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Simple integration

  • Thread starter DeanBH
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  • #1
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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
 
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Answers and Replies

  • #2
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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?
 
  • #3
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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
 
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  • #4
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Re-write your original post, it's very hard to interpret!
 
  • #5
HallsofIvy
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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.
 
  • #6
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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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