🎯⭐ INTERACTIVE LESSON

Partial Fraction Decomposition

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Partial Fraction Decomposition - Complete Interactive Lesson

Part 1: Why, When, and Getting Ready

🧩 Partial Fraction Decomposition

Part 1 of 7 — Why, When, and Getting Ready

Topics in This Part

Section
What "decomposition" means
Proper vs. improper rational functions
Factoring the denominator first

🔑 Key Concept: Adding fractions glues several simple pieces into one complicated fraction. Partial fraction decomposition runs that process backwards — it splits a complicated rational function back into the simple pieces it came from. Those pieces are far easier to integrate, expand into series, or graph.

Running Addition in Reverse

You already know how to combine fractions over a common denominator:

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

Decomposition is the reverse trip. Starting from the messy single fraction, we recover the pieces:

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

Each piece on the right is called a partial fraction. The whole point of this unit is to find those numerators ( $3$ and $2$ here) without being told them in advance.

💡 Why bother? In calculus, $\displaystyle\int \frac{5x-1}{(x-2)(x+1)}\,dx$ looks frightening, but is two easy logarithms. The split is the hard part — and that is exactly what this lesson teaches.

Proper vs. Improper

A rational function is $\dfrac{N(x)}{D(x)}$ , a ratio of two polynomials. Compare the degrees (highest powers):

Type	Condition	Example
Proper	$\deg (N$

Concept Check 🎯

Step Zero: Factor the Denominator

The shape of the decomposition is dictated entirely by how the denominator factors. So before anything else, factor $D(x)$ completely.

Example

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

Factor the Denominator 🧮

Factor each denominator into linear factors of the form $(x \pm a)(x \pm b)$ , then enter the two constants (the numbers, with sign) in increasing order.

1) $x^2 - x - 6 = (x + \,?\,)(x + \,?\,)$ — enter the two constants, smallest first.

Proper or Improper? 🔽

Classify each rational function by comparing numerator and denominator degrees.

Part 2: Distinct Linear Factors

🧩 Partial Fraction Decomposition

Part 2 of 7 — Distinct Linear Factors

🔑 The Core Case: When the denominator factors into different linear factors, each factor gets its own partial fraction with a single unknown constant on top.

The Setup

For each distinct linear factor $(x - r)$ in the denominator, write a term $\dfrac{A}{x - r}$ with an unknown constant :

Part 3: Repeated Linear Factors

🧩 Partial Fraction Decomposition

Part 3 of 7 — Repeated Linear Factors

🔑 The Twist: When a linear factor is repeated (raised to a power), one term is not enough. You need a term for every power from $1$ up to the highest.

The Ladder of Powers

A factor like $(x+1)^2$ contributes a ladder of terms — one for each power up to the exponent:

$\frac{}{}$

Part 4: Irreducible Quadratic Factors

🧩 Partial Fraction Decomposition

Part 4 of 7 — Irreducible Quadratic Factors

🔑 The Rule: Over an irreducible quadratic factor (one that won't factor into real linear pieces, like $x^2 + 1$ ), the numerator is linear: $Bx + C$ , not just a constant.

What "Irreducible" Means

A quadratic is irreducible over the reals when it has no real roots — its discriminant is negative, so it never crosses zero and cannot split into real linear factors.

Part 5: Improper Fractions: Divide First

🧩 Partial Fraction Decomposition

Part 5 of 7 — Improper Fractions: Divide First

🔑 The Gate: If $\deg(N) \ge \deg(D)$ , the fraction is improper and decomposition cannot start. Polynomial long division is the key that unlocks the gate.

Why Division Comes First

Long division rewrites any improper fraction as

$\frac{N (x)}{D (x)} = \underset{}{\underset{⏟}{Q (x)}}$

Part 6: Strategy & Full Synthesis

🧩 Partial Fraction Decomposition

Part 6 of 7 — Strategy & Full Synthesis

You now own every individual move. This part is about choosing the right move and stringing them together start to finish.

The Decision Flowchart

Run every problem through this checklist, in order:

Is it proper? If $\deg(N) \ge \deg(D)$ → long divide first (Part 5), then continue with the remainder.
Factor the denominator completely (Step Zero).
Write one group of terms per factor:

Factor in $D (x)$

Part 7: Mixed Mastery & Exit Quiz

🧩 Partial Fraction Decomposition

Part 7 of 7 — Mixed Mastery & Exit Quiz

You can now (1) test proper vs. improper, (2) factor and set up the right form, (3) handle distinct linear, repeated linear, and irreducible quadratic factors, and (4) divide first when needed. Time to put it all together.

