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Simple integration

  1. May 10, 2008 #1
    i know how to integrate.

    but why is 3/x 3 log x

    3x^-1 3 x^0 X^0 = 1 3*1 = 3.

    i know I'm ignoring the divide by zero

    why does log come into it.

    4e^(1/2) im messing this one up as well.

    4e^(3/2) / (3/2)

    = 8/3e^(3/2) thats like as far as i get it.

    somehow the answers have it at 8e^1/2

    can anyone explain how that is so
    Last edited: May 10, 2008
  2. jcsd
  3. May 10, 2008 #2
    Try differentiating your answer and the correct answer and see what you get (each with respect to x of course). In which case do you get the function you started off with?
  4. May 10, 2008 #3
    I already know that mine is wrong and theirs is right, I just don't know how to get to their answer. >.<

    k : i dont know all rules of integration apparently integral of 1/x = logx

    still need help on 2nd one though
    Last edited: May 10, 2008
  5. May 10, 2008 #4
    Re-write your original post, it's very hard to interpret!
  6. May 10, 2008 #5


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    Well, obviously, your answer is wrong precisely because you are "ignoring that divide by 0".

    One way to show that "log(x)" comes into it is to define log(x) as
    [tex]\int_1^x \frac{dx}{x}[/tex]
    That is done in some textbooks.

    Or you can define log(x) to be the inverse function to ex. Then if y= ln(x), x= ey. Differentiating both sides by x, 1= ey dy/dx or dy/dx= 1/ey. Since ey= x, that is dy/dx= 1/x.

    Since the derivative of log(x) is 1/x, the anti-derivative (indefinite integral) of log(x) is 1/x+ C.
  7. May 10, 2008 #6
    this is a 5 year old exam, i assume the anti-derivatives dont come on my exam now, as the teacher hasnt taught them and they arnt on the last few year exams.

    so il leave it for now. =)
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