Your weak spots

Insights load after your first practice round.

Module 8 · L5 of 20 ~40 min ⚡ +95 XP available

Algebraic Manipulation Before Integration

Before you can integrate, you often need to rewrite the integrand. Dividing a polynomial by a linear factor, expanding brackets, and splitting fractions using partial fractions are the three main techniques. Master these and a whole new family of integrals become routine.

Today's hook, Without doing any algebra, which of these do you think is easier to integrate: $\dfrac{x^2+1}{x+1}$ or $(x+1)^2$? Jot your reasoning. You'll revisit it after card 06.

0/5QUESTS

Recall, your gut answer first

+5 XP warm-up

Consider $\displaystyle\int \frac{x^2+1}{x+1}\,dx$. Without any calculationdo you think this can be integrated directly, or does the integrand need to be rewritten first? Explain your thinking.

auto-saved

The big idea: rewrite before you integrate

+5 XP to read

Most integration problems require the integrand to be in a standard form before you can apply a rule. The three algebraic tools you need are:

Polynomial long division when the degree of the numerator $\geq$ degree of the denominator
Expanding brackets simplify products and powers into individual terms
Partial fractions split a rational expression into simpler fractions with linear denominators

Rewrite $\rightarrow$ Integrate $\rightarrow$ Simplify

Long division first

If degree of numerator $\geq$ degree of denominator, always divide first. The result is a polynomial plus a proper fraction.

Expand everything

Products like $(x+2)(x-3)$ or powers like $(1+x)^3$ can be expanded so each term integrates independently.

Partial fractions

A proper fraction with factorisable denominator splits into simpler terms: $\dfrac{A}{x-a} + \dfrac{B}{x-b}$, each integrating to a log.

What you'll master

Know

Key facts

When to divide before integrating (degree of numerator $\geq$ denominator)
$\int \dfrac{1}{ax+b}\,dx = \dfrac{1}{a}\ln|ax+b| + C$
Partial fractions form: $\dfrac{A}{x-a} + \dfrac{B}{x-b}$ for distinct linear factors

Understand

Concepts

Why algebraic manipulation is needed before standard integration rules apply
How polynomial long division produces a polynomial plus a proper fraction
How the cover-up method or substitution finds partial fraction constants

Can do

Skills

Perform polynomial long division on improper rational integrands
Expand and integrate products and powers of polynomials
Decompose proper fractions into partial fractions and integrate each term

Key terms

Improper fractionA rational expression where the degree of the numerator is greater than or equal to the degree of the denominator. Must be divided before integrating.

Proper fractionA rational expression where the degree of the numerator is less than the degree of the denominator. Can be integrated directly or via partial fractions.

Polynomial long divisionThe process of dividing a polynomial by another polynomial to express an improper fraction as quotient + remainder/divisor.

Partial fractionsA method of writing a proper rational fraction as a sum of simpler fractions. Each simpler fraction has a linear (or irreducible quadratic) denominator.

Cover-up methodA quick way to find partial fraction constants: cover the relevant factor in the denominator and substitute the root of that factor into the remaining expression.

$\int \frac{1}{ax+b}\,dx$$= \dfrac{1}{a}\ln|ax+b| + C$. This is the key integration result for each partial fraction term.

Technique 1, polynomial long division

core concept

When the integrand is an improper fraction (numerator degree $\geq$ denominator degree), you must divide before integrating.

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

How to divide: Ask "what times the leading term of the divisor gives the leading term of the dividend?" Subtract, bring down, and repeat.

Once divided, each term integrates separately:

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

Check your division. Verify by multiplying: $(x+1)(x+2) + 3 = x^2 + 3x + 2 + 3 = x^2 + 3x + 5$. Always check your quotient times divisor plus remainder equals the original numerator.

When the integrand is an improper fraction (numerator degree $\geq$ denominator degree), you must divide before integrating.

Pause, copy the long-division technique: divide numerator by denominator to get quotient $+$ remainder fraction, then integrate term by term into your book.

Quick check: Which of the following must be rewritten using polynomial long division before integrating?

