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Partial Fractions

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Partial Fractions - Complete Interactive Lesson

Part 1: The Concept

Partial Fraction Decomposition

Part 1 of 7 — The Concept & Distinct Linear Factors

Partial fractions is a technique for integrating rational functions (polynomial ÷ polynomial). The idea: break a complex fraction into simpler pieces that are easy to integrate.

Part	Topic
1	Distinct Linear Factors
2	Repeated Linear Factors
3	Irreducible Quadratic Factors
4	Integration with Partial Fractions
5	Long Division First
6	Problem-Solving Workshop
7	Comprehensive Review

When to Use Partial Fractions

Use partial fractions when:

The integrand is a proper rational function (degree of numerator < degree of denominator)
The denominator can be factored
u-substitution doesn’t work directly

Prerequisite: Factor the Denominator

Every polynomial with real coefficients factors into:

Linear factors: $(x - a)$
Irreducible quadratic factors: $(x^2 + bx + c)$ where

Case 1: Distinct Linear Factors

$\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$

Three Distinct Linear Factors

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

Decompose:

Decomposition Practice

Setting Up Decompositions

Cover-Up Computation

Key Takeaways — Part 1

Concept	Details
When to use	Proper rational function with factorable denominator
Distinct linear factors	One constant per factor: $\frac{A}{x-a} + \frac{B}{x-b}$

Part 2: Repeated Linear Factors

Partial Fraction Decomposition

Part 2 of 7 — Repeated Linear Factors

When the denominator has a factor like $(x-a)^n$ , you need $n$ separate terms with increasing powers in the denominator.

Setup for Repeated Factors

$\frac{P (x)}{(x - a)^{n}} = \frac{A_{1}}{}$

Part 3: Integration Practice

Partial Fraction Decomposition

Part 3 of 7 — Irreducible Quadratic Factors

When the denominator contains a quadratic that can’t be factored over the reals (discriminant < 0), the numerator in that partial fraction must be LINEAR, not constant.

Setup Rule

For an irreducible quadratic factor $(ax^2 + bx + c)$ :

$\frac{P (x)}{(x - r)}$

Part 4: Long Division First

Partial Fraction Decomposition

Part 4 of 7 — Integration with Partial Fractions

Now let’s put the decomposition and integration together in complete worked problems from start to finish.

Complete Workflow

Step	Action
1	Check: is it proper? (deg numerator < deg denominator)
2	If improper, do long division first
3	Factor the denominator completely
4	Write the decomposition template
5	Find the constants ( $A$ , $B$ , $C$ , ...)

Part 5: Logistic DE Connection

Partial Fraction Decomposition

Part 5 of 7 — Long Division First (Improper Fractions)

Partial fractions only works on proper rational functions (degree of numerator < degree of denominator). When the fraction is improper, you must do polynomial long division first.

Proper vs. Improper

Fraction	Proper?	Action
$\frac{3x+1}{x^2-4}$

Part 6: Practice Workshop

Partial Fraction Decomposition

Part 6 of 7 — Problem-Solving Workshop

Mixed practice combining all partial fraction techniques. For each problem, decide: Is it proper? What type of factors? Then decompose and integrate.

Decision Guide

Question	If Yes...
Is deg(num) $\geq$ deg(den)?	Long division first
All distinct linear factors?	$\frac{A}{x-a} + \frac{B}{x-b} + \ldots$

Part 7: Final Assessment

Partial Fraction Decomposition — Review

Part 7 of 7 — Comprehensive Review & Assessment

Complete Reference

Denominator Type	Decomposition Form	Integration Result
$(x-a)$ distinct	$\frac{A}{x-a}$

Key Fact: AP Calculus BC only tests partial fractions with linear factors and non-repeating irreducible quadratics. The decomposition setup is the hardest part — once you have the pieces, integration is straightforward.

