不定积分,两道

2025-02-03 11:31:58
推荐回答(2个)
回答(1):

回答(2):

(1) 设√(1+e^(2x))=t x=ln(t^2-1)/2 dx=t/(t^2-1)dt
  ∫dx/√(1+e^(2x)=∫dt/(t^2-1)
  =1/2∫1/(t-1)-1/(t+1)
  =1/2* ln|(t-1)/(t+1)|+C
  =1/2* ln|{√[1+e^(2x)]-1}/(√[1+e^(2x)]+1}|+C
(2) 设arctanx=t 则x=tant dx=sec^2tdt
  ∫e^arctanx/(1+x^2)^(3/2)dx
  =∫e^tsec^2tdt/sec^3t
  =∫coste^tdt
  =coste^t+∫sinte^tdt
  =coste^t+sinte^t-∫coste^tdt
  ∫coste^tdt=1/2(sint+cost)e^t+C
  ∴∫e^arctanx/(1+x^2)^(3/2)dx
  =1/2(sint+cost)e^t+C
  =1/2(1+x)/√(1+x^2)*e^arctanx+C