102 LearnersLast updated on July 17th, 2025
Partial Fraction
Breaking a complex rational expression into a sum of simpler fractions is called partial fractions. It is especially used in integration, Laplace transforms, and to solve differential equations. In this article, we will learn more about the partial fractions.
What is Partial Fraction?
Breaking down a complicated fraction into smaller parts is known as partial fraction. These smaller parts are easier to work with when we add, subtract, or integrate them. Like fractions, partial fractions also have a numerator and a denominator. When the denominator has multiple factors, we can split the fraction into a sum of smaller ones.
What are the Formulas for Partial Fractions?
We use special formulas for solving partial fractions based on the type of denominator in the fraction. These formulas tell us to choose a kind of numerator that can be used for each type of denominator. These formulas help us rewrite difficult fractions, which makes integration and simplification easier. The formulas for partial expressions are mentioned below:
| Type | Form of Rational Expression | Partial Fraction Decomposition |
| Non-Repeated Linear Factor | P(x)/(ax + b) | A/(ax + b) |
| Repeated Linear Factor | P(x)/(ax + b)n | A1/(ax + b) + A2/(ax + b)2 + … + An/(ax + b)n |
| Non-Repeated Quadratic Factor | P(x)/(ax2 + bx + c) | Ax + B/(ax2 + bx + c) |
| Repeated Quadratic Factor | P(x)/(ax2 + bx + c)n | A1x + B1/(ax2 + bx + c) + … + Anx + Bn/(ax2 + bx + c)n |
How to Decompose Partial Fractions?
Breaking a big fraction into smaller or simpler fractions that are easier to work with is called partial fraction decomposition. We can learn easily about the partial fraction decomposition using the following steps:
Step 1: Write the given fraction as a sum of simpler fractions. For example, 7x + 4/(x + 1)(x + 2) = A/x + 1 + B/x + 2.
Step 2: To eliminate the denominator, multiply both sides by (x + 1)(x + 2). Therefore, the equation will become 7x + 4 = A(x + 2) + B(x + 1).
Step 3: Now open the brackets and solve the equation.
7x + 4 = A(x + 2) + B(x + 1)
7x + 4 = Ax + 2A + Bx + B
7x + 4 = (A + B)x + (2A + B)
Coefficients of x: A + B = 7
Constant term: 2A + B = 4
Step 4: Solve for A and B
(1) A + B = 7
(2) 2A + B = 4
Subtract equation (1) from equation (2):
(2A + B = 4) - (A + B = 7) = 4 - 7
When subtracting both equations, the B term gets cancelled, and we will get the value of A as:
A = -3
Substitute A = -3 in the first equation,
A + B = 7
-3 + B = 7
B = 7 + 3
B = 10
Therefore, the final answer becomes: 7x + 4/(x + 1)(x + 2) = -3/x + 1 + 10/x + 2.
Partial Fractions of Improper Fractions
When the numerator in a fraction is bigger than the denominator, then it is called an improper fraction. Convert the improper fraction to a proper one before breaking the fraction into smaller parts. To make the improper fraction into a proper fraction, we use the long division method. Follow the steps given below for the improper fraction of partial fractions.
Step 1: Do the long division method. Divide the numerator by the denominator.
Step 2: After long division, we have to write the numbers in the form of,
Quotient + Remainder / Divisor
So, the improper fraction becomes a polynomial and a smaller proper fraction.
Partial Fractions in Integration
Partial fractions in integration involve breaking a fraction into smaller parts and then integrating them. Follow the steps given below:
Step 1: Break the given fraction into smaller fractions,
3x + 5/(x + 1)(x + 2) = A/x + 1 + B/x + 2.
Step 2: Find the values of A and B.
Step 3: Now the equation will become,
∫ (A/x + 1 + B/x + 2)dx
Step 4: We can integrate this easily using the formula:
∫ 1/x + a dx = ln |x + a| + C
Real Life Applications of Partial Fractions
In real life, partial fractions are used especially in fields such as mathematics, engineering, and science. Listed below are some real-life applications where partial fractions are used.
- Calculus: We can solve complex integrals easily by using partial fractions. If a function is too hard to integrate, using partial fractions, we can break it into smaller parts, integrate each part, and then combine the answers.
- Physics: In physics, partial fractions are used to simplify wave and light equations easily. It is also used in solving equations of motion or systems with damping and oscillation.
- Computer Science: Some problems in computer science, especially in recursive functions, can be solved faster by simplifying them using partial functions.
- Mathematical Modeling: For analyzing population dynamics or for finding the spread of any epidemic, or environmental changes, partial fractions are used.
Common Mistakes and How To Avoid Them in Partial Fractions
Students make mistakes when learning partial fractions. Listed below are some of the common mistakes that they make, and the ways to avoid them help them avoid those mistakes.
Mistake 1
Not checking the fraction.
Forgot to check the fractions before dividing the fraction into smaller parts. Check the fraction before making it in smaller parts. If there is an improper fraction, use the long division method and make the improper fraction to the proper fraction.
Mistake 2
Not multiplying the denominator on both sides.
Not multiplying the denominator on both sides leads to a mistake. To eliminate the entire denominator, multiply both sides of the equation.
For example, 3x + 5/(x + 1)(x + 2) = A/x + 1 + B/x + 2 in this equation, we multiplied both sides by (x + 1)(x + 2) to eliminate the denominator.
Mistake 3
Forgetting the final step.
