Partial Fractions

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.

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 $b^2 - 4c < 0$

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.

Case 1: Distinct Linear Factors

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

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

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

Three Distinct Linear Factors

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

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

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

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

Decomposition Practice

Setting Up Decompositions

Cover-Up Computation

Key Takeaways — Part 1

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

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

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

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

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

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

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

Integration

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

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

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

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

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

Repeated Factors Practice

Decomposition Setup

Finding Constants

Key Takeaways — Part 2

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

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 $(ax^2 + bx + c)$:

$\frac{P(x)}{(x - r)(ax^2 + bx + c)} = \frac{A}{x - r} + \frac{Bx + C}{ax^2 + bx + c}$

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

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:

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

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

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

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

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 $x = 3$: $A = \frac{5(3)-3}{3+2} = \frac{12}{5}$
• Cover-up at $x = -2$: $B = \frac{5(-2)-3}{-2-3} = \frac{-13}{-5} = \frac{13}{5}$

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

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

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

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

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

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

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

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

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

Decision Making

Long Division Practice

Key Takeaways — Part 5

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.

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

Workshop Round 1

Workshop Round 2

Technique Identification

Workshop Challenge

Key Takeaways — Part 6

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

Partial Fraction Decomposition — Review

Part 7 of 7 — Comprehensive Review & Assessment

Complete Reference

Assessment — Conceptual

Assessment — Computational

Method Selection Review

Final Computation

Partial Fractions — Complete! ✅

You’ve mastered:

1. ✔ Distinct linear factor decomposition
2. ✔ Repeated linear factors
3. ✔ Irreducible quadratic factors
4. ✔ Full integration workflow
5. ✔ Long division for improper fractions
6. ✔ Cover-up (Heaviside) method

AP Exam Frequency

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.