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Module 15 · L06 of 16 ~45 min ⚡ +90 XP available

Partial Fractions — Repeated Factors and Quadratics

Not every denominator splits into distinct linear factors. A repeated factor like $(x-a)^2$ needs two constants, not one. An irreducible quadratic like $x^2 + 1$ needs a linear numerator $Ax + B$, and the integration may produce a $\tan^{-1}$ alongside the logarithm. In this lesson you'll master the templates and the completing-the-square trick that handles every quadratic the HSC will throw at you.

Today's hook — Compare two innocent-looking denominators: $(x-1)^2$ and $x^2 + 4$. Neither factors as a product of distinct linear factors. Before reading on, predict what the partial-fraction templates should look like for $\dfrac{3}{(x-1)^2}$ and $\dfrac{5}{x^2+4}$. Which one needs an $Ax + B$ on top? Which one's integral involves $\tan^{-1}$? Compare your guesses after card 05.

0/5QUESTS

Recall — your gut answer first

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From L05 you know $\int \dfrac{1}{x - a}\,dx = \ln|x - a| + C$. What about $\int \dfrac{1}{(x - a)^2}\,dx$? And do you remember the standard form $\int \dfrac{1}{x^2 + a^2}\,dx$? Sketch your reasoning below.

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The two moves for harder partial fractions

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Beyond distinct linear factors you meet two new shapes. A repeated linear factor $(x-a)^k$ contributes $k$ terms with denominators $(x-a), (x-a)^2, \dots, (x-a)^k$. An irreducible quadratic $(x^2 + bx + c)$ contributes one term with a linear numerator $(Ax + B)$ on top. The integration then mixes logs, reciprocals and inverse tangents.

The template-by-factor-shape rule: each irreducible factor of $Q(x)$ contributes its own block to the decomposition. Repeated $(x-a)^k$ → $k$ constants. Irreducible quadratic → one linear numerator. Complete the square to integrate.

$\dfrac{P(x)}{(x-a)^2(x^2+c)} = \dfrac{A}{x-a} + \dfrac{B}{(x-a)^2} + \dfrac{Cx+D}{x^2+c}$

$\displaystyle \int \dfrac{1}{x^2 + a^2}\,dx = \dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + C$

$(x-a)^2$ needs TWO terms

Template: $\dfrac{A}{x-a} + \dfrac{B}{(x-a)^2}$. Omitting the first term loses a degree of freedom — the system becomes unsolvable.

Quadratic → linear top

An irreducible quadratic $(x^2 + bx + c)$ in the denominator gets a numerator of the form $Cx + D$ — one constant for each degree available.

Complete the square

To integrate $\dfrac{1}{x^2 + bx + c}$, rewrite the denominator as $(x + b/2)^2 + (c - b^2/4)$, then use the $\tan^{-1}$ standard form (or log, if the constant is negative).

What you'll master

Know

Key facts

$(x-a)^k$ contributes $\sum_{j=1}^{k} A_j/(x-a)^j$ to the decomposition
Irreducible quadratic $(x^2 + bx + c)$ contributes $(Cx + D)/(x^2 + bx + c)$
$\int 1/(x-a)^2\,dx = -1/(x-a) + C$
$\int 1/(x^2 + a^2)\,dx = (1/a)\tan^{-1}(x/a) + C$

Understand

Concepts

Why a repeated factor needs one term per power up to the multiplicity
Why an irreducible quadratic earns a linear numerator (degree-of-freedom argument)
How completing the square reveals which standard form applies

Can do

Skills

Decompose $P(x)/Q(x)$ with $(x-a)^2$ factors using a mix of cover-up and coefficient matching
Decompose $P(x)/Q(x)$ with an irreducible quadratic factor
Complete the square to integrate to $\ln$ or $\tan^{-1}$ as appropriate

Key terms

Repeated linear factorA factor $(x-a)^k$ with $k \geq 2$ in $Q(x)$. Contributes $k$ partial-fraction terms with denominators $(x-a), (x-a)^2, \dots, (x-a)^k$.

Irreducible quadraticA quadratic $x^2 + bx + c$ with discriminant $b^2 - 4c < 0$ — it cannot be factored further over the reals. Contributes $(Cx + D)/(x^2 + bx + c)$.

Completing the squareRewriting $x^2 + bx + c$ as $(x + b/2)^2 + (c - b^2/4)$. Used to transform an integral into a standard $\tan^{-1}$ or $\ln$ form.

