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Inverse hyperbolic functions watch

1. ??
2. 1) Split it into 2 [2x/sqrt(x2+1)] +1/sqrt(x2+1). Then for the first part let u = x2+1 and the second part let x = sinh x

2) Let x3 = u then let u = cosh y
3. u should get x^2dx=u/3 which cancels the x^2 on the top. This leaves you 1/3 * 1/rt(u^2 - 1)

then use the cosh substituion du = sinhy dy
making the intergal 1/3 (sinh y / rt((sinhy)^2)) = int(1/3)
giving you y/3

then substitute back to get it in terms of x
y/3 = (arccosh u)/3 = 1/3arcosh(x^3)

hope thats ok
4. Yes, but it aids explaination by turning into an integral you can instantly recognise as a standard result.

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Updated: April 12, 2006
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