$\int f(x)\,dx - \int f(x)\,dx$ which is true 
$$\int f(x)\,dx - \int f(x)\,dx = 0$$
or 
$$\int f(x)\,dx - \int f(x)\,dx=c\text{ ?}$$
with $c$ some arbitary constant.
My intuition says that 'something' subtracted by itself is always zero. But if we solve each term separately, we can have different integration constant that will lead the result does not need to be zero.
 A: $\int f(x)\, dx$ generally means the set of all anti-derivatives of $f(x)$.
If you have a set $A$ and a set $B$ then in this context $A-B$ should be interpreted as $\{a-b:\ a\in A, \ b\in B\}$.
Therefore, $\int f(x)\, dx - \int f(x)\, dx = \mathbb R$ (assuming we are talking about real integrable functions). The resulting subtraction is whatever set the "$C$" in the "$+C$" is allowed to come from, as this is the set of all $a-b$ with $a$ and $b$ both antiderivatives of $f(x)$.
But this is more of a question about ambiguity of definitions. You should avoid writing this in practice, if at all possible.
A: Neither is strictly correct: the expression $\displaystyle\int f(x)\,dx$ is not really unambiguously defined as identifying some particular mathematical object except when the context makes it clear.  When it has a precisely defined meaning, it's because of the context that it's precisely defined.
If you mean "What do you get when you subtract an antiderivative of $f$ from another antiderivative of $f$?", then the answer is that you get a constant, or in some cases (where the domain is not connected) a piecewise constant.
This problem comes up sometimes when integrating by parts.  One has
$$
\int \text{whatever} \,dx = \int u\,dv = uv - \int v\,du = (\text{some expression}) - 4\cdot\int\text{same thing}\,dx
$$
where "same thing" mean the very same integral that is expressed on the extreme left above as $\displaystyle\int\text{whatever}\,dx$.  So we add $\displaystyle 4\cdot\int\text{same thing}\,dx$ to both sides, getting
$$
5\int\text{whatever}\,dx = (\text{some expression}) + \text{constant}.
$$
Here the $\text{“}\cdots+\text{constant''}$ cannot be dispensed with.
This occurs in one commonplace method of integrating the cube of the secant function, and in things like $\displaystyle\int e^{ax}\cos(bx)\,dx$.
A: You can combine to get $\int (f(x)-f(x))dx= \int 0 dx=C$.
