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Module 8 · L18 of 20 ~45 min ⚡ +95 XP available

Harder Integration Problems

The hardest HSC integrals rarely yield to a single technique — they demand a chain of moves: substitution to simplify the structure, then perhaps partial fractions to split a rational function, then inverse trig to close it out. In this lesson you'll learn to read an integral, map out the right sequence of techniques, and execute each step cleanly under exam pressure.

Today's hook — You see the integral $\displaystyle\int \dfrac{x}{\sqrt{1-x^4}}\,dx$. Before reading ahead, jot the first substitution you'd try and why. You'll evaluate it fully after card 07.

0/5QUESTS

Recall — your gut answer first

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You want to evaluate $\displaystyle\int \dfrac{x}{\sqrt{1-x^4}}\,dx$. Without computing it — what substitution would you try first, and what form do you expect the answer to take?

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The big idea: chaining techniques

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Harder integrals rarely announce which technique to use. The skill is pattern matching: inspect the integrand, identify its structure, choose the right first move, then reassess after each step.

The five key signals and their techniques:

$\sqrt{a^2-x^2}$ → trig sub $x = a\sin\theta$
$\dfrac{1}{a^2+x^2}$ → inverse tan formula
$\dfrac{f'(x)}{f(x)}$ → $\ln|f(x)|$
Rational function, degree OK → partial fractions
Product of functions → integration by parts

Inspect → Match → Execute → Repeat

Plan before computing

Spend 30 seconds planning the full chain of moves before writing a single integral. This saves time overall by preventing dead-end approaches.

Complete the square first

Quadratics like $x^2 + 4x + 7$ don't match standard forms until completed: $(x+2)^2 + 3$. This unlocks inverse trig or trig substitution immediately.

Substitution changes limits

For definite integrals, always recalculate the limits of integration when you substitute. Forgetting this is one of the most common exam errors.

What you'll master

Know

Key facts

$\displaystyle\int \dfrac{1}{\sqrt{a^2-x^2}}\,dx = \sin^{-1}\dfrac{x}{a} + C$
$\displaystyle\int \dfrac{1}{a^2+x^2}\,dx = \dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + C$
The five pattern-matching signals for technique selection

Understand

Concepts

How substitution can transform a non-standard integrand into a standard form
Why completing the square unlocks inverse trig integration
How multi-step integrals require reassessment after each technique

Can do

Skills

Select and chain two or more integration techniques for a single integral
Evaluate integrals involving inverse trig after completing the square
Handle definite integrals with correct limit changes under substitution

Key terms

Completing the squareRewriting $ax^2+bx+c$ as $a(x+p)^2+q$ to match standard inverse trig or substitution forms.

Partial fractionsDecomposing a proper rational function into simpler fractions whose integrals are known (logarithms or inverse trig).

Trigonometric substitutionReplacing $x$ with $a\sin\theta$ (or $a\tan\theta$) to eliminate a square root of the form $\sqrt{a^2-x^2}$ (or $\sqrt{a^2+x^2}$).

Standard inverse trig forms$\int\frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a}+C$ and $\int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}+C$.

Limit adjustmentWhen substituting $u=g(x)$ in a definite integral, replace $x$-limits $[a,b]$ with $u$-limits $[g(a),g(b)]$.

Numerator decompositionSplitting a numerator like $2x+5$ into a multiple of the denominator's derivative plus a constant, to produce $\ln$ and $\tan^{-1}$ terms separately.

Completing the square for inverse trig

core concept

When you see $\displaystyle\int \dfrac{dx}{x^2+bx+c}$, the quadratic doesn't match the standard form until you complete the square:

$$x^2 + 4x + 7 = (x+2)^2 + 3$$

Now set $u = x+2$, $du = dx$, and the integral becomes:

$$\int\frac{du}{u^2+3} = \frac{1}{\sqrt{3}}\tan^{-1}\frac{u}{\sqrt{3}} + C = \frac{1}{\sqrt{3}}\tan^{-1}\frac{x+2}{\sqrt{3}} + C$$

For $\displaystyle\int\dfrac{dx}{\sqrt{c - x^2 - bx}}$, complete the square in the radicand to get $\sqrt{a^2-(x+p)^2}$, then recognise the $\sin^{-1}$ form:

$$\int\frac{dx}{\sqrt{a^2-(x+p)^2}} = \sin^{-1}\frac{x+p}{a} + C$$

Numerator trick. For $\displaystyle\int\dfrac{2x+5}{x^2+4x+7}\,dx$, write $2x+5 = (2x+4) + 1$, so the integral splits into $\int\dfrac{2x+4}{x^2+4x+7}\,dx + \int\dfrac{1}{x^2+4x+7}\,dx = \ln|x^2+4x+7| + \dfrac{1}{\sqrt{3}}\tan^{-1}\dfrac{x+2}{\sqrt{3}} + C$.

