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Partial Fraction Decomposition | Study Mondo
Topics / Polynomial and Rational Functions / Partial Fraction Decomposition Partial Fraction Decomposition Breaking down rational expressions into simpler fractions for integration and analysis
BC Written and reviewed by Brendan Cusack , Study Mondo Education Team • Last updated April 18, 2026
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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 ) = A x − 1 + B x + 2 \frac{5}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2} ( x − 1 ) ( x + 2 ) 5 = x − 1 A + x + 2 B
We want to find the values of A A A B B 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?
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 ) (x - a_1)(x - a_2)...(x - a_n) ( x − a 1 ) ( x − a 2 ) ... ( x − a n )
P ( x ) ( x − a 1 ) ( x − a 2 ) . . . ( x − a n ) = A 1 x − a 1 + A 2 x − a 2 + . . . + A n x − a n \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} ( x − a 1 ) ( x − a 2 ) ... ( x − a n ) P ( x ) = x − a 1 A 1 + x − a 2 A 2 + ... + x − a n A n
Case 2: Repeated Linear Factors If ( x − a ) (x - a) ( x − a ) n n n
P ( x ) ( x − a ) n = A 1 x − a + A 2 ( x − a ) 2 + . . . + A n ( 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} ( x − a ) n P ( x ) = x − a A 1 + ( x − a ) 2 A 2 + ... + ( x − a ) n A n
Case 3: Distinct Irreducible Quadratic Factors If the denominator has a quadratic that cannot be factored (like x 2 + 1 x^2 + 1 x 2 + 1
P ( x ) ( x 2 + b x + c ) = A x + B x 2 + b x + c \frac{P(x)}{(x^2 + bx + c)} = \frac{Ax + B}{x^2 + bx + c} ( x 2 + b x + c ) P ( x ) = x 2 + b x + c A x + B
Note: The numerator is linear ( A x + B ) (Ax + B) ( A x + B )
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 x x
Solve the system of equations
Method 2: Substitute Convenient Values
Multiply both sides by the common denominator
Substitute strategic values of x x x
Usually choose x x x
Step-by-Step Process
Check if proper : If not, use polynomial division firstFactor the denominator completelySet up partial fraction form based on the factorsClear denominators by multiplying both sidesFind constants using substitution or equating coefficientsWrite final answer as sum of partial fractions
📚 Practice Problems
1 Problem 1medium ❓ Question:Decompose into partial fractions: 7 x − 1 ( x − 2 ) ( x + 3 ) \frac{7x - 1}{(x - 2)(x + 3)} ( x − 2 ) ( x + 3 ) 7 x − 1
💡 Show Solution Solution:
Step 1: Set up the partial fraction form.
Since we have two distinct linear factors:
7 x − 1 ( x − 2 ) ( x + 3 ) = A x − 2 + B x + 3 \frac{7x - 1}{(x - 2)(x + 3)} = \frac{A}{x - 2} + \frac{B}{x + 3} ( x − 2 ) ( x + 3 )
2 Problem 2medium ❓ Question:Decompose into partial fractions: 3 x + 5 ( x + 1 ) 2 \frac{3x + 5}{(x + 1)^2} ( x + 1 ) 2 3
3 Problem 3hard ❓ Question:Decompose into partial fractions: 2 x 2 + 3 x + 4 x ( x 2 + 4 ) \frac{2x^2 + 3x + 4}{x(x^2 + 4)} x ( x 2 + 4 )
4 Problem 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)
5 Problem 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
Explain using: 📝 Simple words 🔗 Analogy 🎨 Visual desc. 📐 Example 💡 Explain
⚠️ Common Mistakes: Partial Fraction DecompositionAvoid these 4 frequent errors
1 Forgetting the constant of integration (+C) on indefinite integrals
▾ 2 Confusing the Power Rule with the Chain Rule
▾ 3 Not checking continuity before applying the Mean Value Theorem
▾ 4 Dropping negative signs when differentiating trig functions
▾ 🌍 Real-World Applications: Partial Fraction DecompositionSee how this math is used in the real world
⚙️ Optimizing Package Design
Engineering
▾ 🏥 Predicting Drug Dosage Decay
Medicine
▾ 🔬 Calculating Distance from Velocity
Physics
▾ 💰 Revenue Optimization
Finance
▾
📝 Worked Example: Related Rates — Expanding CircleProblem: A stone is dropped into a still pond, creating a circular ripple. The radius of the ripple is increasing at a rate of 2 2 2 10 10 10
1 Identify the known and unknown rates Click to reveal →
2 Write the relationship between variables
3 Differentiate both sides with respect to time
🧪 Practice Lab Interactive practice problems for Partial Fraction Decomposition
▾ 📌 Related Topics in Polynomial and Rational Functions❓ Frequently Asked QuestionsWhat is Partial Fraction Decomposition?▾ Breaking down rational expressions into simpler fractions for integration and analysis
How can I study Partial Fraction Decomposition effectively?▾ Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 5 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Partial Fraction Decomposition study guide free?▾ Yes — all study notes, flashcards, and practice problems for Partial Fraction Decomposition on Study Mondo are free to access. No account is needed.