Worked Example: $\int \frac{1}{x^2 - 1}\,dx$

Step	Work
Factor denominator	$x^2-1 = (x-1)(x+1)$
Set up decomposition	$\frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$
Multiply through	$1 = A(x+1) + B(x-1)$
Plug in $x = 1$	$1 = 2A \Rightarrow A = \frac{1}{2}$
Plug in $x = -1$	$1 = -2B \Rightarrow B = -\frac{1}{2}$
Integrate	$\frac{1}{2}\ln

$\boxed{\int \frac{1}{x^2-1}\,dx = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C}$

\frac{A}{x + 1} + \frac{B}{x - 1} + \frac{C}{x + 3}

Multiply through: $x + 7 = A(x-1)(x+3) + B(x+1)(x+3) + C(x+1)(x-1)$

Substitution	Equation	Result
$x = 1$	$8 = B(2)(4) = 8B$	$B = 1$
$x = -1$	$6 = A(-2)(2) = -4A$	$A = -\frac{3}{2}$
$x = -3$	$4 = C(-2)(-4) = 8C$	$C = \frac{1}{2}$

AP Tip: The “cover-up” method (Heaviside) is fastest: to find $A$ , cover $(x+1)$ in the original fraction and evaluate at $x = -1$ .

Coming Up: Part 2 handles repeated linear factors like $(x-a)^2$ or $(x-a)^3$ .

\frac{P ( x )}{( x - a ) ^{n}} = \frac{A _{1}}{x - a} + \frac{A _{2}}{( x - a ) ^{2}} + \dots + \frac{A _{n}}{( x - a ) ^{n}}

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

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

Multiply through: $3x + 5 = A(x+1) + B$

Method	Step	Result
Plug in $x = -1$	$-3+5 = B$	$B = 2$
Compare $x$ -coefficients	$3 = A$	$A = 3$

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

$\int \frac{3x+5}{(x+1)^2}\,dx = 3\ln|x+1| - \frac{2}{x+1} + C$

Key Fact: $\int \frac{A}{(x-a)^n}\,dx = \frac{A}{(1-n)(x-a)^{n-1}} + C$ for $n \geq 2$ .

Mixed: Distinct + Repeated Factors

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

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

$x^2 = A(x+2)^2 + B(x-1)(x+2) + C(x-1)$

Substitution	Equation	Result
$x = 1$	$1 = A(9)$	$A = \frac{1}{9}$

$\boxed{\frac{x^2}{(x-1)(x+2)^2} = \frac{1/9}{x-1} + \frac{8/9}{x+2} - \frac{4/3}{(x+2)^2}}$

Repeated Factors Practice

Decomposition Setup

Finding Constants

Key Takeaways — Part 2

Concept	Details
Repeated factor $(x-a)^n$	Needs $n$ terms: $\frac{A_1}{x-a} + \cdots + \frac{A_n}{(x-a)^n}$
Integration	$\int \frac{A}{(x-a)^n}\,dx = \frac{A}{(1-n)(x-a)^{n-1}} + C$ for
Mixed problems	Combine distinct + repeated factor rules
Finding constants	Substitution at roots + coefficient comparison

Coming Up: Part 3 covers irreducible quadratic factors — denominators like $x^2 + 1$ that can’t be factored further.

\frac{P ( x )}{( x - r ) ( a x ^{2} + b x + c )} = \frac{A}{x - r} + \frac{B x + C}{a x ^{2} + b x + c}

The numerator over the quadratic is $Bx + C$ (not just a constant $B$ ).

Why Linear Numerator?

A quadratic factor has degree 2, so its partial fraction numerator must have degree at most 1 (one less than the factor’s degree). This ensures enough unknowns for a unique decomposition.

Key Fact: “Irreducible” means $b^2 - 4ac < 0$ . Examples: $x^2 + 1$ , $x^2 + 4$ , $x^2 + x + 1$ .

Worked Example: $\int \frac{x+2}{(x-1)(x^2+1)}\,dx$

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

Step 2: Multiply through: $x+2 = A(x^2+1) + (Bx+C)(x-1)$

Method	Result
$x = 1$ : $3 = 2A$	$A = \frac{3}{2}$

Step 3: Integrate each piece

$\int \frac{3/2}{x-1}\,dx = \frac{3}{2}\ln|x-1|$

$\int \frac{-\frac{3}{2}x - \frac{1}{2}}{x^2+1}\,dx = -\frac{3}{4}\ln(x^2+1) - \frac{1}{2}\arctan x$

$\boxed{\frac{3}{2}\ln|x-1| - \frac{3}{4}\ln(x^2+1) - \frac{1}{2}\arctan x + C}$

Integrating $\frac{Bx + C}{x^2 + a^2}$

Split into two integrals:

$\int \frac{Bx + C}{x^2 + a^2}\,dx = B \int \frac{x}{x^2+a^2}\,dx + C\int \frac{1}{x^2+a^2}\,dx$

Integral	Result
$\int \frac{x}{x^2+a^2}\,dx$

AP Tip: Always split the numerator into an $x$ -part (which gives logarithmic) and a constant part (which gives arctangent).

