Posted by Parker on Tuesday, July 3, 2012 at 6:03pm.
Substituting x = sqrt(t) leads to an integral of the form:
Integral of dt/t sqrt(1+t^2)
If you then put t = sinh(u), this becomes:
Integral of cosh^2(u)/sinh(u)du =
Integral of [1/sinh(u) + sinh(u)] du
Then the integral of 1/sinh(u) be evaluated by putting u = Log(v):
du/[exp(u) - exp(u)] =
dv/[v (v - 1/v)] = dv/(v^2 - 1)
which is easily integrated.
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