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Module 15 · L05 of 16 ~40 min ⚡ +90 XP available

Partial Fractions, Distinct Linear Factors

Some rational functions look impossible to integrate, until you split them. Partial fraction decomposition rewrites a single complicated fraction as a sum of simple pieces, each of which integrates to a logarithm. In this lesson you'll master the case where the denominator factors into distinct linear factors: the most common decomposition on the HSC paper, and the foundation for every harder case to come.

Today's hook, The integral $\displaystyle \int \frac{1}{x^2 - 1}\,dx$ looks unfamiliar. But notice $x^2 - 1 = (x-1)(x+1)$. Before reading on, can you find constants $A$ and $B$ so that $\frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$? Once you have them, the integral is just two logarithms. Compare your answer after card 05.
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Recall, your gut answer first

+5 XP warm-up Combine into a single fraction.

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Recall, your gut answer first
+5 XP warm-up

Combine $\dfrac{1}{x-2} + \dfrac{1}{x+3}$ into a single fraction. What is the resulting numerator and denominator? Now reverse the question: if you started with that single fraction, could you recover the two original pieces? Sketch your reasoning below.

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The two moves for partial fractions

+5 XP to read Every partial-fraction decomposition follows the same two-step recipe: factor the denominator into its...

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The two moves for partial fractions
+5 XP to read

Every partial-fraction decomposition follows the same two-step recipe: factor the denominator into its irreducible pieces, then match the right template based on what those factors look like. For distinct linear factors $(x-a)(x-b)\dots$, the template is one constant over each factor.

The factor-template-solve sequence: (1) factor the denominator $Q(x)$ completely, (2) write the partial-fraction template (one $A/(x-a)$ per distinct linear root), (3) solve for the constants via cover-up or equating coefficients.

$\dfrac{P(x)}{(x-a)(x-b)} = \dfrac{A}{x-a} + \dfrac{B}{x-b}$  ·  provided $\deg P < \deg Q$

Sequence for partial-fraction integration flow
$\displaystyle \int \dfrac{1}{x-a}\,dx = \ln|x-a| + C$
Degree check first
Partial fractions only works when $\deg P < \deg Q$. If $\deg P \geq \deg Q$, divide first by long division, then decompose the remainder.
Cover-up = fast track
For a distinct linear factor $(x-a)$, the constant on top equals $\dfrac{P(a)}{Q'(a)}$, found by "covering up" $(x-a)$ in the original and substituting $x = a$ into what remains.
Each piece $\to$ a log
Every $\dfrac{A}{x-a}$ integrates to $A \ln|x-a|$. The absolute value bars are essential, drop them and you lose marks on the HSC.
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What you'll master
Know

Key facts

  • If $Q(x) = (x-a)(x-b)\dots$ has distinct linear roots, the decomposition is $\sum \dfrac{A_i}{x-r_i}$
  • The cover-up method gives each constant directly
  • $\displaystyle \int \dfrac{1}{x-a}\,dx = \ln|x-a| + C$
  • Partial fractions requires $\deg P < \deg Q$
Understand

Concepts

  • Why decomposition is the reverse of finding a common denominator
  • Why the cover-up method works (multiplying both sides by $(x-a)$ and substituting)
  • Why each linear factor contributes exactly one constant
Can do

Skills

  • Decompose a proper rational function with 2 or 3 distinct linear factors
  • Use cover-up to find constants quickly
  • Integrate the decomposition to a sum of logarithms with absolute values
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Key terms
Rational functionA function of the form $P(x)/Q(x)$ where $P$ and $Q$ are polynomials and $Q \neq 0$. Said to be proper when $\deg P < \deg Q$.
Partial fraction decompositionThe unique rewriting of a proper rational function as a sum of simpler fractions, one per irreducible factor of $Q(x)$.
Distinct linear factorA factor of the form $(x - a)$ that appears exactly once in $Q(x)$. Contributes one term $A/(x-a)$ to the decomposition.
Cover-up methodA shortcut for finding the constant over $(x-a)$: cover that factor in $P(x)/Q(x)$ and substitute $x = a$ into what's left.
Equating coefficientsA general method: clear denominators, expand, then match coefficients of like powers of $x$ on both sides to form a linear system.
Log integralThe standard result $\int \dfrac{1}{x-a}\,dx = \ln|x-a| + C$. The basic building block once decomposition is done.
MEX-C1NESA outcome (Further Integration): integrates rational functions using partial fractions, including those with distinct linear factors, repeated linear factors and irreducible quadratic factors.
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Distinct linear factors, template and cover-up

Suppose $Q(x) = (x - a_1)(x - a_2)\cdots(x - a_n)$ with all $a_i$ distinct, and $\deg P < n$.

