不定积分，用第二类换原法做，过程详细，不会的不要凑热闹，谢谢

Question

如题所述

Answer 1

首先，令

则在对原式进行小幅调整后进行换元可得到

对最后一行所得的式子进行积分，并将原变量换回可得到

于是，得到原式积分的结果。

Answer 2

两种换元做法。
(1)设√(1-x²)=t
1-x²=t²
2tdt=-2xdx
tdt=-xdx
xdx=-tdt
∫√(1-x²)dx/x
=∫x√(1-x²)dx/x²
=∫-t²dt/(1-t²)
=∫(1-t²-1)dt/(1-t²)
=∫dt-∫dt/(1-t²)
=∫dt+∫dt/(t²-1)
=t+∫dt/(t-1)(t+1)
=t+(1/2)[∫dt/(t-1)-∫dt/(t+1)]
=t+(1/2)ln|t-1|-(1/2)ln|t+1|+C
=t+(1/2)ln|(t - 1)/(t+ 1)|
再把t=√(1-x²)代入即可。
(2)设x=sint
1-x²=1-sin²t=cos²t
√(1-x²)=cost
dx=costdt
原式=
∫cos²tdt/sint
=∫(1-sin²t)dt/sint
=∫dt/sint-∫sintdt
对于∫dt/sint
= ∫ sint/sin²t dt
= ∫ 1/(cos²t - 1) d(cost)
= (1/2)∫ [(cost + 1) - (cost - 1)]/[(cost + 1)(cost - 1)] d(cost)
= (1/2)∫ [1/(cost - 1) - 1/(cost + 1)] d(cost)
= (1/2)ln|(cost - 1)/(cost + 1)|
所以原积分=(1/2)ln|(cost - 1)/(cost + 1)|+cost+C
再把cost=√(1-x²)代入即可。本回答被提问者和网友采纳

Answer 3

设x＝sint，你自己试试