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Homework Help: Indefinite Integrals

  1. Aug 16, 2007 #1
    1. The problem statement, all variables and given/known data

    Hey guys, I'm trying to teach myself how to integrate an indefinite integral.

    I just am wondering what you can do with something like this:

    2. Relevant equations

    [tex]\int[/tex] 15/(3x+1) dx

    3. The attempt at a solution

    I'm trying to figure out how to go backwards, but I don't see what terms, when derived, give you 15/3x+1 dx.

    Does anyone know a good way to quickly solve these sorts of problems?
     
  2. jcsd
  3. Aug 16, 2007 #2

    Dick

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    What about log(3x+1)?
     
  4. Aug 16, 2007 #3
    Ah. But wouldn't it have to be:

    15 ln(3x+1)

    because its derivative would be:

    15/(3x+1)

    Correct?

    So wouldn't the integral of the original problem be just:

    [tex]\int[/tex] 15/(3x+1) dx = 15ln(3x+1)

    Does it matter whether ln or log is used?
     
  5. Aug 16, 2007 #4

    Dick

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    15 is wrong, do the differentiation again and don't forget the chain rule. it doesn't matter whether it's log or ln because I mean natural logarithm by both. If you want to use logs to a different base then you'll have to adjust the coefficient. log(base a)x=ln(x)/ln(a).
     
  6. Aug 16, 2007 #5
    OK, thanks, lemme work this out.
     
  7. Aug 16, 2007 #6
    Ok, what about this thing:

    [tex]\int[/tex] 15/(3x+1) dx
    and if I factor out the 15:
    15[tex]\int[/tex] (3x+1) dx

    Now, does the constant just disappear?

    and the antiderivative of 1/(3x+1) is just

    log (3x+1)
    ???
     
  8. Aug 16, 2007 #7

    Dick

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    The derivative of log(3x+1) is 3/(3x+1). That's the answer now YOU tell me why.
     
  9. Aug 16, 2007 #8
    Because if you take the derivative as such:

    log(3x+1) dx

    You will get

    d/dx f(g(x)) = f`(g(x))(g`(x))

    Which means that:

    d/dx log(3x+1) = (1/(3x+1)) (3)

    = 3/(3x+1)

    But so now do I have to place a constant to make the derivative 15? I'm wondering if

    [tex]\int[/tex]15/(3x+1) = just log(3x+1)

    Shouldn't it be 5(log(3x+1)), to give it a 15 on top?
     
  10. Aug 16, 2007 #9

    Dick

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    Exactly, 5*log(3x+1).
     
  11. Aug 16, 2007 #10

    learningphysics

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    I think it's safer to use ln... some people use log to refer to base-10 logarithm by default...

    Also, antiderivative of 1/x = ln|x| (absolute value)

    So your answer would be 5*ln|3x+1|
     
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