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

$\frac{5}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$

We want to find the values of $A$ and $B$ .

Why Use Partial Fractions?

Simplify integration (used extensively in Calculus)
Solve differential equations
Simplify complex expressions
Inverse Laplace transforms (engineering)

When Can We Use It?

Requirements:

The expression must be a proper fraction: degree of numerator < degree of denominator
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 $(x - a_1)(x - a_2)...(x - a_n)$ with all different factors:

$\frac{P(x)}{(x - a_1)(x - a_2)...(x - a_n)} = \frac{A_1}{x - a_1} + \frac{A_2}{x - a_2} + ... + \frac{A_n}{x - a_n}$

Case 2: Repeated Linear Factors

If $(x - a)$ appears $n$ times:

$\frac{P(x)}{(x - a)^n} = \frac{A_1}{x - a} + \frac{A_2}{(x - a)^2} + ... + \frac{A_n}{(x - a)^n}$

Case 3: Distinct Irreducible Quadratic Factors

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

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

Note: The numerator is linear $(Ax + B)$ , 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

Multiply both sides by the common denominator
Expand and collect like terms
Equate coefficients of corresponding powers of $x$
Solve the system of equations

Method 2: Substitute Convenient Values

Multiply both sides by the common denominator
Substitute strategic values of $x$ to eliminate variables
Usually choose $x$ values that make factors equal to zero

Step-by-Step Process

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

📚 Practice Problems

1Problem 1medium

❓ Question:

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

💡 Show Solution

Solution:

Step 1: Set up the partial fraction form.

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

Step 2: Clear denominators.

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

Step 3: Find $A$ and $B$ using substitution.

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

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

Step 4: Write the final answer.

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

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

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

2Problem 2medium

❓ Question:

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

💡 Show Solution

Solution:

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

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

Step 2: Clear denominators.

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

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

Step 4: Equate coefficients.

Coefficient of $x$ : $3 = A$ Constant term: $5 = A + B$

Step 5: Solve for $B$ . $5 = 3 + B$ $B = 2$

Step 6: Write the answer.

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

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

3Problem 3hard

❓ Question:

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

💡 Show Solution

Solution:

Step 1: Identify the factors.

We have:

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

Step 2: Set up the partial fraction form.

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

Step 3: Clear denominators.

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

Step 4: Expand. $2x^2 + 3x + 4 = Ax^2 + 4A + Bx^2 + Cx$ $2x^2 + 3x + 4 = (A + B)x^2 + Cx + 4A$

Step 5: Equate coefficients.

$x^2$ : $2 = A + B$ $x^1$ : $3 = C$ $x^0$ : $4 = 4A$ , so $A = 1$

Step 6: Solve for $B$ . $2 = 1 + B$ $B = 1$

Step 7: Write the answer.

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

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

4Problem 4medium

❓ Question:

Decompose into partial fractions: (5x + 2)/[(x + 1)(x - 2)]

💡 Show Solution

Step 1: Set up the partial fraction form: (5x + 2)/[(x + 1)(x - 2)] = A/(x + 1) + B/(x - 2)

Step 2: Multiply both sides by (x + 1)(x - 2): 5x + 2 = A(x - 2) + B(x + 1)

Step 3: Method 1 - Substitution (x = 2): 5(2) + 2 = A(0) + B(3) 12 = 3B B = 4

Step 4: Method 1 - Substitution (x = -1): 5(-1) + 2 = A(-3) + B(0) -3 = -3A A = 1

Step 5: Write the decomposition: (5x + 2)/[(x + 1)(x - 2)] = 1/(x + 1) + 4/(x - 2)

Step 6: Verify by combining: 1/(x + 1) + 4/(x - 2) = [(x - 2) + 4(x + 1)]/[(x + 1)(x - 2)] = [x - 2 + 4x + 4]/[(x + 1)(x - 2)] = (5x + 2)/[(x + 1)(x - 2)] ✓

Answer: 1/(x + 1) + 4/(x - 2)

5Problem 5hard

❓ Question:

Decompose: (3x² + 2x + 1)/[x(x - 1)²]

💡 Show Solution

Step 1: Set up partial fractions (note repeated factor): (3x² + 2x + 1)/[x(x - 1)²] = A/x + B/(x - 1) + C/(x - 1)²

Step 2: Multiply by x(x - 1)²: 3x² + 2x + 1 = A(x - 1)² + Bx(x - 1) + Cx

Step 3: Substitute x = 0: 1 = A(1) + 0 + 0 A = 1

Step 4: Substitute x = 1: 3 + 2 + 1 = 0 + 0 + C(1) C = 6

Step 5: Substitute another value, say x = 2: 3(4) + 2(2) + 1 = 1(1) + B(2)(1) + 6(2) 17 = 1 + 2B + 12 4 = 2B B = 2

Step 6: Write the decomposition: (3x² + 2x + 1)/[x(x - 1)²] = 1/x + 2/(x - 1) + 6/(x - 1)²

Answer: 1/x + 2/(x - 1) + 6/(x - 1)²

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