Of course, I voted Todd Trimble's answer, but I want to address an other aspect of the question.
Somehow, integration is not the inverse of differentiation. If $f:(a,b)\mapsto\mathbb R$ is differentiable, its derivative at each point $x$ is well-understandable, but $f'$ might be ugly as a function. A. Denjoy characterized (or tried to characterize ?) the functions $g:(a,b)\rightarrow\mathbb R$ that are derivatives of everywhere differentiable functions $f$. He used transfinite induction to resconstruct $f$ from $f'$. Of course, integration provides the answer when $g$ is continuous, or integrable, but not in general.
Example. If $\alpha\ge1$, the function $x\mapsto x\sin\frac1{x^\alpha}$ is differentiable everywhere, yet its derivative is not integrable. I am sure that there are much worse examples, where $f'$ is not integrable on any non-trivial interval. The only necessary conditions I see about the derivative $f'$ of an everywhere differentiable function are
- $f'$ is measurable, because it is the pointwise limit of a sequence of continuous functions (the quotients),
- $f'$ satisfies the intermediate value property: if $c\in(f'(x),f'(y))$ then there exists a $z$ between $x$ and $y$ such that $f'(z)=c$. This follows from Rolle's Theorem.