早教吧作业答案频道 -->数学-->
1/(x+根号1-x^2)的积分
题目详情
1/(x+根号1-x^2)的积分_
▼优质解答
答案和解析
设x=sint,dx=d(sint)=costdt
则
S 1/(x+根号1-x^2) dx
=S cost/(sint+cost) dt
故原式=∫cost/(sint+cost)dt=A.
令B=∫sint/(sint+cost)dt,
则A+B=∫dt=t+c1
A-B=∫(cost-sint)/(sint+cost)dt
=∫d(sint+cost)/(sint+cost)
=ln|sint+cost|+c2
所以
A=1/2(t+ln|sint+cost|)+c
故原式=1/2(arcsinx+ln|x+根号(1-x^2)|)+c
则
S 1/(x+根号1-x^2) dx
=S cost/(sint+cost) dt
故原式=∫cost/(sint+cost)dt=A.
令B=∫sint/(sint+cost)dt,
则A+B=∫dt=t+c1
A-B=∫(cost-sint)/(sint+cost)dt
=∫d(sint+cost)/(sint+cost)
=ln|sint+cost|+c2
所以
A=1/2(t+ln|sint+cost|)+c
故原式=1/2(arcsinx+ln|x+根号(1-x^2)|)+c
看了1/(x+根号1-x^2)的积...的网友还看了以下：