Integrating a time derivative let $x$ be a function of time  $$x = x(t)$$ 
what is the integral of $$\int \frac{\dot x}{ \sqrt{1+ \dot x^2}} \, dt$$
I have tried using trigonometric substitution of $$ \dot x = \tan(\theta) $$
but I end up at the same point but with trigonometric terms

You can integrate by parts,
$$\int \frac{\dot x}{\sqrt{1+\dot x^2}}\mathrm d x=\frac{x}{\sqrt{1+\dot x^2}}+\int \frac{\dot x^2 \ddot x}{(1+\dot x ^2)^{\frac32}} \mathrm d x.$$
Since $\mathrm{d} \dot x = \ddot x \; \mathrm{d} t$, plug into the previous integral, 
$$\int \frac{\dot x}{\sqrt{1+\dot x^2}}\mathrm d x=\frac{x}{\sqrt{1+\dot x^2}}+\int \frac{\dot x^2}{(1+\dot x ^2)^{\frac32}} \mathrm d \dot x,$$
which is solvable by the trigonometric substitution you proposed.
