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Exercise 11.5 · Q19

Q.3x−9(x−1)(x+2)(x2+1)\dfrac{3x-9}{(x-1)(x+2)(x^{2}+1)}

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This denominator mixes two linear factors with one irreducible quadratic, so the decomposition needs a linear numerator over x2+1x^2+1.

Step 1. Set up partial fractions. 3x−9(x−1)(x+2)(x2+1)=Ax−1+Bx+2+Cx+Dx2+1\dfrac{3x-9}{(x-1)(x+2)(x^2+1)}=\dfrac{A}{x-1}+\dfrac{B}{x+2}+\dfrac{Cx+D}{x^2+1}, so 3x−9=A(x+2)(x2+1)+B(x−1)(x2+1)+(Cx+D)(x−1)(x+2)3x-9=A(x+2)(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x-1)(x+2).

Step 2. Solve for AA. At x=1x=1: −6=A(3)(2)=6A⇒A=−1-6=A(3)(2)=6A\Rightarrow A=-1.

Step 3. Solve for BB. At x=−2x=-2: −15=B(−3)(5)=−15B⇒B=1-15=B(-3)(5)=-15B\Rightarrow B=1.

Step 4. Solve for C,DC,D using x=ix=i. At x=ix=i (so x2+1=0x^2+1=0): 3i−9=(Ci+D)(i−1)(i+2)3i-9=(Ci+D)(i-1)(i+2). Since (i−1)(i+2)=−3+i(i-1)(i+2)=-3+i, expanding (Ci+D)(−3+i)=(−C−3D)+i(D−3C)(Ci+D)(-3+i)=(-C-3D)+i(D-3C). Matching real and imaginary parts against −9+3i-9+3i: −C−3D=−9-C-3D=-9 and D−3C=3D-3C=3. Solving simultaneously gives D=3D=3, C=0C=0. …

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