Failure of Newton-Leibniz formula

Question 1. If $f=x^2\sin(x^{-2})$, then $f'$ exists everywhere (including $x=0$) but $f'$ is not Lebesgue integrable on $[0,1]$ (precisely because of the singularity at $x=0$).(Taken from here).

Question 2. By this thread, $f$ is absolutely continuous on each segment $[a,b]$. Then the positive answer follows from the fundamental theorem of calculus for Lebesgue integral (see, for instance, [?, Theorem 6.10], [Ba], [Be, Theorem 2]).

References

[Ba] Diómedes Bárcenas, The Fundamental Theorem of Calculus for Lebesgue Integral, Divulgaciones Matemáticas 8:1 (2000), 75-85.

[Ba] Stephen Becker, Absolutely continuous functions, Radon-Nikodym Derivative, APPM 5450 Spring 2016 Applied Analysis 2.

[?] “Section 6.5. Integrating Derivatives: Differentiating Indefinite Integrals”.