∫(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.
Find the indefinite Integral of:
(e^(2x)-e^(-2x))/(e^(2x)+e^(-2x))dx
Thank you very much for your help.
1 answer
1 answer