 Unbounded derivative obstruction to Riemann integrability

A proper Riemann integral on a compact interval requires a bounded integrand, but a derivative can be unbounded. For $f(x)=x^2\sin (1/x^2)$ away from zero and $f(0)=0$, the difference quotient gives $f^{\prime }(0)=0$, whereas $f^{\prime }(x)=2x\sin (1/x^2)-2\cos (1/x^2)/x$ for $x\ne 0$. At $x_n=(2\pi n{)}^{-1/2}$, this derivative is $-2/x_n$ and is unbounded. The fundamental theorem of calculus version that integrates a derivative therefore needs an explicit integrability hypothesis.