Students might stop after finding the answer for A and B. After finding the values, we have to apply them to the partial fractions. If A = 2 and B = 1, substitute the values back into the partial fractions so the final answer becomes,
3x + 5/(x + 1)(x + 2) = 2/x + 1 + 1/x + 2
Mistake 4
Mixing up the signs
Forgetting to write the negative sign or writing the wrong sign when substituting values. Always be careful with the signs.
Mistake 5
Forgetting to add “C” in integration
Students forget to include the constant of integration while integrating partial fractions, which leads to a mistake. Write + C at the end after solving an integration problem. ∫1/x + a dx = ln |x + a| + C.
Solved Examples of Partial Fractions
Problem 1
Break into partial fractions: 5/(x + 1)(x + 2)
5/(x + 1)(x + 2) = 5/(x + 1) - 5/(x + 2)
Explanation
We can write the equation as:
A/x + 1 + B/x + 2 = 5/(x + 1)(x + 2)
Multiply (x + 1)(x + 2) on both sides to remove the denominators.
A(x +2) + B(x + 1) = 5
Ax + 2A + Bx + B = 5
(Ax + Bx) + (2A + B) = 5
(A + B)x + (2A +B) = 5
Comparing both sides, there is no x coefficient, so it can be written as 0.
A + B = 0, 2A + B = 5
B = -A
Substitute the value of A with 2A + B = 5
2A + (-A) = 5
A = 5
Put A = 5, in B = -A
B= -5
Therefore, the final values in
5/(x + 1)(x + 2) = 5/(x + 1) - 5/(x + 2)
Problem 2
Break this into partial fractions: x + 5/(x + 1)2
x + 5/(x + 1)2 = 1/x + 1 - 4/(x + 1)2
Explanation
The equation will become,
x + 5/(x + 1)2 = A/x + 1 - B/(x + 1)2
Multiply (x +1)2 on both sides,
x + 5 = A(x + 1) + B
x +5 = Ax + A + B
x +5 = Ax + (A + B)
Comparing both sides,
A = 1
Substitute A = 1 in A + B = 5
A + B = 5
1 + B = 5
B = 5 - 1
B = 4
Therefore, the answer becomes,
x + 5/(x + 1)2 = 1/x + 1 - 4/(x + 1)2
Problem 3
Break this into partial fractions: 7x + 5/(x + 1)(x + 2)
7x + 5/(x + 1)(x + 2) = -2/x + 1 + 9/x + 2
Explanation
We can write the equation as,
7x + 5/(x + 1)(x + 2) = A/x + 1 + B/x + 2
Multiply (x + 1)(x + 2) on both sides,
7x + 5 = A(x + 2) + B(x + 1)
7x + 5 = Ax + 2A + Bx + B
7x + 5 = (Ax + Bx) + (2A + B)
7x + 5 = (A+ B)x + (2A + B)
Comparing both sides,
A + B = 7, 2A + B = 5
Subtract 2A + B = 5 and A + B = 7
A = -2
Substitute the value of A for A + B = 7
A + B = 7
-2 + B = 7
B = 7 + 2
B = 9
Therefore, the final answer becomes,
7x + 5/(x + 1)(x + 2) = -2/x + 1 + 9/x + 2
Problem 4
Break this into partial fractions: 2x + 3/(x + 1)2
2x + 3/(x + 1)2 = 2/(x + 1) + 1/(x + 1)2
Explanation
The equation will become,
2x + 3/(x + 1)2 = A/(x + 1) + B/(x + 1)2
Multiply both sides by (x + 1)2:
2x + 3 = A(x +1) + B
2x + 3 = Ax +A + B
2x + 3 = Ax + (A + B)
Compare both sides,
A = 2, A + B = 3
Substitute A =2 in A + B = 3
A + B = 3
2 + B = 3
B = 3 - 2
B = 1
The final answer is:
2x + 3/(x + 1)2 = 2/(x + 1) + 1/(x + 1)2
Problem 5
Break this into partial fractions: 5x + 9/(x + 2)(x + 3)
5x + 9/(x + 2)(x + 3) = -1/x + 2 + 6/x + 3
Explanation
The equation is:
5x + 9/(x + 2)(x + 3) = A/x + 2 + B/x + 3
Multiply both the sides by (x +2)(x + 3),
5x + 9 = A(x + 3) + B(x + 2)
5x + 9 = Ax + 3A + Bx + 2B
5x + 9 = (Ax + Bx) + (3A + 2B)
5x + 9 = (A + B)x + (3A + 2B)
Comparing both sides,
A + B = 5, 3A + 2B = 9
Simplify the equation,
A + B = 5
B = 5 - A
Substitute B = 5 - A to 3A + 2B = 9
3A + 2B = 9
3A + 2(5 - A) = 9
3A + 10 - 2A = 9
A + 10 = 9
A = 9 - 10
A = -1
Substitute A = -1 to A + B = 5
A + B = 5
-1 + B = 5
B = 5 + 1
B = 6
Therefore, the equation will become
5x + 9/(x + 2)(x + 3) = -1/x + 2 + 6/x + 3
FAQs of Partial Fractions
1.What are partial fractions?
2.Why do we use partial fractions?
3.When can a fraction be split?
4.Where are partial fractions used in real life?
5.Can partial fractions have more than three parts?
6.How does learning Algebra help students in Canada make better decisions in daily life?
7.How can cultural or local activities in Canada support learning Algebra topics such as Partial Fraction?
8.How do technology and digital tools in Canada support learning Algebra and Partial Fraction?
9.Does learning Algebra support future career opportunities for students in Canada?
Jaskaran Singh Saluja
About the Author
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
Fun Fact
: He loves to play the quiz with kids through algebra to make kids love it.