Splitting the numeratorGiven $(Cx + D)/(x^2 + a^2)$, rewrite the top so part of it is $\tfrac{1}{2}\dfrac{d}{dx}(x^2 + a^2) = x$ — that part integrates to $\tfrac{C}{2}\ln(x^2 + a^2)$, the rest to a $\tan^{-1}$.

Standard form: $\ln$$\displaystyle \int \dfrac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C$. Used when the numerator (after splitting) is the derivative of the denominator.

Standard form: $\tan^{-1}$$\displaystyle \int \dfrac{1}{x^2 + a^2}\,dx = \dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + C$. Used once the denominator is in completed-square form $u^2 + a^2$.

MEX-C1NESA outcome (Further Integration): integrates rational functions using partial fractions, including repeated linear factors and irreducible quadratic factors.

Repeated linear factors and irreducible quadratics

core concept

Repeated factor rule. If $(x - a)^k$ appears in $Q(x)$, the decomposition needs every power from 1 up to $k$:

$$\frac{P(x)}{(x - a)^k \,R(x)} = \frac{A_1}{x - a} + \frac{A_2}{(x - a)^2} + \cdots + \frac{A_k}{(x - a)^k} + \text{(terms for } R(x)\text{)}.$$

Irreducible quadratic rule. If $x^2 + bx + c$ has $b^2 - 4c < 0$, the decomposition includes

$$\frac{Cx + D}{x^2 + bx + c}.$$

Working the hook: $\dfrac{3}{(x-1)^2}$ already is a single partial-fraction term (the second one in the repeated-factor template), so its integral is $\int 3(x-1)^{-2}\,dx = -3/(x-1) + C$. For $\dfrac{5}{x^2 + 4}$, the denominator is already $x^2 + 2^2$, so $\int \dfrac{5}{x^2 + 4}\,dx = \dfrac{5}{2}\tan^{-1}\dfrac{x}{2} + C$.

Completing the square in action. To handle $\int \dfrac{1}{x^2 + 2x + 5}\,dx$, write $x^2 + 2x + 5 = (x+1)^2 + 4$. With $u = x + 1$ this becomes $\int \dfrac{du}{u^2 + 4} = \tfrac{1}{2}\tan^{-1}\dfrac{u}{2} + C = \tfrac{1}{2}\tan^{-1}\dfrac{x+1}{2} + C$.

Repeated factor $(x-a)^k$ → $k$ terms: $A_1/(x-a) + A_2/(x-a)^2 + \dots + A_k/(x-a)^k$ · Irreducible quadratic → numerator $Cx + D$ · $\int 1/(x-a)^2\,dx = -1/(x-a) + C$ · Complete the square then use $\tan^{-1}$ standard form

Pause — copy the repeated-factor template (one term per power up to $k$), the irreducible-quadratic numerator $Cx+D$, and $\int 1/(x-a)^2\,dx = -1/(x-a)+C$ into your book.

Quick check: Which is the correct partial-fraction template for $\dfrac{x + 2}{(x-1)^2 (x^2 + 3)}$?

Integrating $(Cx + D)/(x^2 + a^2)$ — split the numerator

core concept

We just saw that a repeated factor $(x-a)^k$ requires $k$ separate partial-fraction terms $A_1/(x-a)+\cdots+A_k/(x-a)^k$, and an irreducible quadratic needs a numerator $Cx+D$. That raises a question: how do we actually integrate the $(Cx+D)/(x^2+a^2)$ piece once the decomposition is complete? This card answers it → split: write $Cx = \tfrac{C}{2}\cdot 2x$ to get a $\ln$ term, and the constant $D$ gives $\tfrac{D}{a}\arctan(x/a)$.

Once the decomposition gives you $(Cx + D)/(x^2 + a^2)$, the integration uses a two-part trick: split the numerator into a piece proportional to the derivative of the denominator (giving a $\ln$) and a constant piece (giving a $\tan^{-1}$).

Procedure. The derivative of $x^2 + a^2$ is $2x$. So write $Cx + D = \dfrac{C}{2}(2x) + D$, splitting the integral:

$$\int \frac{Cx + D}{x^2 + a^2}\,dx = \frac{C}{2} \int \frac{2x}{x^2 + a^2}\,dx + D \int \frac{1}{x^2 + a^2}\,dx = \frac{C}{2} \ln(x^2 + a^2) + \frac{D}{a} \tan^{-1}\frac{x}{a} + K.$$

If the denominator is $x^2 + bx + c$ with $b \neq 0$, complete the square first ($x^2 + bx + c = (x + b/2)^2 + (c - b^2/4)$) and substitute $u = x + b/2$ to reach the same standard form.

