1/[(x+1)(x+2)]=[A/(x+1)]-[B/(x+2)]=[A(x+2)-B(x+1)]/(x+1)(x+2)]=[(A-B)x+(2A-B]/(x+1)(x+2)]所以有:1/[(x+1)(x+2)]=[(A-B)x+(2A-B]/(x+1)(x+2)]A-B=0 ,2A-B=1解得A=B=1所以有:1/[(x+1)(x+2)]=[1/(x+1)]-[1/(x+2)]