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Here is a statement of the second part of the Fundamental Theorem of Calculus (FTC2), from a well-known calculus text (James Stewart, Calculus, 4th ed):

If $f$ is continuous on $[a,b]$, then $int_a^b f(x) , dx = F(a)-F(b)$, where $F$ is any [emphasis mine] antiderivative of $f$, that is, a function such that $F'=f$.

The following, however, seems to give a counterexample.*

Can someone resolve this for me?:

Let $f(x) = frac{1}{4 sin (x)+5}$.

$f$ is continuous on $[0, 2 pi]$:

Consider two antiderivatives of $f$, $F_1$ and $F_2$:

$$F_1(x) = frac{x}{3}+frac{2}{3} tan^{-1}left(frac{cos (x)}{sin (x)+2}right)$$

$$F_2(x)=frac{1}{3} left(tan ^{-1}left(2-frac{3}{tan left(frac{x}{2}right)+2}right)-tan^{-1}left(2-frac{3}{cot left(frac{x}{2}right)+2}right)right).$$

Using Mathematica, I've confirmed that both $F_1'= f$ and $F_2'= f$. According to my reading of the above statement of FTC(2), $int_0^{2pi} f (x) , dx = F_1(2pi)-F_1(0)= F_2(2pi)-F_2(0)$

However,

$F_1(2pi)-F_1(0)=2pi/3$

$F_2(2pi)-F_2(0)=0$

Note from the plots below that $F_1$ is continuous on $[a,b]$, while $F_2$ is not. Given all of this, it seems the sufficient condition for $int_a^b f (x) , dx = F(a)-F(b)$ is that the antiderivative be continuous on $[a,b]$, not the integrand.

$F_1 =$

$F_2=$

*I've taken this example function from a wolfram.com blog.

A differentiable function is continuous. If your "antiderivative" is not continuous, it is not an antiderivative of a continuous function.

@Martin Argerami posted a succinct and correct answer:

A differentiable function is continuous. If your "antiderivative" is
not continuous, it is not an antiderivative of a continuous function.

To elaborate, $F_2$ is not continuous due to the multi-valued nature of $arctan$. Since $tan$ is a periodic function (with a period of $pi$), this means that $arctan$ has an infinite number of values. Its plot looks like this:

To make it a proper function (one unique output value), we usually pick the primary branch (the one through the origin). However, notice that the primary branch is discontinuous at infinity - if we pass through positive infinity and come back out at negative infinity, we've jumped from $pi/2$ to $-pi/2$. What we really wanted was to move to the next branch up to preserve continuity, but to do that we'd no longer have a single valued function. We wouldn't have a function at all, really, since the common definition of function requires it to be single valued.

The discontinuities of $F_2$ are precisely where the argument of $arctan$ becomes infinite - where $2+cotfrac{x}{2}$ and $2+tanfrac{x}{2}$ become zero - approximately 4.07 and 5.36.