Integrability of derivatives

I believe this answers the question:

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MR0425042 (54 #13000) Goffman, Casper A bounded derivative which is not Riemann integrable. Amer. Math. Monthly 84 (1977), no. 3, 205--206.

In 1881 Volterra constructed a bounded derivative on $[0,1]$ which is not Riemann integrable. Since that time, a number of authors have constructed other such examples. These examples are generally relatively complicated and/or involve nonelementary techniques. The present author provides a simple example of such a derivative $f$ and uses only elementary techniques to show that $f$ has the desired properties.

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The paper is available here:

http://math.uga.edu/~pete/Goffman77.pdf

Open interval $(a,b)$ easy ... make $f'$ unbounded, say $f(x) = \sqrt{x}$ on $(0,1)$.

Requiring differentiability even at the endpoint, the counterexample must be more elaborate. But still an unbounded function is not Riemann integrable, so take some $x^a \sin^b x$.

Even allowing improper Riemann integrals or Lebesgue integral is not enough to avoid the hypothesis that $f'$ is integrable. The Henstock-Kurzweil integral is needed to recover $f$ from $f'$ which exists everywhere on $[a,b]$ in general.

I remember, that there was an example of such a function in the book Counterexamples in Analysis. Just wanted to mention it for the sake of completeness. It can be found in Chapter 8 (Sets and Measure on the Real Axis), Example 35 (A bounded function possessing a primitive on a closed interval but failing to be Riemann-integrable there.)