Common mistake. Students apply $\tan^{-1}$ even when the discriminant is positive (i.e. the quadratic factors over the reals). Always check $b^2 - 4c$ before assuming a quadratic is irreducible. If it factors, you should be using L05 methods, not $\tan^{-1}$.

Split: $Cx + D = \tfrac{C}{2}(2x) + D$ — $2x$ is $\frac{d}{dx}(x^2 + a^2)$ · Result: $\tfrac{C}{2}\ln(x^2 + a^2) + \tfrac{D}{a}\tan^{-1}(x/a) + K$ · Complete the square first if $b \neq 0$ · Check discriminant: if $b^2 - 4c \geq 0$, factor instead

Pause — copy the numerator-split: $Cx+D = \tfrac{C}{2}(2x)+D$, giving $\tfrac{C}{2}\ln(x^2+a^2)+\tfrac{D}{a}\arctan(x/a)+K$; complete-the-square first if $b \neq 0$; check discriminant to decide factor vs arctan into your book.

Did you get this? True or false: $\displaystyle \int \dfrac{2x}{x^2 + 9}\,dx = \ln(x^2 + 9) + C$.

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

PROBLEM 1 · REPEATED LINEAR FACTOR

Decompose $\dfrac{2x + 1}{(x - 2)^2}$ into partial fractions, then find $\displaystyle \int \dfrac{2x + 1}{(x - 2)^2}\,dx$.

Template (repeated factor): $\dfrac{2x + 1}{(x - 2)^2} = \dfrac{A}{x - 2} + \dfrac{B}{(x - 2)^2}$. Multiply by $(x-2)^2$: $2x + 1 = A(x - 2) + B$.

$(x-2)^2$ contributes two terms — one for each power up to 2. Forgetting either constant produces an inconsistent system.

PROBLEM 2 · IRREDUCIBLE QUADRATIC

Decompose $\dfrac{4x + 3}{(x - 1)(x^2 + 4)}$ and find the integral.

$x^2 + 4$ is irreducible (discriminant $= -16$). Template: $\dfrac{4x + 3}{(x - 1)(x^2 + 4)} = \dfrac{A}{x - 1} + \dfrac{Bx + C}{x^2 + 4}$. Clear: $4x + 3 = A(x^2 + 4) + (Bx + C)(x - 1)$.

Irreducible quadratic gets a linear numerator. Three unknowns ($A, B, C$) match the three coefficients we will compare.

PROBLEM 3 · COMPLETE THE SQUARE

Find $\displaystyle \int \dfrac{1}{x^2 + 4x + 13}\,dx$.

Discriminant: $16 - 52 = -36 < 0$, so $x^2 + 4x + 13$ is irreducible — no partial fractions to do, just complete the square. $x^2 + 4x + 13 = (x + 2)^2 + 9$.

Always check the discriminant first. A negative value confirms the quadratic is irreducible and signals the $\tan^{-1}$ route.

Fill the gap: Completing the square: $x^2 + 6x + 13 = (x + $ $)^2 + $ . So $\displaystyle \int \dfrac{dx}{x^2 + 6x + 13} = \dfrac{1}{2}\tan^{-1}\dfrac{x + 3}{2} + C$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Omitting the lower-power term for $(x-a)^2$

Some students write only $\dfrac{B}{(x-a)^2}$ for a repeated factor, skipping $\dfrac{A}{x-a}$. The decomposition then has too few degrees of freedom and you'll be unable to solve for the constants. Always write one term per power from 1 to $k$.

Trap 02

Putting a single constant over a quadratic

For an irreducible quadratic $(x^2 + bx + c)$, the numerator must be $Cx + D$ (linear), not just $C$ (constant). A constant numerator would only describe a special case and miss the $x$-coefficient information.

Trap 03

Forgetting the $1/a$ factor in $\tan^{-1}$

$\int \dfrac{1}{x^2 + a^2}\,dx = \dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + C$ — note BOTH the $1/a$ prefactor AND the $x/a$ inside. Writing $\tan^{-1}(x)$ instead of $\dfrac{1}{a}\tan^{-1}(x/a)$ is a classic dropped-mark error.

Did you get this? True or false: $\displaystyle \int \dfrac{1}{x^2 + 25}\,dx = \tan^{-1}\dfrac{x}{5} + C$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Decompose $\dfrac{3x - 1}{(x + 1)^2}$ into partial fractions.

Decompose $\dfrac{5}{(x - 2)(x + 1)^2}$ into partial fractions.

Find $\displaystyle \int \dfrac{1}{x^2 + 6x + 10}\,dx$ by completing the square.

