1/（-1+x∧2）∧2的不定积分怎么求

Answer 1

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

Answer 2

1/(x^2-1)^2=(1/4)(1/(x+1)+1/(x+1)^2-1/(x-1)+1/(x-1)^2))
所以：
∫1/(x^2-1)^2dx=(1/4)(∫1/(x+1)dx+∫1/(x+1)^2dx-∫1/(x-1)dx+∫1/(x-1)^2)dx)
=(1/4)(ln|x+1|-ln|x-1|-1/(x+1)-1/(x-1))+C