When you see $\displaystyle\int \dfrac{dx}{x^2+bx+c}$, the quadratic doesn't match the standard form until you complete the square:

Pause — copy the completing-the-square route to $\tan^{-1}$: $\int\frac{dx}{x^2+bx+c}\xrightarrow{\text{CTS}}\int\frac{du}{u^2+a^2}=\frac{1}{a}\tan^{-1}(u/a)+C$ into your book.

Quick check: Which is the correct completion of the square for $x^2 - 6x + 13$?

Substitution leading to inverse trig

core concept

We just saw that when the denominator is a quadratic $x^2+bx+c$, complete the square to write it as $(x+p)^2+a^2$, then apply the $\tan^{-1}$ standard integral. That raises a question: sometimes a substitution transforms a non-obvious integrand into a recognisable inverse-trig form — what is the indicator that a substitution (rather than completing the square) is the right move? This card answers it → if a factor of the derivative of the denominator appears in the numerator, try $u=$ denominator; otherwise complete the square.

Sometimes a substitution transforms the integrand into a recognisable inverse trig form. The key is spotting what $u$ makes the radicand or denominator match $a^2 - u^2$ or $a^2 + u^2$.

Example: $\displaystyle\int\dfrac{x}{\sqrt{1-x^4}}\,dx$

Notice $1 - x^4 = 1 - (x^2)^2$. Let $u = x^2$, then $du = 2x\,dx$, so $x\,dx = \dfrac{du}{2}$:

$$\int\frac{x\,dx}{\sqrt{1-x^4}} = \int\frac{du/2}{\sqrt{1-u^2}} = \frac{1}{2}\sin^{-1}u + C = \frac{1}{2}\sin^{-1}(x^2) + C$$

Example 2: $\displaystyle\int\dfrac{e^x}{1+e^{2x}}\,dx$. Let $u = e^x$, $du = e^x\,dx$:

$$\int\frac{e^x\,dx}{1+e^{2x}} = \int\frac{du}{1+u^2} = \tan^{-1}u + C = \tan^{-1}(e^x) + C$$

Hook answer. Using $u = x^2$, $\displaystyle\int\dfrac{x}{\sqrt{1-x^4}}\,dx = \dfrac{1}{2}\sin^{-1}(x^2) + C$. The substitution worked because $x\,dx$ appeared naturally in the numerator, matching the $du = 2x\,dx$ from $u = x^2$.

Sometimes a substitution transforms the integrand into a recognisable inverse trig form. The key is spotting what $u$ makes the radicand or denominator match $a^2 - u^2$ or $a^2 + u^2$.

Pause — copy the decision rule: if numerator is a scalar multiple of the denominator's derivative → substitution; otherwise → complete the square then use $\tan^{-1}$ or $\sin^{-1}$ form into your book.

Did you get this? True or false: The substitution $u = x^2$ transforms $\displaystyle\int\dfrac{x}{\sqrt{1-x^4}}\,dx$ into $\dfrac{1}{2}\displaystyle\int\dfrac{du}{\sqrt{1-u^2}}$.

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

PROBLEM 1 · COMPLETE THE SQUARE + INVERSE TAN

Evaluate $\displaystyle\int\dfrac{3}{x^2-2x+5}\,dx$.

Complete the square: $x^2-2x+5 = (x-1)^2+4$

The quadratic has no real roots; completing the square gives the $u^2+a^2$ form needed for $\tan^{-1}$.

PROBLEM 2 · SUBSTITUTION + PARTIAL FRACTIONS

Find $\displaystyle\int\dfrac{1}{x(x^2+1)}\,dx$ using partial fractions.

$\dfrac{1}{x(x^2+1)} = \dfrac{A}{x} + \dfrac{Bx+C}{x^2+1}$. Multiply through: $1 = A(x^2+1)+(Bx+C)x$

Denominator has a linear factor $x$ and an irreducible quadratic $x^2+1$, so use these partial fraction forms.

PROBLEM 3 · DEFINITE INTEGRAL, MULTI-STEP

Evaluate $\displaystyle\int_0^{\sqrt{3}}\dfrac{1}{\sqrt{3-x^2}}\,dx$.

Recognise the form $\int\dfrac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\dfrac{x}{a}+C$ with $a = \sqrt{3}$

No substitution needed — the integrand already matches the standard form. Identify $a$ first.

