I have a difficult integral...please help!!!?

Question

The integral is: dv / (g-(k/m)v^2). g, k, and m are all constants and obviously v is my variable. it's supposed to end up in the form: du/(a^2-u^2) with a and u as substitutes. any help would be greatly appreciated! I need exact steps as I am writing a paper. Thanks!!

I'm new to this so I don't know how to respond to someone's answer....but Jared, I need to know what steps were involved in getting to that answer please. It looks right to me, I just have to have EVERYTHING step-by-step for this paper. Thanks so much!!

Answer 1

dv/(g-(k/m)v^2) = m/k * dv/(sqrt((gm/k)^2) -v^2)

ok first multiply the top and bottom by m

this gives m*dv/(gm-kv^2)

now, divide the top and bottom by k,

this gives m/k* dv/(gm/k-v^2)

now you need to have the 1st term in the denominator squared, so gm/k = sqrt((gm/k)^2)

that gives you the final answer

Answer 2

Try to get hte denominator in the form you want. What you have is:

g - k/m * v^2
You want the coefficient of v^2 to be 1. So factor the k/m out.
k/m * [mg/k - v^2]

So your expression becomes:
m/k * dv / [mg/k - v^2]

The mg/k is the equivalent of a^2.
The integral of du/(a^2 - u^2) = (1/a) * atanh(u/a)

So your integral will be:
m/k * sqrt(k/mg) * atanh (v*k/mg)

Answer 3

1/(a^2-u^2) = m/(a-u) +n/(a+u)
so
1=m(a+u)+n(a-u) so m= 1/2a and n= -1/2a
and The int is -1/2a (ln Iu-aI-lnIu+aI) = -1/2a lnI (u-a)/(u+a)I