Question

求不定积分,如图所示.

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