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Intergration using logarithms

  1. Jan 13, 2008 #1
    Evaluate the integral.

    [tex]\int3/(3x-2) dx[/tex] from 0 to -1 (top to bottom).

    I change the equation to [tex(1/x - 3/2) dx[/tex]
    then integrated ln x-3/2x, but ln x at 0 is undefined.

    The text book shows it as becoming ln (3x-2), but I'm not completely understanding how to get to that.
     
  2. jcsd
  3. Jan 13, 2008 #2

    olgranpappy

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    Homework Helper

    substitute y = 3x-2
     
  4. Jan 13, 2008 #3
    Thanks, that worked. I'm just a bit rusty I reckon.
     
  5. Jan 14, 2008 #4

    HallsofIvy

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    Staff Emeritus
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    Caution: you are going to find it very difficult to do Calculus if you cannot do basic algebra.

    [itex]a/(b+c)[/itex] is not equal to (a/b)+ (a/c)!

    You had better review your algebra.
     
  6. Jan 14, 2008 #5
    Hmm, guess you're right. I've always used (a+b)/c=a/c+b/c but guess I just assumed it would work vice versa. Is there any combination it does equal?
     
  7. Jan 14, 2008 #6
    My 1st help in answering !!

    1- You have to use change of variables method.
    2- Substitute den with a variable and take derivative of this wrt x
    3- Using this var the limits will change
    4- Once you do 2 and with new limits from 3, the integral will be easy.

    Thanks

    Asif
     
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