Irreducible Quadratic Practice

Integration Technique

Coefficient Finding

Key Takeaways — Part 3

Concept	Details
Irreducible quadratic	$b^2 - 4ac < 0$ ; needs $\frac{Bx+C}{ax^2+bx+c}$
Integration split	Separate $Bx$ part (log) from $C$ part (arctan)
Key formulas	$\int \frac{x}{x^2+a^2}\,dx = \frac{1}{2}\ln(x^2+a^2)$
	$\int \frac{1}{x^2+a^2}\,dx = \frac{1}{a}\arctan\frac{x}{a}$

Coming Up: Part 4 puts it all together — full integration with partial fractions from start to finish.

Key Fact: Each integration produces either $\ln|\text{linear}|$ , $\frac{\text{const}}{\text{linear power}}$ , $\ln(\text{quadratic})$ , or $\arctan$ .

Complete Example 1: $\int \frac{5x-3}{x^2-x-6}\,dx$

Step 1: Factor: $x^2 - x - 6 = (x-3)(x+2)$

Step 2: Decompose: $\frac{5x-3}{(x-3)(x+2)} = \frac{A}{x-3} + \frac{B}{x+2}$

Step 3: Find constants:

Cover-up at $x = 3$ : $A = \frac{5(3)-3}{3+2} = \frac{12}{5}$

Step 4: Integrate:

$\boxed{\int \frac{5x-3}{x^2-x-6}\,dx = \frac{12}{5}\ln|x-3| + \frac{13}{5}\ln|x+2| + C}$

Complete Example 2: $\int \frac{1}{x^2-4}\,dx$

Factor: $x^2 - 4 = (x-2)(x+2)$

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

$A = \frac{1}{2-(-2)} = \frac{1}{4}$

$\boxed{\int \frac{1}{x^2-4}\,dx = \frac{1}{4}\ln\left|\frac{x-2}{x+2}\right| + C}$

AP Tip: This result can also be written as $\frac{1}{4}\ln|x-2| - \frac{1}{4}\ln|x+2| + C$ . Both forms are accepted on the exam.

Integration Practice

Strategy Selection

Definite Integral Computation

Key Takeaways — Part 4

Integration Result	From
$A\ln	x-a
$\frac{A}{(1-n)(x-a)^{n-1}}$	$\frac{A}{(x-a)^n}$ , $n \geq 2$
$\frac{B}{2}\ln(x^2+a^2)$
$\frac{C}{a}\arctan\frac{x}{a}$

Coming Up: Part 5 covers what to do when the fraction is improper — long division before decomposition.

Key Fact: If deg(numerator) $\geq$ deg(denominator), divide first. The result is: quotient + $\frac{\text{remainder}}{\text{denominator}}$ , where the remainder fraction IS proper.

Worked Example: $\int \frac{x^3+2}{x^2-1}\,dx$

Step 1: Long division

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

Check: $x(x^2-1) + (x+2) = x^3 - x + x + 2 = x^3 + 2$ ✔

Step 2: Decompose the remainder: $\frac{x+2}{(x-1)(x+1)}$

$A = \frac{1+2}{1+1} = \frac{3}{2}$ ,

Step 3: Integrate

$\int \left(x + \frac{3/2}{x-1} - \frac{1/2}{x+1}\right)dx$

$\boxed{= \frac{x^2}{2} + \frac{3}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C}$

Improper Fractions Practice

Long Division Practice

Key Takeaways — Part 5

Concept	Details
When to divide	deg(numerator) $\geq$ deg(denominator)
Result format	quotient $+$ $\frac{\text{remainder}}{\text{denominator}}$
Then what	Apply partial fractions to the proper remainder
Common quotients	Often just a polynomial like $x + 1$ or $x^2 - 3$

AP Tip: The AP exam loves to test whether students remember to check for improper fractions. Always compare degrees before starting!