Technique 2, expanding brackets

core concept

We just saw that when the numerator degree $\geq$ denominator degree, polynomial long division rewrites $\frac{p(x)}{q(x)}=d(x)+\frac{r(x)}{q(x)}$ so each term integrates directly. That raises a question: for integrands that are products or powers, expanding brackets before integrating avoids long division entirely, when is this faster, and what is the one-line check? This card answers it → if the integrand is a product of polynomials or a power of a simple expression, expand first; long division is only needed when a fraction is irreducible.

Many integrands involve products or powers that can be expanded so each term is a simple power of $x$.

Example 1, expand then integrate:

$$\int (x+2)(x-3)\,dx = \int (x^2 - x - 6)\,dx = \frac{x^3}{3} - \frac{x^2}{2} - 6x + C$$

Example 2, power expansion:

$$\int \sqrt{x}\,(x+1)\,dx = \int (x^{3/2} + x^{1/2})\,dx = \frac{2x^{5/2}}{5} + \frac{2x^{3/2}}{3} + C$$

Revisit the hook. $(x+1)^2 = x^2 + 2x + 1$ integrates immediately term by term. The fraction $\dfrac{x^2+1}{x+1}$ requires long division first. So expanding is quicker when it's a product or power, but not a fraction.

Many integrands involve products or powers that can be expanded so each term is a simple power of $x$.

Pause, copy the expand-first technique: $(1+x)^3=1+3x+3x^2+x^3$; then $\int(1+x)^3dx=x+\frac{3x^2}{2}+x^3+\frac{x^4}{4}+C$ (or use substitution) into your book.

Did you get this? True or false: $\displaystyle\int (x-1)(x+3)\,dx = \int (x^2 + 2x - 3)\,dx$.

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

PROBLEM 1 · LONG DIVISION

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

Degree of numerator (2) $\geq$ degree of denominator (1) $\Rightarrow$ perform polynomial long division.

Check degrees first. Since 2 ≥ 1, this is an improper fraction and must be divided before integrating.

PROBLEM 2 · EXPANDING

Find $\displaystyle\int \frac{x^3 + 8}{x + 2}\,dx$. (Hint: factorise the numerator first.)

$x^3 + 8 = (x+2)(x^2 - 2x + 4)$ (sum of cubes), so $\dfrac{x^3+8}{x+2} = x^2 - 2x + 4$.

Recognise the sum-of-cubes factorisation $a^3+b^3=(a+b)(a^2-ab+b^2)$ with $a=x$, $b=2$. This cancels the denominator entirely.

PROBLEM 3 · PARTIAL FRACTIONS

Find $\displaystyle\int \frac{5}{(x+1)(x-4)}\,dx$.

Write $\dfrac{5}{(x+1)(x-4)} = \dfrac{A}{x+1} + \dfrac{B}{x-4}$. Multiply both sides by $(x+1)(x-4)$: $5 = A(x-4) + B(x+1)$.

Set up the partial fraction decomposition. The denominator has two distinct linear factors, so we need constants $A$ and $B$.

Fill the gap: $\displaystyle\int \frac{1}{x-3}\,dx = $ .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Skipping the degree check

Many students try to apply partial fractions directly to $\dfrac{x^2+1}{x-2}$ without first performing long division. But this is an improper fraction (degree 2 $\geq$ degree 1), so you must divide first. Partial fractions only apply to proper fractions where degree of numerator $<$ degree of denominator.

Trap 02

Missing the absolute value in logarithms

$\displaystyle\int \frac{1}{x-a}\,dx = \ln|x-a| + C$, not $\ln(x-a) + C$. The absolute value is essential because the logarithm is only defined for positive arguments, $x-a$ can be negative. Losing it will cost marks in any HSC question.

Trap 03

Wrong partial fraction form for repeated factors

If the denominator has a repeated factor $(x-a)^2$, the correct form is $\dfrac{A}{x-a} + \dfrac{B}{(x-a)^2}$, not just $\dfrac{A}{x-a}$. At the Ext 1 level, the HSC typically only tests distinct linear factors, but be aware of this for harder problems.