What course covers Partial Fraction Decomposition?▾ Partial Fraction Decomposition is part of the AP Precalculus course on Study Mondo, specifically in the Polynomial and Rational Functions section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Partial Fraction Decomposition?▾ Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.
💡 Study Tips✓ Work through examples step-by-step ✓ Practice with flashcards daily ✓ Review common mistakes 7
x
−
1
=
x − 2 A +
x + 3 B
Step 2: Clear denominators.
Multiply both sides by ( x − 2 ) ( x + 3 ) (x - 2)(x + 3) ( x − 2 ) ( x + 3 ) 7 x − 1 = A ( x + 3 ) + B ( x − 2 ) 7x - 1 = A(x + 3) + B(x - 2) 7 x − 1 = A ( x + 3 ) + B ( x − 2 )
Step 3: Find A A A B B B
Let x = 2 x = 2 x = 2 (makes the B B B 7 ( 2 ) − 1 = A ( 2 + 3 ) + B ( 0 ) 7(2) - 1 = A(2 + 3) + B(0) 7 ( 2 ) − 1 = A ( 2 + 3 ) + B ( 0 ) 14 − 1 = 5 A 14 - 1 = 5A 14 − 1 = 5 A 13 = 5 A 13 = 5A 13 = 5 A A = 13 5 A = \frac{13}{5} A = 5 13
Let x = − 3 x = -3 x = − 3 (makes the A A A 7 ( − 3 ) − 1 = A ( 0 ) + B ( − 3 − 2 ) 7(-3) - 1 = A(0) + B(-3 - 2) 7 ( − 3 ) − 1 = A ( 0 ) + B ( − 3 − 2 ) − 21 − 1 = − 5 B -21 - 1 = -5B − 21 − 1 = − 5 B − 22 = − 5 B -22 = -5B − 22 = − 5 B B = 22 5 B = \frac{22}{5} B = 5 22
Step 4: Write the final answer.
7 x − 1 ( x − 2 ) ( x + 3 ) = 13 / 5 x − 2 + 22 / 5 x + 3 \frac{7x - 1}{(x - 2)(x + 3)} = \frac{13/5}{x - 2} + \frac{22/5}{x + 3} ( x − 2 ) ( x + 3 ) 7 x − 1 = x − 2 13/5 + x + 3 22/5
Or: = 13 5 ( x − 2 ) + 22 5 ( x + 3 ) = \frac{13}{5(x - 2)} + \frac{22}{5(x + 3)} = 5 ( x − 2 ) 13 + 5 ( x + 3 ) 22
Answer: 13 5 ( x − 2 ) + 22 5 ( x + 3 ) \frac{13}{5(x - 2)} + \frac{22}{5(x + 3)} 5 ( x − 2 ) 13 + 5 ( x + 3 ) 22
x
+
5
💡 Show Solution Solution:
Step 1: Set up the form for repeated linear factors.
