calculus - Why does $frac{sin x}{x}$ have no anti-derivative

I've been messing around with indefinite integrals. Watching some youtube videos and I found the Sinc function and that it has no finite Anti-derivative.
Desmos being my favourite program ever I decided to graph to the equations
$y=\frac{\sin(x)}{x}$ and $y=\int_0^x\frac{\sin(a)}{a}da$ (only way I could do indefinite integrals in Desmos)

If you do this you might notice $y=\frac{\sin(x)}{x}$ looks very similar to $y =\sinh(x)$ and $y=\int_0^x\frac{\sin(a)}{a}da$ looks very similar to $y=\arctan(x)$

I wondered how we know that $\frac{\sin(x)}{x}$ has no finite anti-derivative because these simple functions give seemingly accurate approximations

Best approximation I could get for it in the short time I had was:
$$\int \sinh(x)-\frac{d}{dx}\bigg[\frac{\sin(x)}{x}\bigg]-\frac{\pi}{2}$$

or
$$2\arctan(e^x)+\frac{\sin(x)}{x^2}-\frac{\cos(x)}{x}-\frac{\pi}{2}$$
(In all the examples I'm assume $C=0$ because it's not important and I'm assuming any points with $\frac{0}{0}$ evaluate to their limits as $x$ tends to $0$)