Theory of Intelligence
11TH•Mathematics•Partial Fractions•Question 14

Solution

Resolve into partial fractions: $\dfrac{4x^2-3x+1}{(x+1)(x-1)^2}$

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Solution

Let:

4x2−3x+1(x+1)(x−1)2=Ax+1+Bx−1+C(x−1)2\frac{4x^2-3x+1}{(x+1)(x-1)^2}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{(x-1)^2}

Multiply by (x+1)(x−1)2(x+1)(x-1)^2:

4x2−3x+1=A(x−1)2+B(x+1)(x−1)+C(x+1)4x^2-3x+1=A(x-1)^2+B(x+1)(x-1)+C(x+1)

Solving gives:

A=2,B=2,C=1A=2,\quad B=2,\quad C=1

Hence:

4x2−3x+1(x+1)(x−1)2=2x+1+2x−1+1(x−1)2\boxed{\frac{4x^2-3x+1}{(x+1)(x-1)^2}=\frac{2}{x+1}+\frac{2}{x-1}+\frac{1}{(x-1)^2}}

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