Precalculus, Chapter 7, 7.4, Section 7.4, Problem 41

You need to decompose the fraction in simple irreducible fractions, such that:
(8x-12)/(x^2(x^2+2)^2) = A/x + B/(x^2) + (Cx+D)/(x^2+2) + (Ex+F)/((x^2+2)^2)
You need to bring to the same denominator all fractions, such that:
8x - 12 = Ax(x^2+2)^2 + B(x^2+2)^2 + (Cx+D)*x^2*(x^2+2) + (Ex+F)*x^2
8x - 12 = Ax(x^4 + 4x^2 + 4) + Bx^4 + 4Bx^2 + 4B + (Cx^3 + Dx^2)*(x^2+2) + Ex^3 + Fx^2
8x - 12 = Ax^5 + 4Ax^3 + 4Ax + Bx^4 + 4Bx^2 + 4B + Cx^5 + 2Cx^3 + Dx^4 + 2Dx^2 + Ex^3 + Fx^2
You need to group the terms having the same power of x:
8x - 12 = x^5(A + C) + x^4(B+D) + x^3(4A + 2C + E) + x^2(4B + 2D + F) + x(4A) + + 4B
Comparing the expressions both sides yields:
A + C = 0 => A =-C
B+D = 0 => B = - D
4A + 2C + E = 0 => 2A + E = 0 => E = -2A
4B + 2D + F = 0 => F = -2D
4A = 8 => A = 2 => C = -2 => E = -4
4B=-12 => B = -3 => D = 3 => F = -6
Hence, decomposing the fraction, yields (8x-12)/(x^2(x^2+2)^2) = 2/x- 3/(x^2) + (-2x+3)/(x^2+2) + (-4x-6)/((x^2+2)^2).

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