Coming Up: Part 6 is a mixed practice workshop combining all techniques.

Workshop Round 1

Workshop Round 2

Technique Identification

Workshop Challenge

Key Takeaways — Part 6

Common Mistake	How to Avoid
Forgetting long division	Always check degrees first
Wrong decomposition form	Repeated: need all powers; quadratic: need $Bx+C$
Sign errors in cover-up	Double-check by plugging back in
Missing absolute values	$\ln

Coming Up: Part 7 is the comprehensive review and assessment covering all partial fraction techniques.

Assessment — Conceptual

Assessment — Computational

Method Selection Review

Final Computation

Partial Fractions — Complete! ✅

You’ve mastered:

✔ Distinct linear factor decomposition
✔ Repeated linear factors
✔ Irreducible quadratic factors
✔ Full integration workflow
✔ Long division for improper fractions
✔ Cover-up (Heaviside) method

AP Exam Frequency

PF Topic	Likelihood on AP BC
Simple distinct linear	Very common (MC + FRQ)
Repeated factors	Occasional
Irreducible quadratic	Less common but tested
Improper fraction (division first)	Common trap question

Key Fact: Partial fractions is one of the BC-only integration techniques. Combined with IBP and trig substitution, it’s essential for the full AP BC integration toolkit.

(1−n)(x−a)n−1A+

Cover-up at

x = -2

B = \frac{5(-2)-3}{-2-3} = \frac{-13}{-5} = \frac{13}{5}

B = \frac{1}{-2-2} = -\frac{1}{4}

\frac{C}{x ^{2} + a ^{2}}

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

Partial Fractions

Partial Fractions - Complete Interactive Lesson

Part 1: The Concept

Partial Fraction Decomposition

When to Use Partial Fractions

Prerequisite: Factor the Denominator

Case 1: Distinct Linear Factors

Three Distinct Linear Factors

Key Takeaways — Part 1

Part 2: Repeated Linear Factors

Partial Fraction Decomposition

Setup for Repeated Factors

Part 3: Integration Practice

Partial Fraction Decomposition

Setup Rule

Part 4: Long Division First

Partial Fraction Decomposition

Complete Workflow

Part 5: Logistic DE Connection

Partial Fraction Decomposition

Proper vs. Improper

Part 6: Practice Workshop

Partial Fraction Decomposition

Decision Guide

Part 7: Final Assessment

Partial Fraction Decomposition — Review

Complete Reference

Worked Example: ∫1x2−1 dx\int \frac{1}{x^2 - 1}\,dx∫x2−11​dx

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

Integration

Mixed: Distinct + Repeated Factors

Key Takeaways — Part 2

Why Linear Numerator?

Worked Example: ∫x+2(x−1)(x2+1) dx\int \frac{x+2}{(x-1)(x^2+1)}\,dx∫(x−1)(x2+1)x+2​dx

Integrating Bx+Cx2+a2\frac{Bx + C}{x^2 + a^2}x2+a2Bx+C​

Key Takeaways — Part 3

Complete Example 1: ∫5x−3x2−x−6 dx\int \frac{5x-3}{x^2-x-6}\,dx∫x2−x−65x−3​dx

Complete Example 2: ∫1x2−4 dx\int \frac{1}{x^2-4}\,dx∫x2−41​dx

Key Takeaways — Part 4

Worked Example: ∫x3+2x2−1 dx\int \frac{x^3+2}{x^2-1}\,dx∫x2−1x3+2​dx

Key Takeaways — Part 5

Key Takeaways — Part 6

Partial Fractions — Complete! ✅

AP Exam Frequency

Worked Example: $\int \frac{1}{x^2 - 1}\,dx$

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

Worked Example: $\int \frac{x+2}{(x-1)(x^2+1)}\,dx$

Integrating $\frac{Bx + C}{x^2 + a^2}$

Complete Example 1: $\int \frac{5x-3}{x^2-x-6}\,dx$

Complete Example 2: $\int \frac{1}{x^2-4}\,dx$

Worked Example: $\int \frac{x^3+2}{x^2-1}\,dx$