Quick Reference

Situation	What to do
$\deg(N) \ge \deg(D)$	Long divide first, then decompose the remainder
Distinct linear $(x - r)$

(x−2)(x+1)3(x+1)+2(x−2)=

\displaystyle\int \frac{3}{x-2}\,dx + \int \frac{2}{x+1}\,dx

de g (N) < de g (D)

⚠️ The golden rule: Partial fraction decomposition only works directly on a proper fraction. If the fraction is improper, you must do polynomial long division first (that is Part 5). For Parts 1–4 every fraction will already be proper.

We look for two numbers that multiply to $-2$ and add to $-1$ : those are $-2$ and $+1$ , giving $(x-2)(x+1)$ .

The kinds of factors you can get are:

Factor type	Looks like	Covered in
Distinct linear	$(x-2)(x+1)$	Part 2
Repeated linear	$(x+1)^2$	Part 3
Irreducible quadratic	$x^2 + 1$	Part 4

🔑 You cannot set up the decomposition until the denominator is fully factored. Factoring is Step Zero every single time.

For example, $x^2 + 5x + 6 = (x+2)(x+3)$ , so you would enter $2$ then $3$ .

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

To find $A$ and $B$ , clear the denominators by multiplying both sides by $(x-2)(x+1)$ :

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

This equation must hold for every $x$ . That gives us two reliable ways to find the constants.

Method: Strategic Substitution

Because $5x - 1 = A(x+1) + B(x-2)$ is true for all $x$ , we may plug in any value. The clever move is to choose $x$ -values that make a factor zero, killing one unknown at a time.

Let $x = 2$ (kills the $B$ term, since $x - 2 = 0$ ): $5 (2) - 1 = A (2 + 1) + B (0) \Rightarrow$

Let $x = -1$ (kills the $A$ term, since $x + 1 = 0$ ): $5 (- 1) - 1 = A (0) + B (- 1 - 2$

So the decomposition is:

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

✅ Check: $\dfrac{3}{x-2} + \dfrac{2}{x+1} = \dfrac{3(x+1) + 2(x-2)}{(x-2)(x+1)} = \dfrac{5x - 1}{(x-2)(x+1)}$ ✓

Concept Check 🎯

The Cover-Up Shortcut

Strategic substitution is so reliable for distinct linear factors that it has a one-line version called the cover-up method.

To find the numerator over $(x - r)$ : cover up the factor $(x - r)$ in the original fraction, then evaluate everything that remains at $x = r$ .

Example: $\dfrac{7x - 1}{(x+1)(x-3)}$

Over $(x+1)$ : cover $(x+1)$ , evaluate $\dfrac{7x-1}{x-3}$ at : .

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

💡 The cover-up method is strategic substitution — it just skips writing out the cleared-denominator equation. It works for distinct linear factors only.

Decompose with Cover-Up 🧮

Decompose $\dfrac{x + 7}{(x-1)(x+3)} = \dfrac{A}{x-1} + \dfrac{B}{x+3}$ .

Enter $A$ in the first box and $B$ in the second box.

Trace the Substitution 🎯

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

In general, $(x - r)^k$ contributes:

$\frac{A_1}{x-r} + \frac{A_2}{(x-r)^2} + \cdots + \frac{A_k}{(x-r)^k}$

⚠️ Common mistake: writing only $\dfrac{A}{(x+1)^2}$ and stopping. You must include the $\dfrac{A}{x+1}$ term too — every power from $1$ to $k$ appears.

Worked Example: $\dfrac{3x + 5}{(x+1)^2}$

Clear denominators by multiplying by $(x+1)^2$ :

$3x + 5 = A(x+1) + B$

Let $x = -1$ (kills the $A$ term): $\;3(-1) + 5 = B \;\Rightarrow\; B = 2$ .

Now compare coefficients of $x$ . The right side is $Ax + (A + B)$ , so the coefficient of $x$ gives $A = 3$ .

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

✅ Check: $\dfrac{3}{x+1} + \dfrac{2}{(x+1)^2} = \dfrac{3(x+1) + 2}{(x+1)^2} = \dfrac{3x + 5}{(x+1)^2}$ ✓

💡 With repeated factors, cover-up gives you the highest-power numerator immediately ( $B$ here), but you finish the lower ones by matching coefficients.

Choose the Correct Setup 🔽

Match each denominator to the right decomposition form (use $A, B, C$ for the unknown constants).

Decompose a Repeated Factor 🧮

Decompose $\dfrac{x - 3}{(x-2)^2} = \dfrac{A}{x-2} + \dfrac{B}{(x-2)^2}$ .

Enter $A$ in the first box and $B$ in the second box.

