If $f$ has antiderivative, must $|f|$ also have antiderivative? Suppose function $f:[a,b] \to \mathbb{R}$ has an antiderivative. This means there is some differentiable function $F$ so that $f(x) = F'(x)$ in all of $[a,b]$. 
Must the absolute value $|f|$ have an antiderivative?
What if we also specify that $f$ is Riemann integrable?  I know that if $f$ is Riemann integrable, then $|f|$ must be also, but not every integrable function has an antiderivative.
 A: I believe your understanding of the term antiderivative is correct. A function $F:[a,b] \to \mathbb{R}$ is an antiderivative of $f:[a,b] \to \mathbb{R}$, if it is differentiable and $F'(x) = f(x)$ for all $x \in [a,b]$.
It seems then the question is if $f$ has an antiderivative, then is it necessary that the function $x \mapsto |f(x)|$ has an antiderivative.  Specifically, does there exist differentiable $G:[a,b] \to \mathbb{R}$ such that $G'(x) = |f(x)|$ for all $x \in [a,b]$?
The answer is no  -- it is not necessary.
A counterexample is
$$f(x) = \begin{cases}\cos(x^{-1}), & 0 < x \leqslant   1 \\ 0, & x= 0 \end{cases}$$
Here $f$ has an antiderivative 
$$F(x) = \int_0^x \cos (t^{-1}) \, dt$$
Since $f$ is continuous on $(0,1]$ we have, directly by the FTC, that $F'(x) = \cos(x^{-1})$ for $0 < x \leqslant 1$.  We can also verify that $F(0) = 0$ and
$$\tag{*}F'(0) = \lim_{x \to 0} \frac{F(x) - F(0)}{x} = \lim_{x \to 0}\frac{1}{x}\int_0^x \cos (t^{-1}) \, dt = 0 = f(0) $$
The details of deriving the limit (*) are given here.
On the other hand, we have 
$$g(x) = |f(x)| = \begin{cases}|\cos(x^{-1})|, & 0 < x \leqslant   1 \\ 0, & x= 0 \end{cases}$$
Suppose there exists an antiderivative $G$ such that $G'(x) = g(x)$ for all $x \in [0,1]$. Since $g$ is Riemann integrable (continuous except at one endpoint), it follows by the FTC that
$$G(x) - G(0) = \int_0^x g(t) \, dt$$
By continuity of $g$, clearly have  $G'(x) = |\cos(x^{-1}|$ for $0 < x \leqslant 1$.
However, 
$$\tag{**}G'(0) = \lim_{x \to 0} \frac{G(x) - G(0)}{x} = \lim_{x \to 0}\frac{1}{x}\int_0^x |\cos (t^{-1})| \, dt = \frac{2}{\pi} \neq 0 =  g(0) $$
The details of deriving the limit (**) are given here in robjohn's answer.
This proves that no such antiderivative $G$ exists.
Aside 
The issue of integrability of $f$ is peripheral as there are Riemann integrable functions without antiderivatives.  Furthermore, if we have 
$$F(x) = \int_{[a,x]} f,$$ 
as a Lebesgue integral and $F'(x) = f(x)$ almost everywhere but not for every $x \in [a,b]$, then $F$ does not qualify as an antiderivative.
A: Note the following quote from the wikipedia article on the Henstock–Kurzweil integral (under Properties)

In general, every Henstock–Kurzweil integrable function is measurable,
  and f is Lebesgue integrable if and only if both f and |f| are
  Henstock–Kurzweil integrable. This means that the Henstock–Kurzweil
  integral can be thought of as a "non-absolutely convergent version of
  Lebesgue integral". It also implies that the Henstock–Kurzweil
  integral satisfies appropriate versions of the monotone convergence
  theorem (without requiring the functions to be nonnegative) and
  dominated convergence theorem (where the condition of dominance is
  loosened to g(x) ≤ fn(x) ≤ h(x) for some integrable g, h).
If F is differentiable everywhere (or with countable many exceptions),
  the derivative F′ is Henstock–Kurzweil integrable, and its indefinite
  Henstock–Kurzweil integral is F.

From the highlighted sentences (and the Henstock–Kurzweil differentiation theorem), it follows that if $f$ has an anti-derivative, then $|f|$ will also have an anti-derivative if and only if $f$ is Lesbegue integrable.
