Find the arc lenght

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Find the arc lenght

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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this is where i got stuck..
its a good place to get stuck at for sure ... a trig sub might be a good shot, maybe.
im using this formula..|dw:1442619138404:dw|
we havnt gotten into using trigs yet...
my professor said something about making whats inside the sqrt to a perfect square to be able to cancel the sqrt, but i dont see how i can make it that way...
4x^2 + kx + 1 (2x+1)^2 seems good 4x^2 + 1 + (4x-4x) = (2x+1)^2 -4x
i dont believe that you can just 'make' something into a perfect square, you can rewrite it so that it 'contains' a perfect square
if u = 2x+1, du=2dx and x=(u-1)/2 (2x+1)^2 -4x u^2 -4(u-1)/2 u^2 -2u +2 is still not a perfect square
http://www.wolframalpha.com/input/?i=integrate+sqrt%281%2B4x^2%29+dx notice that the wolfs integral isnt that pretty without knowing some hypertrigs
well thanks for the help ill email my professor.. so she can solve it lol
if y=t, then x^2 = t, and x=sqrt(t) y' = 1, x' = 1/(2sqrt(t)) y'^2 = 1, x'^2 = 1/(4t) but that doesnt make for a pretty integral either
omg i have tried to solve this problem over 100 times and guess what i read the question wrong. the questions sais to use a graphing utility to evaluate it
but again using in getting this https://www.wolframalpha.com/input/?i=arc+length+of+y%3Dsqrt%281%2B4x^2%29+from+x%3D0+to+3
when the the answer is 9.747
arclength of y=x^2
https://www.wolframalpha.com/input/?i=arc+length+of+y%3Dx^2+from+x%3D0+to+3
Are you not able to use the ArcLength formula of vectors to solve this?

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