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Distinct linear factors, template and cover-up
core concept

Suppose $Q(x) = (x - a_1)(x - a_2)\cdots(x - a_n)$ with all $a_i$ distinct, and $\deg P < n$. Then there exist unique constants $A_1, A_2, \dots, A_n$ such that

$$\frac{P(x)}{(x-a_1)(x-a_2)\cdots(x-a_n)} = \frac{A_1}{x-a_1} + \frac{A_2}{x-a_2} + \cdots + \frac{A_n}{x-a_n}.$$

Cover-up method. To find $A_k$: multiply both sides by $(x - a_k)$, then substitute $x = a_k$. On the left, $(x - a_k)$ cancels, leaving $P(a_k)/\prod_{i \neq k}(a_k - a_i)$. On the right, every term except $A_k$ has a factor of $(x - a_k)$ in its denominator and survives, but they vanish when $x = a_k$ is substituted into the cleared form. So

$$A_k = \frac{P(a_k)}{\prod_{i \neq k}(a_k - a_i)}.$$

Working the hook: $\dfrac{1}{(x-1)(x+1)} = \dfrac{A}{x-1} + \dfrac{B}{x+1}$. Cover $(x-1)$ and set $x = 1$: $A = \dfrac{1}{1 + 1} = \dfrac{1}{2}$. Cover $(x+1)$ and set $x = -1$: $B = \dfrac{1}{-1 - 1} = -\dfrac{1}{2}$.

Connecting to integration. Once decomposed, each piece is a standard log integral: $\displaystyle \int \frac{1}{(x-1)(x+1)}\,dx = \tfrac{1}{2} \ln|x-1| - \tfrac{1}{2}\ln|x+1| + C = \tfrac{1}{2}\ln\left|\dfrac{x-1}{x+1}\right| + C$.

Template for distinct linear factors: one $A_i/(x - a_i)$ per root · Cover-up: $A_k = P(a_k) / \prod_{i \neq k}(a_k - a_i)$ · $\int 1/(x-a)\,dx = \ln|x-a| + C$, keep the absolute value · Requirement: $\deg P < \deg Q$ (divide first if not)

Pause, copy the distinct-linear-factor template, the cover-up formula $A_k = P(a_k)/\prod_{i\neq k}(a_k-a_i)$, and the integral $\int 1/(x-a)\,dx = \ln|x-a|+C$ into your book.

Quick check: Decompose $\dfrac{5}{(x-1)(x+4)}$ into partial fractions. Which is correct?

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Equating coefficients, the general workhorse

We just saw the distinct-linear-factor template (one $A_i/(x-a_i)$ per root) and the cover-up rule $A_k = P(a_k)/\prod_{i\neq k}(a_k-a_i)$, integrating to a sum of $\ln|x-a_i|$ terms.

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Equating coefficients, the general workhorse
core concept

We just saw the distinct-linear-factor template (one $A_i/(x-a_i)$ per root) and the cover-up rule $A_k = P(a_k)/\prod_{i\neq k}(a_k-a_i)$, integrating to a sum of $\ln|x-a_i|$ terms. That raises a question: what if cover-up is unavailable, say, for the constant in a sum of two terms? This card answers it → clear denominators, expand, and equate coefficients of each power of $x$ to get a linear system.

Cover-up is fastest for distinct linear factors, but you should also master the equating coefficients method, it generalises to every case (repeated factors, quadratics) you'll meet in L06.

Procedure for $\dfrac{P(x)}{(x-a)(x-b)} = \dfrac{A}{x-a} + \dfrac{B}{x-b}$:

  • Multiply both sides by $(x-a)(x-b)$ to clear denominators: $P(x) = A(x-b) + B(x-a)$.
  • Expand the right side and collect powers of $x$.
  • Match the coefficient of each power of $x$ on the left to that on the right, gives a linear system for $A$, $B$.
  • Or: substitute strategic $x$-values (especially the roots) to isolate one constant at a time, this is the cover-up shortcut in disguise.
$$P(x) = A(x - b) + B(x - a) \quad \Rightarrow \quad \begin{cases} x = a: & P(a) = A(a - b) \\ x = b: & P(b) = B(b - a) \end{cases}$$
Common mistake. Students often forget to check $\deg P < \deg Q$. If $P(x)/Q(x)$ is improper (e.g. $\dfrac{x^3}{x^2 - 1}$), partial fractions will not work directly, you must polynomial-divide first, then decompose the proper remainder.

