求不定积分1/(x^2-1)dx

1/(x^2-1)=1/(x+1)(x-1)
=a/(x+1)+b/(x-1)
=[(a+b)x+(b-a)]/(x+1)(x-1)
所以a+b=0,b-a=1
a=-1/2,b=1/2

所以原式=-1/2∫1/(x+1)dx+1/2∫1/(x-1)dx
=-1/2*ln|x+1|+1/2*ln|x-1|+C
=1/2*ln|(x-1)/(x+1)|+C

=(1/2)ln|x-1/x+1|+C

(1/2)*S[1/(x-1)-1/(x+1)]dx=(1/2)*[ln(x-1)-ln(x+1)]
S是积分号

令x=sest