Integrating a “falling object” equation involving $e$

I'm a programmer and my maths skills were probably never sufficient for this problem. I have a nice curve that describes the velocity ($$v$$) of a "falling" object over time ($$t$$) that accelerates to a terminal velocity (note, this is not a physical simulation, but essentially a sort of "tweening" curve).

$$v = \frac{1}{-e^{2t}} + 1$$

From this, I'm hoping to find two expressions that:

1. calculates the distance at time $$t$$
2. calculates the time at which the object reaches distance $$d$$

My problems are basically that even if I knew how to express the problem correctly mathematically (and frankly I don't), I never completely understood how to go from an integral to a nice, simple expression. And compilers only care for expressions. Oh, and I've chosen an equation that features $$e$$.

• wolframalpha.com/input/?i=integrate+1%2F(-e%5E(2t))+%2B+1 – Matti P. Jul 17 '19 at 10:09
• In addition to the integral, you'd also need the initial conditions (location and velocity in the beginning) – Matti P. Jul 17 '19 at 10:09
• @MattiP I'm assuming starting with $v = 0$ and $d = 0$ – Andrew Jul 17 '19 at 10:18

We have $$v(t) = \frac{dx}{dt} = 1-\frac{1}{e^{2t}}$$ So $$x(t) = \int 1-{e^{-2t}} dt = t + \frac{1}{2}{e^{-2t}} + K$$ where $$K$$ is a constant that you can find solving $$x(0)=x_0$$.
The time at which the object is at the position $$x$$ is given by $$x = t + \frac{1}{2}{e^{-2t}} + K$$ This is a bit tricky for me but Wolfram gives $$t = \frac{1}{2}(W_n(-e^{2K-2x}) + x - K)$$ where $$W_n$$ is the analytic continuation of the product log function.