First we need to find the linear factors of the denominator (note that 2 of these factors only exist as complex values):
x^3+x=x(x^2+1)=color(blue)(x(x+i)(x-i))x3+x=x(x2+1)=x(x+i)(x−i)
We need to find values A, B, and CA,B,andC such that
(5x^2+7x+3)/(x^3+x)=A/x +B/(x+i)+C/(x-i)5x2+7x+3x3+x=Ax+Bx+i+Cx−i
color(white)("XXXXX")=(A(x+i)(x-i)+B(x)(x-i)+C(x)(x+i))/(x(x+i)(x-i))XXXXX=A(x+i)(x−i)+B(x)(x−i)+C(x)(x+i)x(x+i)(x−i)
color(white)("XXXXX")=(Ax^2+A+Bx^2-Bix+Cx^2-Cix)/(x^3+x)XXXXX=Ax2+A+Bx2−Bix+Cx2−Cixx3+x
rarr5x^2+7x+3=((A+B+C)x^2+(-Bi+Ci)x+A→5x2+7x+3=((A+B+C)x2+(−Bi+Ci)x+A
rarrcolor(white)("XXXXX"){([1]color(white)("X")A+B+C=5),([2]color(white)("X")-Bi+Ci=7),([3]color(white)("X")A=3):}
Combining [1] and [3] we have
color(white)("XXXXX")[4]color(white)("X")B+C=2
and from [2] we can
color(white)("XXXXX")[5]color(white)("X")B-C=7i
Adding [4] and [5] then dividing by 2:
color(white)("XXXXX")[6]color(white)("X")B=1+7/2i
Then substituting back into [4]
color(white)("XXXXX")C=1-7/2i
