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

Solution

Resolve into partial fractions: $\dfrac{x^2+x}{(x^2-1)^2}$

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Solution

Factor:

(x2−1)2=(x−1)2(x+1)2(x^2-1)^2=(x-1)^2(x+1)^2

Let:

x2+x(x−1)2(x+1)2=Ax−1+B(x−1)2+Cx+1+D(x+1)2\frac{x^2+x}{(x-1)^2(x+1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}+\frac{D}{(x+1)^2}

Solving gives:

A=14,B=12,C=−14,D=0A=\frac14,\quad B=\frac12,\quad C=-\frac14,\quad D=0

Hence:

x2+x(x2−1)2=14(x−1)+12(x−1)2−14(x+1)\boxed{\frac{x^2+x}{(x^2-1)^2}=\frac{1}{4(x-1)}+\frac{1}{2(x-1)^2}-\frac{1}{4(x+1)}}

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