求∫1/(e^x+e^-x)dx

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求∫1/(e^x+e^-x)dx求∫1/(e^x+e^-x)dx求∫1/(e^x+e^-x)dx∫[1/(e^x+e^-x)]dx=∫[e^x/(e^2x+1)]=∫[1/(e^2x+1)]d(e^x

求∫1/(e^x+e^-x)dx
求∫1/(e^x+e^-x)dx

求∫1/(e^x+e^-x)dx
∫[1/(e^x+e^-x)]dx
=∫[e^x/(e^2x+1)]
=∫[1/(e^2x+1)]d(e^x)
=arctan(e^x)+C.