Case IV: Repeated Irreducible Quadratic Factors
If Q(x) contains (ax2+bx+c)n with n≥2, then
Q(x)P(x)=ax2+bx+cA1x+B1+(ax2+bx+c)2A2x+B2+⋯+(ax2+bx+c)nAnx+Bn
Example 8
Resolve (x2+1)2(x−1)4x2 into partial fractions.
Solution
(x2+1)2(x−1)4x2=x2+1Ax+B+(x2+1)2Cx+D+x−1E
4x2=(Ax+B)(x2+1)(x−1)+(Cx+D)(x−1)+E(x2+1)2
x=1 ⇒ E=1. Equating coefficients: A=−1, B=−1, C=2, D=2.
Hence
x2+1−x−1+(x2+1)22x+2+x−11