If $f(x)=x^2 \sin(1/x^2)$ for $x\neq 0$, $f(0)=0$ then $\int_0^1 |f'(x)| dx = \infty$

I am trying to prove if $f(x)=x^2 \sin(1/x^2)$ for $x\neq 0$, $f(0)=0$ then $\int_0^1 |f'(x)| dx = \infty$

so $f'\notin L^1$

This came up in Rudin's Real and Complex Analysis Section 7.16 (a).

I tried showing;

$\int_0^1 \frac{2}{x}| x^2 \sin(1/x^2)- \cos(1/x^2)| dx=\infty$

with not much luck.

It seems like it should be simple but I am having trouble proving the integral is infinite.

I found some comments on stackexchange of people saying this integral is infinite but I couldn't find a proof.

Things I tried include showing $|\int_0^1 f'(x) dx|=\infty$ which didn't work because I believe this integral is finite, I tried using Jensen's inequality but it is not valid I think in this case, I also tried bounding $|x^2 \sin(1/x^2)-\cos(1/x^2)|$ from below. I noticed $x^2 \sin(1/x^2)\to 0$ as $x \to 0$ so perhaps this could help.