A classical and simple example is given by the function $x \mapsto \sin\frac{1}{x}$ extended by $0\mapsto 0$, on (say) $[-1,1]$:
It is discontinuous at $0$.
It is a derivative (= has an antiderivative) because it differs from the derivative of the differentiable function $x \mapsto x^2 \cos\frac{1}{x}$ by a continuous function (everything extended to $0$ by $0$).
It is Riemann-integrable on $[-1,1]$ by Lebesgue's criterion, as it is bounded with a set of discontinuities $\{0\}$ having Lebesgue measure zero.
Examples with more interesting sets of discontinuity points can be constructed by summing appropriately convergent series of suitably scaled translates of this function.
It is also worth emphasizing that the fundamental theorem of calculus is valid in the following form: $F$ is differentiable and $f := F'$ is Riemann-integrable on $[a,b]$ then the Riemann integral of $f$ from $a$ to $x$ equals $F(x)-F(a)$ (note that this is with no continuity assumption on $f$, so it applies to the above example). I don't have a reference for this at hand, but it's probably in many textbooks.
A classical reference concerning functions that are derivatives (= have an antiderivative) is Bruckner's Differentiation of Real Functions (2d ed 1994), which has various things to say about derivatives that are bounded and/or continuous a.e. (which is what you're asking about).
