1/[x(x2-1)]
=1/x(x+1)(x-1)
设a/x+b/(x+1)+c/(x-1)
=[a(x2-1)+bx(x-1)+cx(x+1)]/x(x+1)(x-1)
=(ax2-a+bx2-bx+cx2+cx)/x(x+1)(x-1)
=[(a+b+c)x2+(c-b)x-a]/x(x+1)(x-1) =1/x(x+1)(x-1)
则有
a+b+c=0
c-b=0
-a=1
解得
a=-1 b=1/2 c=1/2
∴1/[x(x2-1)]=-1/x +1/2(x+1)+1/2(x-1)
∴∫1/[x(x2-1)]dx
=-∫dx/x +∫dx/2(x+1)+∫dx/2(x-1)
=-lnx +1/2ln(x+1)+1/2ln(x-1) +C
=ln√[(x+1)(x-1)]-lnx+c
=|ln[√(x2-1)]/x|+c