Partial Fraction Decomposition

Breaking down rational expressions into simpler fractions for integration and analysis

Partial Fraction Decomposition

What is Partial Fraction Decomposition?

Partial fraction decomposition is the process of breaking a complex rational expression into a sum of simpler fractions.

Think of it as the reverse of adding fractions!

Example Concept

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

We want to find the values of AA and BB.

Why Use Partial Fractions?

  1. Simplify integration (used extensively in Calculus)
  2. Solve differential equations
  3. Simplify complex expressions
  4. Inverse Laplace transforms (engineering)

When Can We Use It?

Requirements:

  1. The expression must be a proper fraction: degree of numerator < degree of denominator
  2. If improper, use long division first to get: quotient + proper fraction

Cases for Denominator Factorization

Case 1: Distinct Linear Factors

If the denominator factors as (x−a1)(x−a2)...(x−an)(x - a_1)(x - a_2)...(x - a_n) with all different factors:

P(x)(x−a1)(x−a2)...(x−an)=A1x−a1+A2x−a2+...+Anx−an\frac{P(x)}{(x - a_1)(x - a_2)...(x - a_n)} = \frac{A_1}{x - a_1} + \frac{A_2}{x - a_2} + ... + \frac{A_n}{x - a_n}

Case 2: Repeated Linear Factors

If (x−a)(x - a) appears nn times:

P(x)(x−a)n=A1x−a+A2(x−a)2+...+An(x−a)n\frac{P(x)}{(x - a)^n} = \frac{A_1}{x - a} + \frac{A_2}{(x - a)^2} + ... + \frac{A_n}{(x - a)^n}

Case 3: Distinct Irreducible Quadratic Factors

If the denominator has a quadratic that cannot be factored (like x2+1x^2 + 1):

P(x)(x2+bx+c)=Ax+Bx2+bx+c\frac{P(x)}{(x^2 + bx + c)} = \frac{Ax + B}{x^2 + bx + c}

Note: The numerator is linear (Ax+B)(Ax + B), not just a constant!

Case 4: Repeated Quadratic Factors

Similar to Case 2, but with linear numerators for each power.

Method for Finding Constants

Method 1: Clear Denominators and Equate Coefficients

  1. Multiply both sides by the common denominator
  2. Expand and collect like terms
  3. Equate coefficients of corresponding powers of xx
  4. Solve the system of equations

Method 2: Substitute Convenient Values

  1. Multiply both sides by the common denominator
  2. Substitute strategic values of xx to eliminate variables
  3. Usually choose xx values that make factors equal to zero

Step-by-Step Process

  1. Check if proper: If not, use polynomial division first
  2. Factor the denominator completely
  3. Set up partial fraction form based on the factors
  4. Clear denominators by multiplying both sides
  5. Find constants using substitution or equating coefficients
  6. Write final answer as sum of partial fractions

📚 Practice Problems

1Problem 1medium

❓ Question:

Decompose into partial fractions: 7x−1(x−2)(x+3)\frac{7x - 1}{(x - 2)(x + 3)}

💡 Show Solution

Solution:

Step 1: Set up the partial fraction form.

Since we have two distinct linear factors: 7x−1(x−2)(x+3)=Ax−2+Bx+3\frac{7x - 1}{(x - 2)(x + 3)} = \frac{A}{x - 2} + \frac{B}{x + 3}

Step 2: Clear denominators.

Multiply both sides by (x−2)(x+3)(x - 2)(x + 3): 7x−1=A(x+3)+B(x−2)7x - 1 = A(x + 3) + B(x - 2)

Step 3: Find AA and BB using substitution.

Let x=2x = 2 (makes the BB term disappear): 7(2)−1=A(2+3)+B(0)7(2) - 1 = A(2 + 3) + B(0) 14−1=5A14 - 1 = 5A 13=5A13 = 5A A=135A = \frac{13}{5}

Let x=−3x = -3 (makes the AA term disappear): 7(−3)−1=A(0)+B(−3−2)7(-3) - 1 = A(0) + B(-3 - 2) −21−1=−5B-21 - 1 = -5B −22=−5B-22 = -5B B=225B = \frac{22}{5}

Step 4: Write the final answer.

7x−1(x−2)(x+3)=13/5x−2+22/5x+3\frac{7x - 1}{(x - 2)(x + 3)} = \frac{13/5}{x - 2} + \frac{22/5}{x + 3}

Or: =135(x−2)+225(x+3)= \frac{13}{5(x - 2)} + \frac{22}{5(x + 3)}

Answer: 135(x−2)+225(x+3)\frac{13}{5(x - 2)} + \frac{22}{5(x + 3)}

2Problem 2medium

❓ Question:

Decompose into partial fractions: 3x+5(x+1)2\frac{3x + 5}{(x + 1)^2}

💡 Show Solution

Solution:

Step 1: Set up the form for repeated linear factors.

Since (x+1)(x + 1) appears twice: 3x+5(x+1)2=Ax+1+B(x+1)2\frac{3x + 5}{(x + 1)^2} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2}

Step 2: Clear denominators.

Multiply by (x+1)2(x + 1)^2: 3x+5=A(x+1)+B3x + 5 = A(x + 1) + B

Step 3: Expand. 3x+5=Ax+A+B3x + 5 = Ax + A + B

Step 4: Equate coefficients.

Coefficient of xx: 3=A3 = A Constant term: 5=A+B5 = A + B

Step 5: Solve for BB. 5=3+B5 = 3 + B B=2B = 2

Step 6: Write the answer.

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

Answer: 3x+1+2(x+1)2\frac{3}{x + 1} + \frac{2}{(x + 1)^2}

3Problem 3hard

❓ Question:

Decompose into partial fractions: 2x2+3x+4x(x2+4)\frac{2x^2 + 3x + 4}{x(x^2 + 4)}

💡 Show Solution

Solution:

Step 1: Identify the factors.

We have:

  • One linear factor: xx
  • One irreducible quadratic: x2+4x^2 + 4 (cannot factor over reals)

Step 2: Set up the partial fraction form.

For the quadratic factor, use a linear numerator: 2x2+3x+4x(x2+4)=Ax+Bx+Cx2+4\frac{2x^2 + 3x + 4}{x(x^2 + 4)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 4}

Step 3: Clear denominators.

Multiply by x(x2+4)x(x^2 + 4): 2x2+3x+4=A(x2+4)+(Bx+C)(x)2x^2 + 3x + 4 = A(x^2 + 4) + (Bx + C)(x)

Step 4: Expand. 2x2+3x+4=Ax2+4A+Bx2+Cx2x^2 + 3x + 4 = Ax^2 + 4A + Bx^2 + Cx 2x2+3x+4=(A+B)x2+Cx+4A2x^2 + 3x + 4 = (A + B)x^2 + Cx + 4A

Step 5: Equate coefficients.

x2x^2: 2=A+B2 = A + B x1x^1: 3=C3 = C x0x^0: 4=4A4 = 4A, so A=1A = 1

Step 6: Solve for BB. 2=1+B2 = 1 + B B=1B = 1

Step 7: Write the answer.

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

Answer: 1x+x+3x2+4\frac{1}{x} + \frac{x + 3}{x^2 + 4}

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