Question

Although this was at first a physics exercise, turned out to be a pretty hard calculus problem:
How do I solve this integral

Answer 1

\[\int\limits \frac{ v }{ f _{0} + kv } dv \]

Answer 2

Are fo and k constants? If so, integration by parts should make it easy.

Answer 3

by parts ?
adjust the numerator,
v = kv/k
\(\large \int\limits \frac{ v }{ f _{0} + kv } dv = (1/k)\int \frac{ kv }{ f _{0} + kv } dv \\ = \large (1/k)\int \frac{ kv+f_0-f_0 }{ f _{0} + kv } dv \)

Answer 4

\(=\large (1/k)\int \frac{ kv+f_0 }{ f _{0} + kv }- \frac{f_0 }{ f _{0} + kv } dv \)
got this ?

Answer 5

i've brought it down to \[\frac{ 1 }{ k² } \int\limits\limits \frac{ u }{ f _{0} + u } du => u = kv\]

Answer 6

the first term is just 1
and the 2nd term is easy to integrate, can you ?

Answer 7

yeah, that is pretty clear!

Answer 8

although I am not used to this method. Is it just logic or are there mechanisms to know when to apply this?

Answer 9

Integration by parts is pretty simple here... you'll have an integral of an ln function, which is relatively easy to integrate.

Answer 10

you can think of such thing 'adjusting the numerator' that way, if you have enough practice....now since you've seen this method, maybe next time you'll be able to think of it on your own.

Answer 11

oh, actually it is even easier to see after changing variables u = kv. yeah, I can see that clearer now thanks!!

Answer 12

welcome ^_^

Answer 13

\[\int\limits\limits \frac{ v }{ f _{0} + kv } dv \]
let f = v so f' = 1
and let
g' = 1/(fo+kv) so g=(1/k)ln(fo+kv)

Then you just have to integrate g again to solve by parts.

Answer 14

i feel thats a long way out(integrating ln also requires by parts again)
Also, i think of by parts when we have 2 different functions, here both are algebraic.

Answer 15

Integrating the ln can be done with a table of integrals with a basic substitution (you can do it by parts again if you like, it's pretty easy)

Answer 16

yes yes, its easy, my point was its just longer....

Answer 17

Idk, it takes about the same effort as adjusting the numerator. But it also depends how familar you are with integration by parts, or if you're comfortable integrating ln functions.