Equating coefficients: clear denominators, match powers of $x$ · Or substitute roots to isolate constants (cover-up reinterpreted) · If improper ($\deg P \geq \deg Q$), divide first · Check by recombining your decomposition over a common denominator

Pause, copy the equating-coefficients method (clear denominators, match powers of $x$), the improper-fraction check ($\deg P \geq \deg Q$ → divide first), and the verification by recombining into your book.

Did you get this? True or false: the rational function $\dfrac{x^3 + 1}{x^2 - 4}$ can be decomposed directly into partial fractions of the form $\dfrac{A}{x-2} + \dfrac{B}{x+2}$.

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Worked examples · 3 in a row, reveal as you go

Worked examples · 3 in a row, reveal as you go

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · 2 DISTINCT LINEAR FACTORS

Express $\dfrac{3x + 5}{(x-1)(x+2)}$ as partial fractions.

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Write the template: $\dfrac{3x + 5}{(x-1)(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{x+2}$. Multiply by $(x-1)(x+2)$: $3x + 5 = A(x+2) + B(x-1)$.
One constant per distinct linear factor. Clearing denominators always converts the identity into a polynomial equation valid for all $x$.
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PROBLEM 2 · 3 DISTINCT LINEAR FACTORS

Decompose $\dfrac{x^2 + 1}{x(x-1)(x+2)}$ into partial fractions.

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Template: $\dfrac{x^2 + 1}{x(x-1)(x+2)} = \dfrac{A}{x} + \dfrac{B}{x-1} + \dfrac{C}{x+2}$. Clear denominators: $x^2 + 1 = A(x-1)(x+2) + Bx(x+2) + Cx(x-1)$.
Three distinct linear factors give three constants. Check $\deg P = 2 < \deg Q = 3$, proper, so decomposition is valid.
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PROBLEM 3 · DECOMPOSE THEN INTEGRATE

Find $\displaystyle \int \dfrac{2x + 3}{x^2 - x - 6}\,dx$.

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Factor the denominator: $x^2 - x - 6 = (x - 3)(x + 2)$. Template: $\dfrac{2x + 3}{(x-3)(x+2)} = \dfrac{A}{x-3} + \dfrac{B}{x+2}$. Clear: $2x + 3 = A(x+2) + B(x-3)$.
Always factor first. Recognising $x^2 - x - 6$ as $(x-3)(x+2)$ converts the problem into the distinct-linear-factor template.
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Fill the gap: Once decomposed, $\displaystyle \int \dfrac{A}{x - a}\,dx = $ $\ln$ $+\, C$. The absolute value bars are essential.

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Misconceptions to fix · the 3 traps that cost marks

Misconceptions to fix · the 3 traps that cost marks

Misconceptions to fix · the 3 traps that cost marks

Trap 01
Skipping the degree check
Partial fractions assumes $\deg P < \deg Q$. If you try to decompose $\dfrac{x^3}{x^2 - 1}$ directly with $\dfrac{A}{x-1} + \dfrac{B}{x+1}$, you'll get an inconsistent system. Divide first to get a polynomial plus a proper remainder.
Trap 02
Forgetting absolute values on $\ln$
$\int \dfrac{1}{x - a}\,dx$ equals $\ln|x - a| + C$, NOT $\ln(x - a) + C$. The integrand is defined for $x < a$ as well, where $x - a$ is negative. Markers deduct for missing bars.
Trap 03
Sign slips during cover-up
When you cover $(x + 2)$ and substitute $x = -2$, watch the signs in what remains. $(-2 - 3) = -5$, not $5$. Many lost marks come from missing one negative when subtracting the other roots.

Did you get this? True or false: $\displaystyle \int \dfrac{1}{x - 5}\,dx = \ln(x - 5) + C$ is a fully correct HSC-style answer.

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Revisit your thinking

Activities · practice with the ideas

Activities · practice with the ideas

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Decompose $\dfrac{7}{(x-2)(x+5)}$ into partial fractions using the cover-up method.

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Express $\dfrac{4x - 1}{(x-3)(x+1)}$ as a sum of partial fractions.

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Find $\displaystyle \int \dfrac{1}{x^2 - 9}\,dx$ via partial fractions.

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Decompose $\dfrac{6}{x(x-1)(x+1)}$ and evaluate $\displaystyle \int \dfrac{6}{x(x-1)(x+1)}\,dx$.

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Evaluate $\displaystyle \int_2^3 \dfrac{1}{x(x-1)}\,dx$ exactly.

Odd one out: Three of these rational functions are immediately ready for partial-fraction decomposition into distinct linear factors. Which one is NOT?

