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Algebra II - Why Does sqrt(x+1/x+2)=(x+1)/sqrt(x)?

  1. Aug 11, 2011 #1
    Hi, I'm actually trying to find the length of the arc

    y=x^(3/2)/3-x^(1/2); [4,16]

    And when I was trying to evaluate

    integral[4,16] sqrt(1/2+x/4+1/(4x))dx

    I had no idea were to begin so I plugged this into wolfram alpha
    http://www.wolframalpha.com/input/?i=integral+sqrt(1/2+x/4+1/(4x))dx
    and followed it up to this step
    1/2*integral[4,16] sqrt(x+1/x+2)dx

    It then says it simplified the powers and that

    1/2*integral[4,16] sqrt(x+1/x+2)dx = 1/2*integral[4,16] (x+1)/sqrt(x)dx

    I don't follow how this was done at all, apparently
    sqrt(x+1/x+2) = (x+1)/sqrt(x)
    and I don't see how these are equal

    I figuered I would post this here because I'm not having problems with the actual calculus but the algebra II, (I think what ever was done to establish that these two expressions are equal to each other is an algebra II topic.)

    Thanks for any help you can proivde
     
  2. jcsd
  3. Aug 11, 2011 #2

    Hurkyl

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    Staff Emeritus
    Science Advisor
    Gold Member

    To simplify in-place, wouldn't the obvious thing to do be to collect the fractions into a single term?


    That said, you could always prove they have the same sign and the same square. (squaring to get rid of the square root) or you could cross multiply, or other things.
     
  4. Aug 11, 2011 #3
    I don't see how you could collect the fractions into a single term though
    inside the square root we have
    x+1/x+2
    I could write that like this

    x^2/x+1/x+(2x)/x
    (x^2+1+2x)/x
    (x+1)^2/x
    ah man thanks =O lol i feel dumb
     
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