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April 21, 2014

April 21, 2014

Posted by **Jake** on Sunday, February 27, 2011 at 8:29pm.

(e^(2x)-e^(-2x))/(e^(2x)+e^(-2x))dx

Thank you very much for your help.

- Calculus AB -
**MathMate**, Sunday, February 27, 2011 at 10:41pm∫(e^(2x)-e^(-2x))/(e^(2x)+e^(-2x))dx

substitute

sinh(2x)=(e^(2x)-e^(-2x))/2

cosh(2x)=(e^(2x)+e^(-2x))/2

=∫sinh(2x)/cosh(2x)dx

Substitute

u=cosh(2x)

du=2sinh(2x)dx

(1/2)du = sinh(2x)dx

to get

=(1/2)∫du/u

=(1/2)log(u)+C

=(1/2)log(cosh(2x))+C

=(1/2)log((e^(2x)+e^(-2x))/2)+C

Alternately, you could also substitute u=e^(2x)+e^(-2x)

and do integration manually, with the same results.

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