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Help me find the antiderivates

  1. Aug 31, 2014 #1

    RJLiberator

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    f(x) = 1/xln(x)
    g(x) = e^x sqrt(1+e^x)

    Hm.

    I guess, my first question is: Do I need to use U substitution?
     
  2. jcsd
  3. Aug 31, 2014 #2

    PeroK

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    What substitutions are you thinking of using?
     
  4. Aug 31, 2014 #3

    RJLiberator

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    Well, that's where I am kinda stuck.

    Let's take the first one for example.

    If I were to substitute U as ln(x) I would get:

    U = ln(x)
    du = 1/xdx

    Working that out, I get 1/u*du

    I suppose that works. Seems to check out with me.

    Then it becomes ln(u) and the answer becomes ln(ln(x))

    Um, wow. That is the right answer, that was a LOT easier then I made it out on my paper 0_0. It's as if I went on this site and magic happens.

    OKAY, Cool, let's try example two:

    e^x sqrt(1+e^x)

    My two possibilities for U seem to be either e^x or sqrt(1+e^x)
    Lets try U=e^x and du = e^x dx
    Hm, on first impression, this does not make sense, but let's see:
    du sqrt(1+u)
    (2(1+u)^3/2)/3
    Inputing back for e^x

    (2(1+e^x)^3/2)/3

    That seems to be the correct answer...

    Is my reasoning correct?
     
  5. Aug 31, 2014 #4

    PeroK

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    Yes, looks like you got it. Don't forget the constant of integration, though.

    You may want to take a quick look at using Latex to post mathematics. E.g:

    [tex]\int e^x \sqrt{1 + e^x}dx = \int \sqrt{1 + u}du = \frac{2}{3}(1+u)^{\frac{3}{2}} + C[/tex]
    If you "quote" my post, you'll get the Latex I used and see how it's done.
     
  6. Aug 31, 2014 #5

    RJLiberator

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    PeroK, thank you kindly for all of your help here. I will look at using Latex for future posts now that I will likely be posting more here due to semester starting :).

    Kind regards.
     
  7. Aug 31, 2014 #6

    RJLiberator

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    Okay, one more quick question on a different question (I don't feel it is warranted to create a thread):

    True or False?
    For any real numbers a and b:
    [itex]\sqrt{(a+b)^2} = a+b[/itex]

    My problem with this problem is that (a+b)^2 = (a+b)(a+b) = a^2+2ba+b^2 OR does it equal a^2+b^2

    I am leaning towards thinking this statement is TRUE as 3+6 = 9 raised to the second power = 81. Square root of 81 = 9 and 3+6 = 9 on the other side.

    Correct?
     
  8. Aug 31, 2014 #7

    PeroK

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    What about negative numbers?
     
  9. Aug 31, 2014 #8

    RJLiberator

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    Oh, I am sorry Perok - I mean the absolute values of a and b on the right side, let me rewrite.

    True or False?
    For any real numbers a and b:
    [itex]\sqrt{(a+b)^2} = |a|+|b|[/itex]
     
  10. Aug 31, 2014 #9

    micromass

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    Try some examples. Try various combinations of positive and negative numbers. For example, ##a## positive and ##b## positive, ##a## negative and ##b## positive, etc.
     
  11. Aug 31, 2014 #10

    PeroK

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    No. For example, a = 1, b = -1.

    Taking the square root of a square quite often seems to cause trouble. Note that:

    [tex]If \ x \ge 0, \ \ then \ \ \sqrt{x^2} = x; \ \ and \ if \ x < 0, \ then \ \sqrt{x^2} = -x[/tex]

    So, now you can apply this to (a + b).
     
  12. Aug 31, 2014 #11

    RJLiberator

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    I have tried many examples now, and they all check out, so it must be true.
    AH.

    I get it now. (a+b)(a+b) = (a+b)^2
    (a+b)^2 = a^2+b^2+2ab

    Ah... silly me... thanks.

    Aha.
     
  13. Aug 31, 2014 #12

    Ray Vickson

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    Have you tried a = 1 and b = -2?
     
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