Integration of a velocity function by partial fractions

[dustbin]

Homework Statement

I need to integrate

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

to show that the position function is given by

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

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:

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

and then performing long division to get

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

When I put that back into the integral I get

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

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?

 

[SammyS]

Homework Statement

I need to integrate

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

to show that the position function is given by

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

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.

...

It looks as if the substitution $\displaystyle u=e^{-2g\,t/V}$ should work just fine.

Show how you are trying to implement that.

 

[HallsofIvy]

I think I would be inclined to use the substitution $u= 1+ e^{-2gt/V}$ but it can be done either way.

 

[SammyS]

I think I would be inclined to use the substitution $u= 1+ e^{-2gt/V}$ but it can be done either way.

I agree.

This will make life easier.

 

[dustbin]

Okay, so I got the answer using Sammy's suggestion (though I did it again with HoI's and got it as well). Here is what I have:

let $$u = e^{-2gt/V}.$$ Then $$du = -\frac{Vdu}{2g}\frac{1}{u}$$ and $$lnu = -\frac{2gt}{V}.$$

Substituting gives $$-\frac{V^2}{2g}\int\frac{1-u}{u(1+u)}du.$$

By partial fractions (on the integrand) I got

$$-\frac{V^2}{2g}\int\frac{du}{u} + \frac{V^2}{2g}\int\frac{2du}{1+u}$$

which gives

$$= -\frac{V^2}{2g}lnu + \frac{V^2}{g}ln(1+u) + C'$$

where C' is C - ln2. Subsituting for lnu and u gives

$$= Vt + \frac{V^2}{g}ln(1+e^{-2gt/V}) - ln2 + C$$

which gives the position function

$$= Vt - \frac{V^2}{g}ln( \frac{1 + e^{-2gt/V}}{2} ) + C.$$

Since s(0) = 0, we have C = 0 and the position function is obtained.
Sorry if there are some errors with the Tex stuff... still learning.

 

[SammyS]

Okay, so I got the answer using Sammy's suggestion (though I did it again with HoI's and got it as well). Here is what I have:

let $$u = e^{-2gt/V}.$$ Then $$du = -\frac{Vdu}{2g}\frac{1}{u}$$ and

Good !

Of course that should be $\displaystyle dt = -\frac{Vdu}{2g}\frac{1}{u}\ .$

 

[dustbin]

Yes, sorry . I get lost in all the tex stuff :-)

Thank you for your help. Greatly appreciated!