Need help on a simple but frustrating integral

[Deuterium2H]

I am reading Paul J. Nahn's excellent "Mrs. Perkins's Electric Quilt: and other Intriquing Stories of Mathematical Physics".

Unfortunately, while I can understand much of the complex analysis, I sometimes get hung up over the what I believe are rather elementary integrals which are never solved in a step by step fashion...but are just shown and then (with no explanation) given a solution. This can be a bit frustrating, especially if it has been years (decades) since one had advanced Calc, and has forgotten many of the common integral forms/solutions.

Anyways, I didn't even get through the damn Preface without encountering a problem.

We are given the following differential equation (integral):

ds = [ v₀ / (1 + kv₀t) ] dt

k is a Constant of proportionality, and v₀ is initial velocity, which obviously itself is a function of time.

The solution is given as:

s = ln[(1 + kv₀t)^(1/k)] + Z

How does one integrate the right side of the equation? I can get the sense that the solution will be a natural logarithm, due to the fact that we have a form INTEGRAL (dt/t).
Would someone be so kind as to break this down step by step for me. I am almost embarrassed to ask, but oh well. Thanks in advance!

 

[Deuterium2H]

Ah sugar...I think I might have remembered something. Is this one of those instances where we have an integral in which the numerator is a derivative of the denominator, and evaluated by the formula:

INT [ f'(x) / f(x) ] dx = ln |f(x)| + C ??

So,

s = (1/k) INT [ (kv₀)/(1 + kv₀t) ] dt

 

[Deuterium2H]

On further thought, I guess that does solve it. Would appreciate someone else's confirmation, however. Thanks again

 

[Char. Limit]

Yup. That's just about right.

 

[Deuterium2H]

Yup. That's just about right.

Thanks Char. Limit,

The only thing troubling me (still) is the fact that v₀ is itself a funtion of "t"...and it almost seems like there is a partial derivative involved. However, if I just treat
d(kv₀t)/dt = kv₀, then it all works out simply.

 

[Char. Limit]

Are you sure that v₀ is a function of t?

EDIT: Initial velocity is actually NOT a function of time. Indeed, it's not a function at all! It's a constant value.

 

[Deuterium2H]

Are you sure that v₀ is a function of t?

EDIT: Initial velocity is actually NOT a function of time. Indeed, it's not a function at all! It's a constant value.

AHHHHH Yes!

Once again, thanks Char. Limit. You are absolutely right, and I had totally overlooked that.