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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.

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Recall, your gut answer first

+5 XP warm-up From L05 you know $\int \dfrac{1}{x - a}\,dx = \ln|x - a| + C$.

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

+5 XP to read Beyond distinct linear factors you meet two new shapes.

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.

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

Recognise the denominator's factor shape to choose the right integration template.

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

$(x-a)^2$ 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

Repeated factor rule.

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

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 $Cx+D$.

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: $\displaystyle \int \dfrac{1}{x^2 + 25}\,dx = \tan^{-1}\dfrac{x}{5} + C$.

Revisit your thinking

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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Show what you have learned

Multiple choice, then short answer under exam conditions.

Multiple choice

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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].

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Module overview · Extension 2