Theory of Intelligence
11TH•Mathematics•Partial Fractions•Question 3

Solution

Resolve into partial fractions: $\dfrac{2x+32}{(x-2)(x^2+2)^2}$

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Solution

Let:

2x+32(x−2)(x2+2)2=Ax−2+Bx+Cx2+2+Dx+E(x2+2)2\frac{2x+32}{(x-2)(x^2+2)^2}=\frac{A}{x-2}+\frac{Bx+C}{x^2+2}+\frac{Dx+E}{(x^2+2)^2}

Clearing denominators and solving gives:

A=1,B=−1,C=−2,D=−6,E=−10A=1,\quad B=-1,\quad C=-2,\quad D=-6,\quad E=-10

Hence:

2x+32(x−2)(x2+2)2=1x−2−x+2x2+2−6x+10(x2+2)2\boxed{\frac{2x+32}{(x-2)(x^2+2)^2}=\frac{1}{x-2}-\frac{x+2}{x^2+2}-\frac{6x+10}{(x^2+2)^2}}

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