题目

求不定积分1题,
1/[x+√(1-x∧2)]

提问时间：2021-03-23

答案

代换x=sint π/2>t>-π/2 cost>0
dx=costdt
∫1/[x+√(1-x^2)]dx
=∫cost/(sint+cost)dt
I=∫cost/(sint+cost)dt
J=∫sint/(sint+cost)dt
I+J=∫(sint+cost)/(sint+cost)dt=∫dt=t=arcsinx+C
J-I=∫(cost-sint)/(sint+cost)dt=
=∫d(sint+cost)/(sint+cost)dt
=ln(sint+cost)
=ln(x+√(1-x^2))+C
I=1/2*((I+J)-(J-I))
=1/2arcsinx-1/2ln(x+√(1-x^2))+C