把 -1/[(cosx+2)*(cosx+1)*(-cosx+1)]看成:A/(cosx+2)+B/(cosx+1)+C/(cosx+1)三个多项式之和
-1=[A/(cosx+2)+B/(cosx+1)+C/(cosx+1)]*[(cosx+2)+(cosx+1)+(-cosx+1)]
-1=A*[(cosx+1)(-cosx+1)]+B*[(cosx+2)(-cosx+1)]+C*[(cosx+2)(cosx+1)]
令cosx+1=0,
cosx=-1,-1=B*[(cosx+2)(-cosx+1)] = B*[1*(1+1)]=2B,B=-1/2;
令-cosx+1=0,
cosx=1,-1=C*[(cosx+2)(cosx+1)] = C*[3*(1+1)]=6C,C=-1/6;
从而,-1/[(cosx+2)*(cosx+1)*(-cosx+1)]看成:
A/(cosx+2)+(-1/2)/(cosx+1)+(-1/6)/(-cosx+1)
进而求A,cosx+2∈[1,3],取cosx=0
A/(2)+(-1/2)/(1)+(-1/6)/(1)=-1/[(cosx+2)*(cosx+1)*(-cosx+1)]=-1/2
A=1/3
从而,-1/[(cosx+2)*(cosx+1)*(-cosx+1)]看成:
(1/3)/(cosx+2)+(-1/2)/(cosx+1)+(-1/6)/(-cosx+1).
---注意积分号前面的“ - ”负号.