第1个回答 2017-05-09
设x/(x+1)(x+2)(x+3)=A/(x+1)+B/(x+2)+C/(x+3)右边通分得[A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2)]/(x+1)(x+2)(x+3)=[A(x²+5x+6)+B(x²+4x+3)+C(x²+3x+2)]/(x+1)(x+2)(x+3)=[(A+B+C)x²+(5A+4B+3C)x+(6A+3B+2C)]/(x+1)(x+2)(x+3)∴A+B+C=05A+4B+3C=16A+3B+2C=0解得A=-1/2,B=2,C=-3/2∴x/(x+1)(x+2)(x+3)=-1/2(x+1)+2/(x+2)-3/2(x+3)∴原式=-1/2*∫dx/(x+1)+2*∫dx/(x+2)-3/2*∫dx/(x+3)=-1/2*ln|x+1|+2*ln|x+2|-3/2*ln|x+3|+C本回答被网友采纳