设1/u(u^2-1)=A/(u-1)+B/u+C/(u+1) A+B+C=0 A-C=0 -B=1 所以A=1/2 B=-1 C=1/2 所以1/u(u^2-1)=(1/2)/(u-1)-1/u+(1/2)/(u+1) ∫1/u(u^2-1)du=1/2*∫du/(u-1)-∫du/u+1/2*∫du/(u+1) =1/2*ln|u-1|-ln|u|+1/2*ln|u+1|+C
