First we need to finish factoring the denominator into factors that are irreducible using Real coefficients:
x/((x+1)(x^2 -1)) = x/((x+1)(x+1)(x-1))x(x+1)(x2−1)=x(x+1)(x+1)(x−1)
= x/((x+1)^2(x-1))=x(x+1)2(x−1)
So we need:
A/(x+1) + B/(x+1)^2 + C/(x-1) = x/((x+1)^2(x-1))Ax+1+B(x+1)2+Cx−1=x(x+1)2(x−1)
Clear the denominator (multiply by ((x+1)^2(x-1))((x+1)2(x−1))on both sides), to get:
A(x^2-1)+B(x-1)+C(x+1)^2 = xA(x2−1)+B(x−1)+C(x+1)2=x
Ax^2-A + Bx -B +Cx^2+2Cx+C = xAx2−A+Bx−B+Cx2+2Cx+C=x
Ax^2+Cx^2 +Bx +2Cx -A-B+C = 0x^2 +1x+0Ax2+Cx2+Bx+2Cx−A−B+C=0x2+1x+0
So we need to solve:
A+C = 0A+C=0
B+2C = 1B+2C=1
-A-B+C=0−A−B+C=0
From the first equation, we get: A = -CA=−C and we can substitue in the third equation to get
Eq 2: B+2C = 1B+2C=1
and: -B+2C = 0−B+2C=0
Adding gets us C = 1/4C=14, so A = -1/4A=−14 and subtracting gets us B = 1/2B=12
A/(x+1) + B/(x+1)^2 + C/(x-1) = (-1/4)/(x+1) + (1/2)/(x+1)^2 + (1/4)/(x-1) Ax+1+B(x+1)2+Cx−1=−14x+1+12(x+1)2+14x−1
= 1/4[(-1)/(x+1) + (2)/(x+1)^2 + 1/(x-1)]=14[−1x+1+2(x+1)2+1x−1]