Decompose $\dfrac{x^2 + 1}{x(x^2 + 1)}$. Hint: simplify first if possible.

Find $\displaystyle \int \dfrac{2x + 5}{x^2 + 9}\,dx$ by splitting the numerator.

Odd one out: Three of these denominators are irreducible quadratics (so need the $(Cx+D)/(\text{quadratic})$ template). Which one factors over the reals and so does NOT?

Revisit your thinking

Earlier you predicted templates for $\dfrac{3}{(x-1)^2}$ and $\dfrac{5}{x^2+4}$ — and which integral would produce $\tan^{-1}$.

$\dfrac{3}{(x-1)^2}$ is already a single partial-fraction term (the second one in the repeated-factor template) and integrates by the power rule to $-3/(x-1) + C$. $\dfrac{5}{x^2 + 4}$ has an irreducible quadratic denominator and integrates to $\tfrac{5}{2}\tan^{-1}(x/2) + C$. Together with L05, you can now decompose and integrate ANY proper rational function that appears on the HSC Extension 2 paper.

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Multiple choice

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

Short answer

ApplyBand 32 marks

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

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ApplyBand 43 marks

Q2. Find $\displaystyle \int \dfrac{1}{x^2 - 4x + 13}\,dx$ by completing the square. (3 marks)

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Analyse4 marks

Q3. Find $\displaystyle \int \dfrac{3x + 1}{(x + 1)(x^2 + 1)}\,dx$. (4 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $3x - 1 = A(x+1) + B$. $x = -1$: $-4 = B$. Coefficient of $x$: $3 = A$. So $\dfrac{3x-1}{(x+1)^2} = \dfrac{3}{x+1} - \dfrac{4}{(x+1)^2}$.

2. $A = 5/9$, $C = -5/3$. Equate $x^2$ coefficients: $0 = A + B$, so $B = -5/9$. Therefore $\dfrac{5}{(x-2)(x+1)^2} = \dfrac{5}{9(x-2)} - \dfrac{5}{9(x+1)} - \dfrac{5}{3(x+1)^2}$.

3. $x^2 + 6x + 10 = (x+3)^2 + 1$. So $\displaystyle \int \dfrac{dx}{(x+3)^2 + 1} = \tan^{-1}(x + 3) + C$.

4. Simplification: $\dfrac{x^2 + 1}{x(x^2 + 1)} = \dfrac{1}{x}$, so integral is $\ln|x| + C$. (Partial fractions also gives $A/x + (Bx + C)/(x^2 + 1)$ with $A = 1$, $B = 0$, $C = 0$.)

5. Split: $\dfrac{2x + 5}{x^2 + 9} = \dfrac{2x}{x^2 + 9} + \dfrac{5}{x^2 + 9}$. First piece integrates to $\ln(x^2 + 9)$; second to $\dfrac{5}{3}\tan^{-1}(x/3)$. Total: $\ln(x^2 + 9) + \dfrac{5}{3}\tan^{-1}\dfrac{x}{3} + C$.

Q1 (2 marks): Template $\dfrac{A}{x-1} + \dfrac{B}{(x-1)^2}$ with $x + 3 = A(x - 1) + B$ [1]. $x = 1$: $B = 4$. Coefficient of $x$: $A = 1$. So $\dfrac{1}{x-1} + \dfrac{4}{(x-1)^2}$ [1].

Q2 (3 marks): $x^2 - 4x + 13 = (x - 2)^2 + 9$ [1]. With $u = x - 2$: $\displaystyle \int \dfrac{du}{u^2 + 9}$ [1]. Apply standard form: $\dfrac{1}{3}\tan^{-1}\dfrac{x - 2}{3} + C$ [1].

Q3 (4 marks): Template $\dfrac{A}{x+1} + \dfrac{Bx + C}{x^2 + 1}$ with $3x + 1 = A(x^2 + 1) + (Bx + C)(x + 1)$ [1]. $x = -1$: $-2 = 2A$, so $A = -1$. Equate $x^2$: $0 = A + B$, so $B = 1$. Equate constant: $1 = A + C$, so $C = 2$ [1]. Decomposition: $\dfrac{-1}{x+1} + \dfrac{x + 2}{x^2 + 1}$ [1]. Integrate: $-\ln|x + 1| + \dfrac{1}{2}\ln(x^2 + 1) + 2\tan^{-1} x + C$ [1].

Boss battle · The Quadratic Tamer

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Five timed questions on repeated factors, irreducible quadratics and completing the square. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

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Climb platforms by answering quick partial-fraction and standard-form questions. Lighter alternative to the boss.

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Module overview · Extension 2 · Checkpoint 1