Are all functions that have a primitive differentiable?
Solution 1:
A primitive for $f(x)=x^{1/3}$ is $F(x)=\frac{3}{4}x^{4/3}$, but $f'(0)$ doesn't exist, because $$\lim\limits_{x\to0}\dfrac{f(x)-f(0)}{x-0}=\lim\limits_{x\to0}\dfrac{x^{1/3}}{x}=\lim\limits_{x\to0}\dfrac{1}{x^{2/3}}=\infty$$
Solution 2:
Even more simply: $f(x) := |x|$.
Solution 3:
Any bounded function that has discontinuity at a single point is integrable but of course that function will be non differentiable.