Let us consider in one dimension. Let $a>0$ be a given constant, $$ C[0,a]:=\{f:[0,a]\to\mathbb{R} \mid \text{$f$ is continuous in $[0,a]$}\}, $$ $$ C^{1}(0,a]:=\{f:(0,a]\to\mathbb{R} \mid \text{$f$ is once differentiable in $(0,a]$ & $f'$ is conti. in $(0,a]$}\} $$ and $$ L^{1}(0,a):=\{f:(0,a)\to\mathbb{R} \mid \text{$f$ is Lebesgue integrable in $(0,a)$}\}. $$ Here $f'$ denotes the first derivative of $f$.
My question : Is there function $f$ satisfying $f\in C^{1}(0,a]\cap C[0,a]$ but $f'\notin L^{1}(0,a)$?
For example, functions like square root $f(x)=x^{b}$, where $b\in(0,1)$, are failed since these belong to not only $C^{1}(0,a]\cap C[0,a]$ but also $L^{1}(0,a)$. The function
・$f(x)=x\sin(1/x)$ if $x\neq0$, $0$ if $x=0$
is also failed.
To begin with, I think there is not since $$ \left|\int_{0}^{a}f'(x)dx\right|=|f(a)-f(0)|<\infty. $$
I'm glad if you give examples as many as possible.
Your last function, $f(x) = x\sin (1/x),$ actually provides the example you seek. To prove this you need to show $[\cos (1/x)]/x \not \in L^1(0,1).$ Think about the integral of this function over $[1/(2n\pi+\pi/4), 1/2n\pi].$