I want to parametrize this curve $\psi(t) = (t \cos\frac{1}{t}, t\sin\frac{1}{t})$, $t\in[-1,0)$ by its arc length. Now $\|\psi'(t)\|=\sqrt{1+\frac{1}{t^2}}$ is $\ne0, \forall t\in[-1,0)$, so this curve is regular.
Now the arc length of this curve is $s(t)=\int_{-1}^{t}\|\psi'(u)\|\,du$ and now I should solve for $t$ as $t=t(s)$, but the integral's solution it's very difficult and I can't find any way to solve for t.
What am I doing wrong?
As @Michael Hoppe already commented, you cannot solve for $t$ in the equation $$s=\tanh ^{-1}\left(\sqrt{t^2+1}\right)-\sqrt{t^2+1}$$ but you can have approximations.
For convenience, let $x=\sqrt{t^2+1}$ to make the problem $$s=\tanh ^{-1}(x)-x=\sum_{n=1}^\infty \frac {x^{2n+1}}{2n+1}$$ If we truncate to some order, we have the Taylor series that we can inverse to obtain for example $$x=y-\frac{y^3}{5}+\frac{3 y^5}{175}+\frac{2 y^7}{1575}-\frac{16 y^9}{202125}-\frac{362 y^{11}}{9384375}-\frac{49711 y^{13}}{12415528125}+O\left(y^{15}\right)$$ where $y=\sqrt[3]{3s}$.
Edit
In fact, there is a formal analytical solution.
Define $k=\tanh(s)$ to make $$k=-\tanh \left(x-\tanh ^{-1}(x)\right)=\frac{x \cosh (x)-\sinh (x)}{\cosh (x)-x \sinh (x)}$$ which leads to $$e^{-2x}=\frac{(k+1) }{(k-1) }\frac{ (x-1)}{ (x+1)}$$ the solution of which being given in terms of the generalized Lambert function (have a look at equation $(4)$ in the linked paper).