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追问
可不可以从这一步直接求啊？【∫cost/(sint+cost)dt】不用令A，B了，常规一点的！
追答直接不容易积,这样要简单些!
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