It doesn't even match the endonym of "Bodwadmi" either,,,

*the manipulation of infinitesimals can actually be made rigorous by using something like measure theory (iirc) or another vaguely similar theory as a foundation for the definition thereof

I think it annoys me because it feels like the difficulty is not from the calculus, but from the final algebraic step. Like ln|1/sin x - 1/cot x| only involves functions you learn by high school, but solving for x in this equation is also the biggest difficulty in trying to solve f'(x)=sin(f(x))

Similarly, if we do the case where g=sin, f'(x) = sin(f(x)) gets us h(x) = ∫csc(x) dx. This integral is not difficult, and it comes to simply ln|csc x - cot x|, but inverting this function seems like a pain

a very complicated inverse function, so if you want to use an explicit form for it in f(x) = li⁻¹(x + C), it's quite painful. (more importantly, it's the inverse of an important function and so

...a non-elementary function, but that is actually (in a way) convenient for the integration step, since it's a very important one so it gets its own symbol and name: li, the logarithmic integral. So now we have li(f(x)) = x+C. Unfortunately, the logarithmic integral has a very complicated invers...

This then solves to f(x)=h⁻¹(x+C). However, if 1/g(x) is difficult to integrate or h is difficult to invert that'll make this method significantly more. The two most obvious examples to me are g=ln and g=sin (or g=cos):
For g=ln, f'(x) = ln(f(x)) gets us h(y)=∫dy/ln(y). This is a non-elementary f...

...for f'(x)=g(f(x)), it's quite easy to get a non-differential form by rearranging to f'(x)/g(f(x)), integrating with a u-sub of u=f(x) (or equivalently manipulating infinitesimals to get df/g(f) = dx and integrating ∫df/g(f) = ∫dx*) which gives h(f(x))=x+C, where h(y)=∫dy/g(y). This then solves...

Like something of the form f'(f(x)) + g(f(x)) = g(x)^f(x) - f(x)^g(x) is probably a bitch to solve because of the form (I mean fucking LOOK at it), and many equations in physics are a pain because of the number of variables (although the form is usually also a massive issue). But for f'(x)=g(x)f(x),