Did you get this? True or false: $\displaystyle\int \frac{x+5}{x-1}\,dx$ requires polynomial long division before integrating, because the numerator and denominator have the same degree.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Perform polynomial long division to find $\dfrac{x^2+5x+9}{x+2}$, then integrate.

Expand and integrate $\displaystyle\int (2x-1)(x+3)\,dx$.

Use partial fractions to find $\displaystyle\int \frac{4}{(x+2)(x-2)}\,dx$.

Find $\displaystyle\int \frac{x^3-1}{x-1}\,dx$ by factorising the numerator first.

Integrate $\displaystyle\int \frac{3x+1}{(x+1)(x+2)}\,dx$ using partial fractions.

Odd one out: Three of these integrals can be evaluated by expanding brackets alone. Which one requires a different technique?

Revisit your thinking

At the start you considered $\displaystyle\int \dfrac{x^2+1}{x+1}\,dx$. Using long division: $\dfrac{x^2+1}{x+1} = x - 1 + \dfrac{2}{x+1}$, so the integral is $\dfrac{x^2}{2} - x + 2\ln|x+1| + C$.

Did your intuition about needing to rewrite it prove correct? What was the key step?

auto-saved

Multiple choice

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

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

Short answer

ApplyBand 32 marks

Q1. Find $\displaystyle\int \frac{x^2+2}{x+1}\,dx$. (2 marks)

auto-saved

ApplyBand 43 marks

Q2. Use partial fractions to find $\displaystyle\int \frac{6}{(x-1)(x+2)}\,dx$. (3 marks)

auto-saved

AnalyseBand 53 marks

Q3. Find $\displaystyle\int_0^1 \frac{x^2-x-2}{x-2}\,dx$. (3 marks)

auto-saved

Comprehensive answers (click to reveal)

Activity answers:

1. $\dfrac{x^2+5x+9}{x+2} = x+3+\dfrac{3}{x+2}$. Integral: $\dfrac{x^2}{2}+3x+3\ln|x+2|+C$.

2. $(2x-1)(x+3) = 2x^2+5x-3$. Integral: $\dfrac{2x^3}{3}+\dfrac{5x^2}{2}-3x+C$.

3. $A=1, B=-1$. Integral: $\ln|x+2|-\ln|x-2|+C = \ln\left|\dfrac{x+2}{x-2}\right|+C$.

4. $x^3-1=(x-1)(x^2+x+1)$, so simplified to $x^2+x+1$. Integral: $\dfrac{x^3}{3}+\dfrac{x^2}{2}+x+C$.

5. $\dfrac{3x+1}{(x+1)(x+2)} = \dfrac{-2}{x+1}+\dfrac{5}{x+2}$ (check: $A=-2$, $B=5$). Integral: $-2\ln|x+1|+5\ln|x+2|+C$.

Q1 (2 marks): $\dfrac{x^2+2}{x+1} = x-1+\dfrac{3}{x+1}$ [1]. Integral: $\dfrac{x^2}{2}-x+3\ln|x+1|+C$ [1].

Q2 (3 marks): $\dfrac{6}{(x-1)(x+2)} = \dfrac{A}{x-1}+\dfrac{B}{x+2}$; $A=2$, $B=-2$ [1]. $= \dfrac{2}{x-1}-\dfrac{2}{x+2}$ [1]. Integral: $2\ln|x-1|-2\ln|x+2|+C = 2\ln\left|\dfrac{x-1}{x+2}\right|+C$ [1].

Q3 (3 marks): $\dfrac{x^2-x-2}{x-2} = x+1+\dfrac{0}{x-2} = x+1$ (since $x^2-x-2=(x-2)(x+1)$) [1]. $\int_0^1(x+1)\,dx = \left[\dfrac{x^2}{2}+x\right]_0^1 = \dfrac{1}{2}+1 - 0 = \dfrac{3}{2}$ [2].

Boss battle · The Algebraic Manipulator

earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering algebraic manipulation and integration questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 4 · Trigonometric Substitutions Lesson 6 · Integrating sin² nx and cos² nx →

Module overview · Extension 1 · Checkpoint 1