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Just a quicky, which one of these two is correct? Logarithmic integrals.

  1. May 5, 2009 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations
    None really......I suppose [tex]\int\frac{a}{bx + c}[/tex] = [tex]\frac{1}{b}ln.a.(bx + c)[/tex]

    3. The attempt at a solution

    Pretty much, using the formula above gives you:


    But I was thinking, if you rearrange [tex]\frac{ds}{s$^2$}[/tex] to [tex]s$^{-1}$[/tex] then presumable it integrates to -[tex]\frac{1}{s}[/tex] + c?

    4. My own comments:
    I didn't know that this BB would support LaTeX, that's awesome! I hope it's formatted correctly, i'm not too good with latex.

    edit: it didn't format correctly, original question is integration of ds/s^2
    My two possible answers are: ln (s^2) OR -1/s^2.

    I hope that makes sense.

    Thanks for any help given :)
    Last edited: May 5, 2009
  2. jcsd
  3. May 5, 2009 #2


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    Homework Helper

    Both answers are wrong. What is the derivative of each function you have calculated so far? If it's not 1/s^2 the answer is incorrect. Write 1/s^2 as [itex]s^{-2}[/itex]. Now use the normal power rule for integration/differentiation, do you see why it breaks down for 1/s? (important)
  4. May 5, 2009 #3


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    Science Advisor

    No, you cannot do
    [itex]\int \frac{1}{s^2}ds[/itex]
    as a logarithme. While it is true that [irtex]\int (1/x) dx= ln|x|+ C[/itex], because, of course, (ln|x|+ C)'= 1/x, it is NOT true that [itex]\int (1/f(x)) dx is ln|f(x)|+ C[/itex]. It is not true because, if you were to differentiate that, you would have to use the chain rule:
    (ln|f(x)|+ C)'= [1/f(x)]f'(x) which is not the same as 1/f(x).

    The only correct way to [itex]\int (1/s^2)ds[/itex] is to write it as [/itex]\int s^{-2} ds[/itex] and use the power rule.
  5. May 5, 2009 #4
    That's what I was trying to do, write it as [tex]\int s^{-2}[/tex] which surely then goes to [tex]-s^{-1}[/tex] ?

    Thanks by the way, very speedy responses.
  6. May 5, 2009 #5


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    Homework Helper

    Yes that's correct. Now try to use that same rule for [itex]s^{-1}[/itex] so you understand why 1/s is a very special case.
  7. May 5, 2009 #6
    Ahhh yes, silly me. It would be [tex]\frac{-1}{s}[/tex] not -1/s^2

    I have that answer written down so I don't know why I wrote it differently on here :P

    As for [tex]\frac{1}{s}[/tex] that would integrate to [tex]0s^{0}[/tex].

    And anything to the power zero is 1. But multipled by zero would be zero. So it's nothingness. Or is it like a divide by zero? You can write it, but you can never really calculate it?

    I can't this one to format correctly, hope you understand.
    Last edited: May 5, 2009
  8. May 5, 2009 #7
    Oh wait, am I being stupid?

    integral of 1/s wouldn't be the same as 1/f(x) so you'd then use that rule? So then instead of getting zero, which is wrong because surely it isn't zero. That where you use logs?

    because if you get the answer zero you can't differentiate zero to get back to 1/s. And that's why you use that log rule?
  9. May 5, 2009 #8


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    Homework Helper

    Well the power rule for integration basically states that you have to raise the power by one and then divide the function by its new power. So for the function s^-1 we raise it by one which yields s^0 and then divide through its new exponent which is 0, which yields 1/0 s^0, which in turn is an indeterminate expression.
  10. May 5, 2009 #9
    Yup sorry, I'd like to blame my stupidity on the hour of the morning, but it's afternoon.
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