- #1

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## Homework Statement

I need to integrate

[tex] v(t) = V( \frac{1- e^{-2gt/V}}{1+ e^{-2gt/V}}) [/tex]

to show that the position function is given by

[tex] s(t) = Vt + \frac{V^2}{g}ln(\frac{1 + e^{-2gt/V}}{2}) [/tex]

## Homework Equations

g is the acceleration due to gravity

V is the terminal velocity

## The Attempt at a Solution

I've tried a few different approaches. I've tried letting u = e^(2g/V) but I never get anywhere by doing this after finding du and substituting that in. My closest answer has been:

[tex]

-V(\frac{e^{-2gt/V} - 1}{e^{-2gt/V} + 1})

[/tex]

and then performing long division to get

[tex]

-V(1 - \frac{2}{1 + e^{-2gt/V}})

[/tex]

When I put that back into the integral I get

[tex]

-V\int dt + 2V\int\frac{dt}{1 + e^{-2gt/V}}

[/tex]

but I don't get the position function and have that negative at the start of the expression... can I get some hints on how to approach this?