Concept Check 🎯

Quadratic	Discriminant	Status
$x^2 + 1$	$0 - 4 = -4 < 0$	irreducible
$x^2 + x + 1$	$1 - 4 = -3 < 0$
$x^2 - 4$	$0 + 16 = 16 > 0$	reducible → $(x -$

🔑 Why a linear numerator? A partial fraction's numerator always has degree one less than its denominator. Over a linear factor (degree $1$ ) that means a constant (degree $0$ ); over a quadratic (degree $2$ ) it means a linear $Bx + C$ (degree $1$ ).

Worked Example: $\dfrac{2x^2 - x + 4}{x(x^2 + 1)}$

The denominator has a linear factor $x$ and an irreducible quadratic $x^2 + 1$ , so:

$\frac{2x^2 - x + 4}{x(x^2+1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$

Clear denominators:

$2x^2 - x + 4 = A(x^2 + 1) + (Bx + C)\,x$

Let $x = 0$ : $\;4 = A(1) \Rightarrow A = 4$ .

Expand and match: the right side is $(A + B)x^2 + Cx + A$ . Comparing with $2x^2 - x + 4$ :

$x^2$ : $\;A + B = 2 \Rightarrow 4 + B = 2 \Rightarrow B = -2$

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

✅ Check $x=1$ : LHS $= \frac{2-1+4}{1\cdot 2} = \frac{5}{2}$ ; RHS ✓

Concept Check 🎯

Find the Quadratic-Factor Numerator 🧮

Decompose $\dfrac{x^2 + 2x + 3}{(x+1)(x^2 + 1)} = \dfrac{A}{x+1} + \dfrac{Bx + C}{x^2 + 1}$ .

Enter $A$ , then $B$ , then $C$ .

Match the Numerator Form 🔽

A partial fraction's numerator always has degree one less than its own denominator. Choose the correct numerator form for each piece.

\frac{N ( x )}{D ( x )} = polynomial Q (x) + proper remainder D ( x )

where the remainder $R(x)$ has degree smaller than $D(x)$ . The quotient $Q(x)$ is left as-is; the proper remainder is then decomposed with the methods from Parts 2–4.

Example: $\dfrac{x^2 + 1}{x - 1}$

Dividing $x^2 + 1$ by $x - 1$ gives quotient $x + 1$ and remainder $2$ :

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

✅ Check: $(x+1)(x-1) + 2 = x^2 - 1 + 2 = x^2 + 1$ ✓ — quotient times divisor plus remainder rebuilds the numerator.

Worked Example: $\dfrac{2x^2 + 3}{x^2 - x}$

Here $\deg(N) = \deg(D) = 2$ , so the fraction is improper — divide first.

Dividing $2x^2 + 3$ by $x^2 - x$ gives quotient $2$ and remainder :

$\frac{2x^2 + 3}{x^2 - x} = 2 + \frac{2x + 3}{x^2 - x}$

Now the remainder is proper. Factor $x^2 - x = x(x - 1)$ and decompose:

$\frac{2x + 3}{x(x-1)} = \frac{A}{x} + \frac{B}{x-1}$

Cover-up: $A = \dfrac{2(0)+3}{0-1} = -3$ , and $B =$ . Altogether:

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

💡 Don't forget to carry the quotient $2$ all the way into the final answer.

Concept Check 🎯

Divide, Then Read Off the Quotient 🧮

For $\dfrac{x^2 + 1}{x - 1} = Q(x) + \dfrac{R}{x - 1}$ , the quotient is $Q(x) = x + q_0$ and the remainder is the constant $R$ .

Enter $q_0$ (the constant term of the quotient) in the first box and $R$ in the second box.

Degree of the Quotient 🔽

When you long-divide an improper fraction, the quotient's degree equals $\deg(N) - \deg(D)$ . Pick the quotient's degree for each.

Solve for the unknowns by clearing denominators and using strategic substitution and/or matching coefficients.

🔑 The numerator of each piece always has degree one less than that piece's denominator — that's why linear factors get constants and quadratics get $Bx + C$ .

Diagnose Each Problem 🔽

For each fraction, choose the first correct action or setup.

Full Synthesis: $\dfrac{3x - 4}{x^2 - x - 6}$

Step 1 — Proper? Degree $1$ over degree $2$ . ✓ Proper.

Step 2 — Factor: $x^2 - x - 6 = (x - 3)(x + 2)$ .

