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

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  • #1
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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
 

Answers and Replies

  • #2
Hurkyl
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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.
 
  • #3
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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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