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Integration, u substitution, 1/u

  1. Sep 9, 2007 #1
    [SOLVED] Integration, u substitution, 1/u

    -- +C at the end of the integral solutions, I can't seem to add it in the LaTeX thing --

    1. The problem statement, all variables and given/known data
    #1 [tex]\int\frac{1}{8-4x}dx[/tex]
    #2 [tex]\int\frac{1}{2x}dx[/tex]

    3. The attempt at a solution
    Rewrite algebraically:

    Pull out constant:

    Usub, u=x-2, du=1*dx,

    [tex]\frac{-ln(u)}{4}[/tex] -> Answer: [tex]\frac{-ln(x-2)}{4}[/tex]

    Pull out constant: [tex]\frac{1}{2}*\int\frac{1}{x}dx[/tex]

    Answer: [tex]\frac{ln(x)}{2}[/tex]

    --OR, if I don't pull out constants--

    Usub, u=8-4x, du=-4dx,

    [tex]\frac{-ln(u)}{4}[/tex] -> Answer: [tex]\frac{-ln(8-4x)}{4}[/tex], which =/= [tex]\frac{-ln(x-2)}{4}[/tex]

    Usub, u=2x, du = 2dx,

    [tex]\frac{ln(u)}{2}[/tex] -> Answer: [tex]\frac{ln(2x)}{2}[/tex], which =/= [tex]\frac{ln(x)}{2}[/tex]

    I know the answers are -1/4*ln(8-4x) and ln(x)/2 respectively, but I don't understand why I cannot pull out the constant for the first one, yet I am required to pull it out for the second?

    Bonus question: Why does my TI-89 integrate #1 incorrectly, but #2 correctly? :(
    Last edited: Sep 9, 2007
  2. jcsd
  3. Sep 9, 2007 #2
    (8-4x),x ... did you mean dx? cuz i'm confused.

    #2 is correct for sure, let me go back to # 1.

    i'm getting dizzy reading your text but from what i see, your algebra-manipulation should have been for #1

    Last edited: Sep 9, 2007
  4. Sep 9, 2007 #3
    Yeah, by integrate(4x,x) I would mean 4x*dx. It's standard notation when typing integrals for calculators/CAS's etc, I did it that way because I was thinking the forums might automatically convert it to the proper mathematical symbols.
  5. Sep 9, 2007 #4
    oh sorry, i did not know that! :)
  6. Sep 9, 2007 #5
    Ah hah, that's how you make it pretty!

    Anyway yeah, I have [tex]\frac{-1}{4}\int\frac{1}{x-2}[/tex]
    which is equivalent and still valid, right?
  7. Sep 9, 2007 #6
    Introducing LaTeX Math Typesetting - https://www.physicsforums.com/showthread.php?t=8997 it gets addicting!!!

    yep, don't forget + C ;) .. -1 or maybe the whole problem will be counted wrong, lol :D
    Last edited: Sep 9, 2007
  8. Sep 9, 2007 #7
    when you do your final integrations the ln shouldn't be in the denominator of the fraction.

    It's (-1/4)*(lnx-2)+c which gives -ln(x-2)/4 + C the same goes for the second one.
  9. Sep 9, 2007 #8
    Oh sorry, just an error while converting to [tex] images I think.

    -ln(x-2)/4+c is incorrect though, which is my problem.
  10. Sep 9, 2007 #9
    I'm not sure why you pulled the constant out in the first place. I would go u = 8-4x du= -4dx and then write (-1/4)*integral 1/u du.

    Integrate and get (-1/4)ln(8-4x)

    the second one I would make u = 2x du = 2dx and then write the integral as 1/2 * integral 1/u * du and the integrate and get (1/2) ln (2x)
  11. Sep 9, 2007 #10
    Right, I did both of them both ways. If you do not pull out the constants, you get the correct answer for #1. However, #2 is not (1/2)*ln(2x), it is ln(x)/2.

    Pull out constants:
    #1 wrong, #2 right

    Do NOT pull out constants:
    #1 right, #2 wrong
    Last edited: Sep 9, 2007
  12. Sep 9, 2007 #11
    there really is no benefit in simplifying this problem, i would just go at what you're given.
  13. Sep 9, 2007 #12
    I'm not sure what you're saying. I shouldn't pull out constants?

    Why is it wrong to pull them out then?
    And how do you correctly do #2 without pulling out constants first?
  14. Sep 9, 2007 #13
    for 1, integrate by u subst., but 2 you have to pull out the constant.

    i just don't see a point in pulling out a constant for 1 b/c you run the risk of making algebraic mistakes, that's all really.
  15. Sep 9, 2007 #14
    I agree, but why is it wrong when I do? I think we can agree that there are no algebraic mistakes there. Why do I HAVE to pull them out for #2?
  16. Sep 9, 2007 #15


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    ln(2x) = ln(x) + ln(2) = ln(x) + C, with C = ln(2). Since indefinite integrals are only defined up to an additive constant, your two answers are the same. This is true for both problems.
  17. Sep 10, 2007 #16
    Ah hah! Thanks :) [hits self on head]
  18. Sep 10, 2007 #17


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    Science Advisor

    You're welcome! :) (It's always the little things that cause trouble ...)
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