Question

1/[(e^x+e^-x)^2]不定积分

Answer 1

令t=e^x,得

原式=∫1/[t(t+t^(-1))^2]dt

=∫t/(t^4+2t^2+1)dt

=(1/2)∫1/(t^2+1)^2d(t^2+1)

=-1/[2(t^2+1)]+C

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