Step 3 — Setup: two distinct linear factors: $\frac{3x - 4}{(x-3)(x+2)} = \frac{A}{x-3} + \frac{B}{x+2}$

Step 4 — Solve (cover-up):

$A = \dfrac{3(3) - 4}{3 + 2} = \dfrac{5}{5} = 1$

$\frac{3x - 4}{x^2 - x - 6} = \frac{1}{x-3} + \frac{2}{x+2}$

✅ Check $x=0$ : LHS $= \frac{-4}{-6} = \frac{2}{3}$ ; RHS ✓

Full Decomposition 🧮

Decompose $\dfrac{x + 4}{x(x + 2)} = \dfrac{A}{x} + \dfrac{B}{x + 2}$ from scratch.

Enter $A$ in the first box and $B$ in the second box.

Strategy Check 🎯

⚠️ Three classic traps: decomposing an improper fraction without dividing first; forgetting the lower-power terms of a repeated factor; and putting a constant (instead of $Bx + C$ ) over an irreducible quadratic.

Mixed Practice 🎯

One More from Scratch 🧮

Decompose $\dfrac{5x + 1}{(x - 1)(x + 2)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 2}$ .

Enter $A$ in the first box and $B$ in the second box.

Exit Quiz ✅

Answer all three to finish the lesson.

5 (2) - 1 = A (2 + 1) + B (0) \Rightarrow 9 = 3 A \Rightarrow A = 3

5 (- 1) - 1 = A (0) + B (- 1 - 2) \Rightarrow - 6 = - 3 B \Rightarrow B = 2

(x−2)(x+1)3(x+1)+2(x−2)=

\;\dfrac{7(-1)-1}{-1-3} = \dfrac{-8}{-4} = 2

Over $(x-3)$ : cover

(x-3)

, evaluate

\dfrac{7x-1}{x+1}

x = 3

\;\dfrac{7(3)-1}{3+1} = \dfrac{20}{4} = 5

= \frac{4}{1} + \frac{-3}{2} = \frac{5}{2}

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

B = \dfrac{3(-2) - 4}{-2 - 3} = \dfrac{-10}{-5} = 2

= \frac{1}{-3} + \frac{2}{2} = -\frac{1}{3} + 1 = \frac{2}{3}

Distinct linear $(x - r)$	$\dfrac{A}{x - r}$
Repeated linear $(x - r)^k$	$\dfrac{A_1}{x-r} + \cdots + \dfrac{A_k}{(x-r)^k}$
Irreducible quadratic $(x^2 + bx + c)$	$\dfrac{Bx + C}{x^2 + bx + c}$

Partial Fraction Decomposition

Partial Fraction Decomposition - Complete Interactive Lesson

Part 1: Why, When, and Getting Ready

🧩 Partial Fraction Decomposition

Topics in This Part

Running Addition in Reverse

Proper vs. Improper

Step Zero: Factor the Denominator

Example

Part 2: Distinct Linear Factors

🧩 Partial Fraction Decomposition

The Setup

Part 3: Repeated Linear Factors

🧩 Partial Fraction Decomposition

The Ladder of Powers

Part 4: Irreducible Quadratic Factors

🧩 Partial Fraction Decomposition

What "Irreducible" Means

Part 5: Improper Fractions: Divide First

🧩 Partial Fraction Decomposition

Why Division Comes First

Part 6: Strategy & Full Synthesis

🧩 Partial Fraction Decomposition

The Decision Flowchart

Part 7: Mixed Mastery & Exit Quiz

🧩 Partial Fraction Decomposition

Quick Reference

Method: Strategic Substitution

The Cover-Up Shortcut

Example: 7x−1(x+1)(x−3)\dfrac{7x - 1}{(x+1)(x-3)}(x+1)(x−3)7x−1​

Worked Example: 3x+5(x+1)2\dfrac{3x + 5}{(x+1)^2}(x+1)23x+5​

Worked Example: 2x2−x+4x(x2+1)\dfrac{2x^2 - x + 4}{x(x^2 + 1)}x(x2+1)2x2−x+4​

Example: x2+1x−1\dfrac{x^2 + 1}{x - 1}x−1x2+1​

Worked Example: 2x2+3x2−x\dfrac{2x^2 + 3}{x^2 - x}x2−x2x2+3​

Full Synthesis: 3x−4x2−x−6\dfrac{3x - 4}{x^2 - x - 6}x2−x−63x−4​

Example: $\dfrac{7x - 1}{(x+1)(x-3)}$

Worked Example: $\dfrac{3x + 5}{(x+1)^2}$

Worked Example: $\dfrac{2x^2 - x + 4}{x(x^2 + 1)}$

Example: $\dfrac{x^2 + 1}{x - 1}$

Worked Example: $\dfrac{2x^2 + 3}{x^2 - x}$

Full Synthesis: $\dfrac{3x - 4}{x^2 - x - 6}$