Theory of Intelligence
11TH•Mathematics•Partial Fractions•Question 9

Solution

Resolve into partial fractions: $\dfrac{x^2}{(x^2+a)(x^2+b)(x^2+c)}$

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Solution

Let t=x2t=x^2. Then:

x2(x2+a)(x2+b)(x2+c)=t(t+a)(t+b)(t+c)\frac{x^2}{(x^2+a)(x^2+b)(x^2+c)}=\frac{t}{(t+a)(t+b)(t+c)}

Assume:

t(t+a)(t+b)(t+c)=At+a+Bt+b+Ct+c\frac{t}{(t+a)(t+b)(t+c)}=\frac{A}{t+a}+\frac{B}{t+b}+\frac{C}{t+c}

So:

t=A(t+b)(t+c)+B(t+a)(t+c)+C(t+a)(t+b)t=A(t+b)(t+c)+B(t+a)(t+c)+C(t+a)(t+b)

Put t=−at=-a:

−a=A(b−a)(c−a)⇒A=a(a−b)(a−c)-a=A(b-a)(c-a)\Rightarrow A=\frac{a}{(a-b)(a-c)}

Put t=−bt=-b:

−b=B(a−b)(c−b)⇒B=b(b−a)(b−c)-b=B(a-b)(c-b)\Rightarrow B=\frac{b}{(b-a)(b-c)}

Put t=−ct=-c:

−c=C(a−c)(b−c)⇒C=c(c−a)(c−b)-c=C(a-c)(b-c)\Rightarrow C=\frac{c}{(c-a)(c-b)}

Now replace t=x2t=x^2:

x2(x2+a)(x2+b)(x2+c)=a(a−b)(a−c)x2+a+b(b−a)(b−c)x2+b+c(c−a)(c−b)x2+c\boxed{\frac{x^2}{(x^2+a)(x^2+b)(x^2+c)}=\frac{\dfrac{a}{(a-b)(a-c)}}{x^2+a}+\frac{\dfrac{b}{(b-a)(b-c)}}{x^2+b}+\frac{\dfrac{c}{(c-a)(c-b)}}{x^2+c}}

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