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Revisit your thinking

Earlier you tried to find $A$ and $B$ so that $\dfrac{1}{(x-1)(x+1)} = \dfrac{A}{x-1} + \dfrac{B}{x+1}$.

The cover-up method gives $A = \tfrac{1}{2}$ and $B = -\tfrac{1}{2}$, so $\displaystyle \int \dfrac{dx}{(x-1)(x+1)} = \tfrac{1}{2}\ln|x-1| - \tfrac{1}{2}\ln|x+1| + C = \tfrac{1}{2}\ln\left|\dfrac{x-1}{x+1}\right| + C$. The technique that converts an unfamiliar fraction into two standard log integrals is the most useful idea in Module 15, and in L06 you'll extend it to repeated and quadratic factors.

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Show what you have learned

Multiple choice, then short answer under exam conditions.

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Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer
ApplyBand 32 marks

Q1. Express $\dfrac{x + 7}{(x-1)(x+3)}$ in partial-fraction form. (2 marks)

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ApplyBand 43 marks

Q2. Find $\displaystyle \int \dfrac{3}{x^2 + x - 2}\,dx$ using partial fractions. (3 marks)

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AnalyseBand 54 marks

Q3. Evaluate $\displaystyle \int_3^5 \dfrac{x + 5}{x(x-2)}\,dx$, giving your answer in exact logarithmic form. (4 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. Cover $(x-2)$ at $x = 2$: $A = 7/7 = 1$. Cover $(x+5)$ at $x = -5$: $B = 7/(-7) = -1$. So $\dfrac{7}{(x-2)(x+5)} = \dfrac{1}{x-2} - \dfrac{1}{x+5}$.

2. $x = 3$: $11 = 4A$, $A = 11/4$. $x = -1$: $-5 = -4B$, $B = 5/4$. So $\dfrac{4x-1}{(x-3)(x+1)} = \dfrac{11}{4(x-3)} + \dfrac{5}{4(x+1)}$.

3. $\dfrac{1}{(x-3)(x+3)} = \dfrac{1/6}{x-3} - \dfrac{1/6}{x+3}$. Integral $= \dfrac{1}{6}\ln|x-3| - \dfrac{1}{6}\ln|x+3| + C = \dfrac{1}{6}\ln\left|\dfrac{x-3}{x+3}\right| + C$.

4. $A = -6$, $B = 3$, $C = 3$. So $\dfrac{6}{x(x-1)(x+1)} = -\dfrac{6}{x} + \dfrac{3}{x-1} + \dfrac{3}{x+1}$. Integral $= -6\ln|x| + 3\ln|x-1| + 3\ln|x+1| + C$.

5. $\dfrac{1}{x(x-1)} = -\dfrac{1}{x} + \dfrac{1}{x-1}$. Antiderivative $\ln|(x-1)/x|$. At $x=3$: $\ln(2/3)$. At $x=2$: $\ln(1/2)$. Difference: $\ln(2/3) - \ln(1/2) = \ln(4/3)$.

Q1 (2 marks): $\dfrac{x+7}{(x-1)(x+3)} = \dfrac{A}{x-1} + \dfrac{B}{x+3}$ [1]. $x=1$: $8 = 4A$ so $A = 2$. $x=-3$: $4 = -4B$ so $B = -1$. Therefore $\dfrac{2}{x-1} - \dfrac{1}{x+3}$ [1].

Q2 (3 marks): Factor $x^2 + x - 2 = (x-1)(x+2)$ [1]. Decompose: $\dfrac{3}{(x-1)(x+2)} = \dfrac{1}{x-1} - \dfrac{1}{x+2}$ [1]. Integrate: $\ln|x-1| - \ln|x+2| + C = \ln\left|\dfrac{x-1}{x+2}\right| + C$ [1].

Q3 (4 marks): $A = -5/2$, $B = 7/2$ [1]. Antiderivative: $-\tfrac{5}{2}\ln|x| + \tfrac{7}{2}\ln|x-2|$ [1]. At $x=5$: $-\tfrac{5}{2}\ln 5 + \tfrac{7}{2}\ln 3$. At $x=3$: $-\tfrac{5}{2}\ln 3 + \tfrac{7}{2}\ln 1 = -\tfrac{5}{2}\ln 3$ [1]. Difference: $-\tfrac{5}{2}\ln 5 + \tfrac{7}{2}\ln 3 + \tfrac{5}{2}\ln 3 = -\tfrac{5}{2}\ln 5 + 6\ln 3 = \ln\dfrac{3^6}{5^{5/2}} = \ln\dfrac{729}{5^{5/2}}$ [1].