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Integration and natural logs

  1. Sep 19, 2014 #1
    1. The problem statement, all variables and given/known data
    Integrate (-m-kx)^-1 dx from a to b.





    3. The attempt at a solution

    So using the integral form I get the integral of (-m-kx)^-1 dx is (-1/k)*[ln(-m-kx)] with the bracketed expression being evaluated between a and b.

    (-1/k)*[ln(-m-ka) - ln(-m-kb)]
    My first problem with this is that if we assume m,k,a,b are all positive then the negative argument of the natural log function will lead the function to not have a proper result, but if we use properties of log's to adjust the expression we arrive at an equivalent expression:

    (-1/k)*ln((-m-ka)/(-m-kb))
    which is equivalent to
    (-1/k)*ln((m+ka)/(m+kb))
    which DOES have a proper answer since the argument of the log function is now positive... what voodoo is going on here!?

    Furthermore and most frusturating of all is that the back of my book says the integral should come out to (-1/k)*ln((m+kb)/(m+ka)). However this book is infamous for typo's so the answers can't be trusted. Any mathamagical mavericks out there wanna spread some light on my dilemma's?
     
  2. jcsd
  3. Sep 19, 2014 #2

    statdad

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    Homework Helper

    Why not start the integration as
    [tex]
    \int_a^b \dfrac{1}{-m-kx} \, dx = \int_a^b \dfrac{-1}{m+kx} \, dx = -\int_a^b \dfrac 1 {m+kx} \, dx
    [/tex]

    and see how your integral turns out?
     
  4. Sep 20, 2014 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Whether the numbers are positive or not is not directly relevant: [tex]\int \frac{du}{u}= ln|u|+ C[/tex].

    Of course, you will want to let u= -m- kx.
     
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