Question

根号下1+e的x次方的积分?

Answer 1

令根号下1+e^x=t 则有1+e^x=t^2
dx=[2t/(t^2-1)]dt
原式=2∫t^2/(t^2-1)dt
=2∫1+1/(t^2-1)dt
=2t+ln|(t-1)/(t+1)|+c