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Toolkit 43

Partial Fraction Decomposition

5x+1(x−1)(x+2)=2x−1+3x+2\frac{5x+1}{(x-1)(x+2)} = \frac{2}{x-1} + \frac{3}{x+2}

Partial Fraction Decomposition is a technique for rewriting a rational expression as a sum of simpler fractions.

Two Ways to Calculate the Numerators

Start with

5x+1(x−1)(x+2)=Ax−1+Bx+2\frac{5x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}

Approach 1 — Comparing Coefficients

Multiply both sides by (x−1)(x+2)(x-1)(x+2):

5x+1=A(x+2)+B(x−1)5x+1 = A(x+2)+B(x-1)

Expand:

5x+1=(A+B)x+(2A−B)5x+1 = (A+B)x+(2A-B)

Compare coefficients:

A+B=5A+B=5
2A−B=12A-B=1

Therefore,

A=2,B=3.A=2,\qquad B=3.

Approach 2 — Heaviside Cover-Up Method

Let

f(x)=5x+1(x−1)(x+2)=Ax−1+Bx+2.f(x) = \frac{5x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}.

Multiply both sides by (x−1)(x-1):

(x−1)f(x)=5x+1x+2=A+B(x−1)x+2.(x-1)f(x) = \frac{5x+1}{x+2} = A+\frac{B(x-1)}{x+2}.

Substitute x=1x=1:

A=5(1)+11+2=63=2.A = \frac{5(1)+1}{1+2} = \frac{6}{3} = 2.

Similarly, multiply both sides by (x+2)(x+2):

(x+2)f(x)=5x+1x−1=A(x+2)x−1+B.(x+2)f(x) = \frac{5x+1}{x-1} = \frac{A(x+2)}{x-1}+B.

Substitute x=−2x=-2:

B=5(−2)+1−2−1=−9−3=3.B = \frac{5(-2)+1}{-2-1} = \frac{-9}{-3} = 3.
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