Theory of Intelligence
11TH•Mathematics•Partial Fractions•5.4 Repeated Quadratic Factors

Case IV: Repeated Irreducible Quadratic Factors

Partial fractions when the denominator has repeated irreducible quadratic factors.

Case IV: Repeated Irreducible Quadratic Factors

If Q(x)Q(x) contains (ax2+bx+c)n(ax^2+bx+c)^n with n≥2n \ge 2, then

P(x)Q(x)=A1x+B1ax2+bx+c+A2x+B2(ax2+bx+c)2+⋯+Anx+Bn(ax2+bx+c)n\dfrac{P(x)}{Q(x)} = \dfrac{A_1 x + B_1}{ax^2+bx+c} + \dfrac{A_2 x + B_2}{(ax^2+bx+c)^2} + \cdots + \dfrac{A_n x + B_n}{(ax^2+bx+c)^n}

Example 8

Resolve 4x2(x2+1)2(x−1)\dfrac{4x^2}{(x^2+1)^2(x-1)} into partial fractions.

Solution

4x2(x2+1)2(x−1)=Ax+Bx2+1+Cx+D(x2+1)2+Ex−1\dfrac{4x^2}{(x^2+1)^2(x-1)} = \dfrac{Ax+B}{x^2+1} + \dfrac{Cx+D}{(x^2+1)^2} + \dfrac{E}{x-1} 4x2=(Ax+B)(x2+1)(x−1)+(Cx+D)(x−1)+E(x2+1)24x^2 = (Ax+B)(x^2+1)(x-1) + (Cx+D)(x-1) + E(x^2+1)^2

x=1x = 1 ⇒\Rightarrow E=1E = 1. Equating coefficients: A=−1A = -1, B=−1B = -1, C=2C = 2, D=2D = 2.

Hence

−x−1x2+1+2x+2(x2+1)2+1x−1\dfrac{-x-1}{x^2+1} + \dfrac{2x+2}{(x^2+1)^2} + \dfrac{1}{x-1}

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