can take any real value we can have
$$0=1$$.
And i want to know, where is the mistake?
 A: Good question!
The problem has to do with vague language about what $\int f(x) dx + C$ means. It is not a particular function, but an equivalence class of functions, where two functions are equivalent iff they differ by a constant iff their derivatives are equal.
Unfortunately, that's not the most readable wikipedia page if you are unfamilair with this notion -- perhaps you can look here or at any 'introduction to mathematical proofs' kind of text. For now, you can think of it as a partitioning of the set of all functions into sets of 'equivalent functions,' where two functions are equivalent if they have the same derivative. An equivalence relation on a set is essentially the same notion as a partition of a set, so there's nothing lost by taking this point of view.
In particular, $0$ and $1$ refer to the same equivalence class, namely that of all constant functions. There is no contradiction, but you have understand that when you write equality between two such indefinite integrals you mean an equality between the corresponding equivalence classes.
Picking a value for $C$ amounts to choosing a representative for an equivalence class; and it is not contradiction to say that two equal equivalence classes have representatives that are not equal.
As the comments point out, the confusion at various points in your calculation will disappear when you define the indefinite integral this way. It's really a problem of ambiguity that will disappear when you internalize the correct definition.
For the calculus student, usually the end goal of indefinite integration is to do a definite integration. When you use the fundamental theorem of calculus to compute $\int_a^b f(x) dx$, as long you use the same value of $C$ for both endpoints then any choice of $C$ will give the same answer. So this is why intro calculus courses typically tell students to 'pick any C' -- but, as you've discovered, this can be conceptually misleading.
(This also explains why students get docked for not writing $+C$ -- a decision that would make more sense if we told them it was there to indicate that one was thinking of the equivalence class, and not a particular function...)
Also, you can shorten your paradox by considering $F(x) + C = \int f(x) dx = F(x) + C'$.
(See here for some further discussion: https://math.stackexchange.com/a/306927/54092 )
A: Derivative of a constant is zero.
$C$ is a constant, and  $C+C_1$ is an another constant
It will give different constants without a certain $(x;y)$ point
$$g_1(x)=\int f(x)dx \not= \int f(x)dx=g_2(x)$$ their primitive function are not equal
