不定积分,两道
发布网友
发布时间：2022-04-24 08:44
共2个回答
热心网友
时间：2022-06-18 03:27
热心网友
时间：2022-06-18 03:27
(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
