Question

求原函数x=(2s*s+12s+6)%s(s+2)(s+3)

Answer 1

设实数A、B、C,使得x=(2s^2+12s+6)/s(s+2)(s+3)=A/s+B/(s+2)+C/(s+3)

2s^2+12S+6=A(s+2)(s+3)+Bs(s+3)+Cs(s+2)=(A+B+C)s^2+(5A+3B+2C)s+6A

即：

①A+B+C=2

②5A+3B+2C=12

③6A=6

所以A=1 B=5 C=-4

所以x=1/s+5/(s+2)-4/(s+3)

所以x的原函数=∫ds/s+∫5ds/(s+2)-∫4ds/(s+3)=ln|s|+5ln|s+2|-4ln|s+3|+C