(6x^2+16x+18)/(x+1)(x+2)(x+3)

分解部分分式,一般采用待定系数法.
设(6x^2+16x+18)/(x+1)(x+2)(x+3)＝a/(x+1)+b/(x+2)+c/(x+3)
＝[(a+b+c)x^2+(5a+4b+3c)x+(6a+3b+2c)]/(x+1)(x+2)(x+3)
因此有：a+b+c＝6,5a+4b+3c＝16,6a+3b+2c＝18
解这个三元一次方程组得：a=4,b=-10,c=12
所以(6x^2+16x+18)/(x+1)(x+2)(x+3)＝4/(x+1)-10/(x+2)+12/(x+3)