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March 31, 2015

March 31, 2015

Posted by **flo** on Friday, December 9, 2011 at 2:42pm.

- integral calculus -
**Steve**, Friday, December 9, 2011 at 3:25pmNote 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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