Posted by flo on Friday, December 9, 2011 at 2:42pm.
Note that you have Integral(tanh(x) dx)
= Integral (sinh(x)/cosh(x) dx)
let u = cosh(x)
du = sinh(x) dx
and you have Integral(du/u) = ln(u) = ln(cosh(x)) = ln(e^x + e^-x) + C
Equivalently, let u = e^x + e^-x
Then du = e^x - e^-x du
and you have du/u
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