How can I find the antiderivative of dx/dv^2 in my physics book?

[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.

<br /> \int \frac {dv} {v^2} = - \frac {1} {v} <br /> [\tex]<br /> <br /> how does this happen??<br /> <br /> looking at the basic antiderivative formulas section in my maths book, it says that: <br /> <br /> &lt;br /&gt; \int \frac {dv} {v} = ln v&lt;br /&gt; [\tex}&lt;br /&gt; &lt;br /&gt; but nowwhere do i find info on how to do the problem in my physics book.&lt;br /&gt; &lt;br /&gt; would reallly appreciate the help! :)&lt;br /&gt; &lt;br /&gt; +edit+&lt;br /&gt; &lt;br /&gt; sorry, very new to the forums.. why doesn&amp;#039;t the latex work? &lt;img src=&quot;https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f615.png&quot; class=&quot;smilie smilie--emoji&quot; loading=&quot;lazy&quot; width=&quot;64&quot; height=&quot;64&quot; alt=&quot;:confused:&quot; title=&quot;Confused :confused:&quot; data-smilie=&quot;5&quot;data-shortname=&quot;:confused:&quot; /&gt; &lt;br /&gt; &lt;br /&gt; just in case others can&amp;#039;t view it as well.. what i meant was.. &lt;br /&gt; &lt;br /&gt; how do you integrate dv/v^2? i know the answer is -1/v but don&amp;#039;t know the rules for this. the closest rule i can find is that integrating dv/v is ln v..&lt;br /&gt; &lt;br /&gt; incidentally, i don&amp;#039;t think I&amp;#039;ve activated this account properly.. how can i activate my account? (i deleted the activate account e-mail i think)

 

[inha]

the closing tag should be [/tex]

 

[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.

 

[whozum]

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

\int \frac{dv}{v} = ln|v|

 

[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.

 

[kape]

ah thanks a lot! thank you inha for that [/tex]! 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} <br /> = \int \frac{1}{v^2} dv<br /> = \int (v^-2) dv<br /> = \frac{v^-1}{-1}<br /> = \frac {\frac{1}{v}} {\frac{-1}{1}}<br /> = \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}?

 

[whozum]

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

a^{apples} = a^{apples}

 

[kape]

thanks! :)

 

[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.