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arcsin __ = arctan (x/x+1)
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x/(2x^2+2x+1)^2
Put it in that blank
Sqrt?
Yes
the opposite side is x, adjacent side is x+1 and so, hypotenuse is (2x^2+2x=1)^2
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sqrt(2x^2+2x+1)
haha its square root not squared
Oh yes..its square root
so \[\frac{ x }{ \sqrt{2x ^{2}+2x+1}}\] is the answer?
Yup @ambernicole u r correct
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