求不定积分，如图所示。要详细过程，谢谢

Question

如题所述

Answer 1

令x=根号(1-t)
dx=(-1/2)(1-t)^(-1/2)dt
(1-t)^(-1/2)dt=-2dx
x^2=1-t
t=1-x^2
原积分=积分(-2dx)/(1-x^2)
注意2/(x^2-1)=1/(x-1)-1/(x+1)
=积分dx/(x-1)-积分dx/(x+1)
=ln|x-1|-ln|x+1|+C
=ln|(x-1)/(x+1)|+C
=ln|(根号(1-t)-1)/(根号(1-t)+1)|+C

Answer 2