Fill the gap: $\displaystyle\int\dfrac{e^x}{1+e^{2x}}\,dx$. Let $u = e^x$. Then the integral becomes $\displaystyle\int\dfrac{du}{1+u^2} =$ .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting to change limits under substitution

When you substitute $u = g(x)$ in a definite integral, you MUST replace $x = a$ with $u = g(a)$ and $x = b$ with $u = g(b)$. Leaving the original limits while integrating with respect to $u$ gives the wrong answer. Always write the new limits explicitly before integrating.

Trap 02

Improper partial fractions form

For an irreducible quadratic factor like $x^2+1$, the numerator must be $Bx+C$ (degree one less than the quadratic), not just $B$. Writing $\frac{B}{x^2+1}$ misses the $Bx$ term, leading to a wrong decomposition. Always use $\frac{Bx+C}{\text{quadratic}}$.

Trap 03

Applying $\tan^{-1}$ form when $a^2$ is not correctly identified

$\int\frac{du}{u^2+a^2} = \frac{1}{a}\tan^{-1}\frac{u}{a}+C$. Students frequently write $\tan^{-1}\frac{u}{a}$ without the $\frac{1}{a}$ factor out front. Always identify $a^2$ first, extract $a$, and include $\frac{1}{a}$ in the result.

Did you get this? True or false: $\displaystyle\int\dfrac{du}{u^2+9} = \tan^{-1}\dfrac{u}{3} + C$ (without any numerical prefactor).

Work mode · how are you completing this lesson?

Activities · practice with the ideas

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

Find $\displaystyle\int\dfrac{e^{2x}}{1+e^{2x}}\,dx$. (Hint: what is the derivative of $1+e^{2x}$?)

Use partial fractions to evaluate $\displaystyle\int\dfrac{1}{x^2-1}\,dx$. (Hint: factorise the denominator.)

Evaluate $\displaystyle\int_0^1\dfrac{2x}{1+x^4}\,dx$. (Hint: let $u = x^2$.)

A student evaluates $\displaystyle\int_0^2\dfrac{1}{\sqrt{4-x^2}}\,dx$ and gets $\sin^{-1}(2)-\sin^{-1}(0) = \sin^{-1}(2)$. Explain the error and give the correct value.

Odd one out: Three of these antiderivatives are correct. Which one is NOT?

Revisit your thinking

Earlier you guessed a substitution for $\displaystyle\int\dfrac{x}{\sqrt{1-x^4}}\,dx$.

The key insight was spotting $1-x^4 = 1-(x^2)^2$: the substitution $u = x^2$ maps the integrand to $\dfrac{1}{2}\displaystyle\int\dfrac{du}{\sqrt{1-u^2}} = \dfrac{1}{2}\sin^{-1}(x^2)+C$. The $x$ in the numerator was the signal — it matched the $du = 2x\,dx$ we needed. Any time you see $x\cdot f(x^2)$, try $u = x^2$.

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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.

Short answer

ApplyBand 32 marks

Q1. Evaluate $\displaystyle\int\dfrac{1}{x^2-4x+13}\,dx$. (2 marks)

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ApplyBand 43 marks

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

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AnalyseBand 53 marks

Q3. Evaluate $\displaystyle\int_0^1\dfrac{x}{\sqrt{4-x^4}}\,dx$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers: 1. $(x+3)^2+1$; $\tan^{-1}(x+3)+C$ · 2. $\frac{1}{2}\ln(1+e^{2x})+C$ (reverse chain rule) · 3. $\frac{1}{2}\ln|\frac{x-1}{x+1}|+C$ · 4. $\tan^{-1}(x^2)\big|_0^1 = \pi/4$ · 5. Error: standard form is $\sin^{-1}(x/a)$ not $\sin^{-1}(x)$; correct value is $[\sin^{-1}(x/2)]_0^2 = \pi/2$.

Q1 (2 marks): $x^2-4x+13=(x-2)^2+9$ [1]; $\int\frac{du}{u^2+9} = \frac{1}{3}\tan^{-1}\frac{x-2}{3}+C$ [1].

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

Q3 (3 marks): $u=x^2$, $du=2x\,dx$, limits $0$ to $1$ [1]; $\frac{1}{2}\int_0^1\frac{du}{\sqrt{4-u^2}} = \frac{1}{2}\left[\sin^{-1}\frac{u}{2}\right]_0^1$ [1]; $= \frac{1}{2}\sin^{-1}\frac{1}{2} = \frac{1}{2}\cdot\frac{\pi}{6} = \frac{\pi}{12}$ [1].

Boss battle · The Integration Gauntlet

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 harder integration questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

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Module overview · Extension 1 · Checkpoint 4