Question

求解不定积分：根号下e的2x次幂减1

Answer 1

令：√（e^2x-1) = t ,x=1/2*ln(t^2+1) ,dx= t/(t^2+1)
∫ √（e^2x-1) dx
=∫ t^2 /(t^2+1) dt
=∫ (t^2+1-1) /(t^2+1) dt
=∫ 1 dt -∫ 1/(t^2+1) dt
= t-arctant +C
= √（e^2x-1) -arctan(√（e^2x-1)) +C