有理函數(shù)積分主要是部分分式的分
設(shè)Q(x)=c(x-a)^α...(x-b)^β(x^2+px+q)^λ...(x^2+rx+s)^μ
(其中p^2-4q<0,...,r^2-4s<0.).
那么真分式P(x)/Q(x)可以分解成如下部分分式之和：
P(x)/Q(x)=A1/(x-a)^α+A2/(x-a)^(α-1)+...+A[α]/(x-a)+...+
+B1/(x-b)^β+B2/(x-b)^(β-1)+...+B[β]/(x-b)+
(M1x+N1)/(x^2+px+q)^λ+...+(M[λ]x+N[λ])/(x^2+px+q)+.+
(R1x+S1)/(x^2+rx+s)^μ+...+(R[μ]x+S[μ])/(x^2+rx+s).
x/[(x+1)(x+2)(x+3)]=A/(x+1)+B/(x+2)+C/(x+3),
x=A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2).
令x=-1,得A=-1/2,
令x=-2,得B=2,
令x=-3,得C=-3/2,
x/[(x+1)(x+2)(x+3)]=(-1/2)*1/(x+1)+2/(x+2)-(3/2)*1/(x+3),
或由x=(A+B+C)x^2+(5A+4B+3C)x+(6A+3B+2C),
比較系數(shù)得A+B+C=0,5A+4B+3C=1,6A+3B+2C=0,
解出A,B,C.
3/(x^3+1)=1/(x+1)(x^2-x+1)=A/(x+1)+(Mx+N)/(x^2-x+1),
3=A(x^2-x+1)+(Mx+N)(x+1).
令x=-1,得A=1,
(Mx+N)(x+1)=3-A(x^2-x+1)=-x^2+x-2=-(x-2)(x+1),
Mx+N=-x+2,M=-1,N=2.
3/(x^3+1)=1/(x+1)-(x-2)/(x^2-x+1).