Since ( x + 1 ) (x + 1) ( x + 1 ) 3 x + 5 ( x + 1 ) 2 = A x + 1 + B ( x + 1 ) 2 \frac{3x + 5}{(x + 1)^2} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} ( x + 1 ) 2 3 x + 5 = x + 1 A + ( x + 1 ) 2 B
Step 2: Clear denominators.
Multiply by ( x + 1 ) 2 (x + 1)^2 ( x + 1 ) 2 3 x + 5 = A ( x + 1 ) + B 3x + 5 = A(x + 1) + B 3 x + 5 = A ( x +
Step 3: Expand.
3 x + 5 = A x + A + B 3x + 5 = Ax + A + B 3 x + 5 = A x + A + B
Step 4: Equate coefficients.
Coefficient of x x x 3 = A 3 = A 3 = A 5 = A + B 5 = A + B 5 = A + B
Step 5: Solve for B B B 5 = 3 + B 5 = 3 + B 5 = 3 + B B = 2 B = 2 B = 2
Step 6: Write the answer.
3 x + 5 ( x + 1 ) 2 = 3 x + 1 + 2 ( x + 1 ) 2 \frac{3x + 5}{(x + 1)^2} = \frac{3}{x + 1} + \frac{2}{(x + 1)^2} ( x + 1 ) 2 3 x + 5 =
Answer: 3 x + 1 + 2 ( x + 1 ) 2 \frac{3}{x + 1} + \frac{2}{(x + 1)^2} x + 1 3 + ( x + 1 )
2 x 2 + 3 x + 4
💡 Show Solution Solution:
Step 1: Identify the factors.
We have:
One linear factor: x x x
One irreducible quadratic: x 2 + 4 x^2 + 4 x 2 + 4
Step 2: Set up the partial fraction form.
For the quadratic factor, use a linear numerator:
2 x 2 + 3 x + 4 x ( x 2 + 4 ) = A x + B x + C x 2 + 4 \frac{2x^2 + 3x + 4}{x(x^2 + 4)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 4} x ( x 2 + 4 ) 2 x 2 + 3 x + 4 =
Step 3: Clear denominators.
Multiply by x ( x 2 + 4 ) x(x^2 + 4) x ( x 2 + 4 ) 2 x 2 + 3 x + 4 = A ( x 2 + 4 ) + ( B x + C ) ( x ) 2x^2 + 3x + 4 = A(x^2 + 4) + (Bx + C)(x) 2 x 2
Step 4: Expand.
2 x 2 + 3 x + 4 = A x 2 + 4 A + B x 2 + C x 2x^2 + 3x + 4 = Ax^2 + 4A + Bx^2 + Cx 2 x 2 + 3 x + 4 = A x 2 + 4
Step 5: Equate coefficients.
x 2 x^2 x 2 2 = A + B 2 = A + B 2 = A + B x 1 x^1 x 1
Step 6: Solve for B B B 2 = 1 + B 2 = 1 + B 2 = 1 + B B = 1 B = 1 B = 1
Step 7: Write the answer.
2 x 2 + 3 x + 4 x ( x 2 + 4 ) = 1 x + x + 3 x 2 + 4 \frac{2x^2 + 3x + 4}{x(x^2 + 4)} = \frac{1}{x} + \frac{x + 3}{x^2 + 4} x ( x 2 + 4 ) 2 x
Answer: 1 x + x + 3 x 2 + 4 \frac{1}{x} + \frac{x + 3}{x^2 + 4} x 1 + x 2 +
1
)
+
B
x + 1 3 +
( x + 1 ) 2 2
2
2
x
A
+
x 2 + 4 B x + C
+
3 x +
4 =
A ( x 2 +
4 ) +
( B x +
C ) ( x )
A
+
B x 2 +
C x
2 x 2 + 3 x + 4 = ( A + B ) x 2 + C x + 4 A 2x^2 + 3x + 4 = (A + B)x^2 + Cx + 4A 2 x 2 + 3 x + 4 = ( A + B ) x 2 + C x + 4 A 3 = C 3 = C 3 = C
2
+
3
x
+
4
=
x 1 +
x 2 + 4 x + 3
4
x + 3