Theory of Intelligence
11TH•Mathematics•Partial Fractions•Question 6

Solution

Resolve into partial fractions: $\dfrac{6x^4+40x^2}{(4-x^2)(x^2+4)^2}$

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Solution

6x4+40x2(4−x2)(x2+4)2\frac{6x^4+40x^2}{(4-x^2)(x^2+4)^2}

Note 4−x2=−(x2−4)=−(x−2)(x+2)4-x^2=-(x^2-4)=-(x-2)(x+2). So:

6x4+40x2(4−x2)(x2+4)2=−6x4+40x2(x−2)(x+2)(x2+4)2\frac{6x^4+40x^2}{(4-x^2)(x^2+4)^2}=-\frac{6x^4+40x^2}{(x-2)(x+2)(x^2+4)^2}

Let:

−6x4+40x2(x−2)(x+2)(x2+4)2=Ax−2+Bx+2+Cx+Dx2+4+Ex+F(x2+4)2-\frac{6x^4+40x^2}{(x-2)(x+2)(x^2+4)^2} =\frac{A}{x-2}+\frac{B}{x+2}+\frac{Cx+D}{x^2+4}+\frac{Ex+F}{(x^2+4)^2}

Solving gives:

A=−1,B=1,C=0,D=−2,E=0,F=−8A=-1,\quad B=1,\quad C=0,\quad D=-2,\quad E=0,\quad F=-8

Hence:

6x4+40x2(4−x2)(x2+4)2=−1x−2+1x+2−2x2+4−8(x2+4)2\boxed{\frac{6x^4+40x^2}{(4-x^2)(x^2+4)^2}=-\frac{1}{x-2}+\frac{1}{x+2}-\frac{2}{x^2+4}-\frac{8}{(x^2+4)^2}}

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