Existence theorem for planar curves (II)

We follow with the above entry taking an 'easy' function $k$: find a planar curve as a   $y=f(x)$ where $k(x)=2x$.

By the given method in the last entry, we integrate $k$ between $a$ and $x$. Take $a=0$, obtaining  $g(x)=x^2$ (here $m=0$). Then the solution is
$$y(x)=\int_0^x \frac{g(t)}{\sqrt{1-g(t)^2}}dt.$$
In this case,
$$\int_0^x\frac{t^2}{\sqrt{1-t^4}}dt.$$
This integral is not possible to compute!

Other example. Take $k(x)=e^x$. Now
$$\int_0^x e^t dt=e^x-1.$$
If we take $m=-1$, then $g(x)=e^x$ and finally $$f(x)=\int:0^x\frac{e^t}{\sqrt{1-e^{2t}}}dt=\mbox{arc}\sin (e^x).$$In order to make sense in the integrand, we are assuming that  $x<0$. The solution is $y(x)=\mbox{arc}\sin (e^x)$ whose graphic is