# Integrating dx/dv^2 ?

1. May 9, 2005

### kape

Integrating dx/dv^2 ??

i'm trying to figure out an example in my physics book but i don't quite understand the maths.

$$\int \frac {dv} {v^2} = - \frac {1} {v} [\tex] how does this happen?? looking at the basic antiderivative formulas section in my maths book, it says that: $$\int \frac {dv} {v} = ln v [\tex} but nowwhere do i find info on how to do the problem in my physics book. would reallly appreciate the help! :) +edit+ sorry, very new to the forums.. why doesn't the latex work? just in case others can't view it as well.. what i meant was.. how do you integrate dv/v^2? i know the answer is -1/v but don't know the rules for this. the closest rule i can find is that integrating dv/v is ln v.. incidentally, i don't think i've activated this account properly.. how can i activate my account? (i deleted the activate account e-mail i think) Last edited: May 9, 2005 2. May 9, 2005 ### inha the closing tag should be$$ 3. May 9, 2005 ### Nylex I can't use LaTeX properly, so I'll just write it out again. The integral of x^n dx = [x^(n + 1)]/(n + 1) + C for all n (positive or negative), except n = -1, in which case, it is ln x + C. So, in your case you have the integral of dx/x^2 = x^-2 dx and just apply the rule above. 4. May 9, 2005 ### whozum $$\int \frac{dv}{v^2} = \frac{-1}{v}$$ $$\int \frac{dv}{v} = ln|v|$$ 5. May 9, 2005 ### dextercioby Does it make sense to have $\frac{1}{v^{2}}$ in one member and $\frac{1}{v}$ in the other...?I think not. Daniel. 6. May 10, 2005 ### kape ah thanks a lot! thank you inha for that$$! and thank you nylex for that explanation!! i never knew that it didn't apply when n = -1.. now that i look in my textbook, it does specify that.. though wrongly.. it says when n = 1 (checking in another book, shows that it is indeed when n = -1)

but what do you mean to have $$\frac {1} {v^2}$$ in one member and $$\frac {1} {v}$$ in another? i don't understand at all. (what's a member, first of all?)

and thanks whozum for the proper rendition in latex! but why is it $$\int \frac{dv}{v^2} = \frac{-1}{v}$$ and not $$\int \frac{dv}{v^2} = - \frac{1}{v}$$? i know that both actually mean the same thing but when i do the calculation i reach $$\int \frac{dv}{v^2} = \frac{1}{-v}$$.. which means that i may doing something wrong so, just in case.. i'll write out how i do it and if you could be so kind, could you point out where i went wrong?

$$\int \frac{dv}{v^2} = \int \frac{1}{v^2} dv = \int (v^-2) dv = \frac{v^-1}{-1} = \frac {\frac{1}{v}} {\frac{-1}{1}} = \frac{-1}{v}$$

i guess this is where the difference is? $$\frac{1}{-1}$$ instead of $$\frac{-1}{1}$$? does it matter? i haven't done any maths for a long long time and i have no idea if it does or not..

** by the way, why is it that the "1" is not in superscript in $$\frac{v^-1}{-1}$$?

Last edited: May 10, 2005
7. May 10, 2005

### whozum

The answers you gave are identical and both correct. To superscript it, put { } about the exponential argument

a^{apples} = $$a^{apples}$$

8. May 10, 2005

### kape

thanks! :)

9. May 10, 2005

### HallsofIvy

The general formula, by the way is that the anti-derivative of $$x^n$$ is
$$\frac{1}{n+1}x^{n+1}+ C$$ unless n+1= 0 (since we can't have 1/0) . That happens, of course, when n= -1. The anti-derivative of $$x^{-1}= \frac{1}{x}$$ is ln|x|. There are a variety of ways to show that- the simplest is to define ln|x| in that way.