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Module 15 · L14 of 16 ~40 min ⚡ +90 XP available

Mixed Integration Problems I

By now you have five techniques in your kit — substitution, by parts, partial fractions, trig substitution, and the $t$-formula. The HSC exam doesn't tell you which one to use. This lesson is about identification: reading the structural features of an integrand and choosing the technique that will work, before committing pen to paper.

Today's hook — Without working any of them out, decide which technique you would use for each: (a) $\int x e^x\, dx$, (b) $\int \dfrac{2x+1}{x^2 + x + 1}\, dx$, (c) $\int \sqrt{4 - x^2}\, dx$, (d) $\int \dfrac{1}{x^2 - 1}\, dx$. Compare after card 05.

0/5QUESTS

Recall — your gut answer first

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List the five integration techniques you have seen so far and write one structural feature of the integrand that signals each. Before checking — which technique does each of these scream to you: $\int \dfrac{1}{\sqrt{9 - x^2}}\, dx$, $\int x \cos x\, dx$, $\int \dfrac{1}{(x-1)(x+2)}\, dx$?

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The two moves for technique identification

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Choosing a technique rewards two habits: read the integrand's shape (rational, product, radical, trig in sin/cos), then check for a quick win — is the numerator the derivative of the denominator? Is there a $u$ whose derivative is sitting right there? Only after those quick checks fail do you reach for the heavier tools.

The shape-check-technique reading: (1) classify the integrand by gross shape, (2) try the cheapest move first (recognition, then substitution), (3) escalate to parts / partial fractions / trig sub / $t$-sub only when the cheap moves fail.

Order of cost: recognition < substitution < by parts < partial fractions < trig sub < $t$-sub

$\text{shape} \;\longrightarrow\; \text{cheap check} \;\longrightarrow\; \text{technique}$

Cheapest move first

Always ask: is this an $f'(x)/f(x)$ pattern, a chain-rule reversal, or a standard form? A 10-second check saves 10 minutes of partial-fractions algebra.

Match the shape to the tool

Proper rational with factorable denominator → partial fractions. Product of unrelated types → by parts. $\sqrt{a^2 - x^2}$ → trig sub. Rational in $\sin, \cos$ → $t$-formula.

Don't be afraid to switch

If a technique starts to look ugly after one step, stop. The exam never wants three pages — if your method is producing a mess, you've probably picked the wrong tool.

What you'll master

Know

Key facts

The five MEX-C1 techniques and their signature integrand shapes
Substitution standard forms: $\int \frac{f'(x)}{f(x)}\, dx$, $\int f'(x) e^{f(x)}\, dx$
By-parts LIATE priority for choosing $u$
Trig sub triggers: $\sqrt{a^2 - x^2} \Rightarrow x = a\sin\theta$; $\sqrt{a^2 + x^2} \Rightarrow x = a\tan\theta$
$t$-substitution: $t = \tan(x/2)$ for rational functions of $\sin x, \cos x$

Understand

Concepts

Why technique selection precedes computation in HSC questions
Why each technique exploits a specific algebraic structure
Why some integrals admit several valid techniques (e.g., substitution vs by parts)

Can do

Skills

Identify the appropriate technique by inspection in under 30 seconds
Justify the choice using a structural feature of the integrand
Carry the chosen technique through to a final answer

Key terms

SubstitutionReplace $u = g(x)$, $du = g'(x)\, dx$ when the integrand has $g'(x)$ multiplied by a function of $g(x)$. Cheapest technique.

Integration by parts$\int u\, dv = uv - \int v\, du$. Use for products of unrelated types (polynomial $\times$ $\ln$, polynomial $\times$ exp, polynomial $\times$ trig). LIATE picks $u$.

Partial fractionsDecompose a proper rational $\tfrac{P(x)}{Q(x)}$ into simpler fractions. Use when $Q(x)$ factorises (and the numerator is not already $Q'(x)$).

Trigonometric substitutionChoose $x = a\sin\theta$ for $\sqrt{a^2 - x^2}$; $x = a\tan\theta$ for $\sqrt{a^2 + x^2}$ or $a^2 + x^2$; $x = a\sec\theta$ for $\sqrt{x^2 - a^2}$.

$t$-substitution (Weierstrass)$t = \tan(x/2)$ converts $\sin x = \frac{2t}{1+t^2}$, $\cos x = \frac{1-t^2}{1+t^2}$, $dx = \frac{2}{1+t^2}\, dt$. Use for rational functions of $\sin x, \cos x$.

Proper rationalA rational function $\tfrac{P(x)}{Q(x)}$ where $\deg P < \deg Q$. Partial fractions require this; if not, divide first.

MEX-C1NESA outcome (Further Integration): selects and applies appropriate techniques of integration including substitution, by parts, partial fractions, trigonometric substitutions, and the $t$-formula.

Decision rules — reading the integrand

core concept

Run through this checklist top-to-bottom. The first match wins.

Standard form / recognition. Is this $\int \frac{1}{1+x^2}\, dx$, $\int \frac{1}{\sqrt{1-x^2}}\, dx$, $\int e^x\, dx$ etc.? Write it down.
$f'(x)/f(x)$ or substitution. Is the numerator (a constant multiple of) the derivative of the denominator? Or is there an inner function $g(x)$ with $g'(x)$ sitting in the integrand? Substitute.
By parts. Is the integrand a product of two unrelated functions (polynomial $\times$ log, polynomial $\times$ exp, polynomial $\times$ trig)? Apply LIATE.
Partial fractions. Is it a proper rational with a factorable denominator (and not in $f'/f$ form)? Decompose.
Trig substitution. Does the integrand contain $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$ or $\sqrt{x^2 - a^2}$? Substitute with sin, tan or sec respectively.
$t$-substitution. Is the integrand a rational function of $\sin x$ and $\cos x$ that doesn't yield to anything above? Use $t = \tan(x/2)$.

Worked through the hook:

(a) $\int x e^x\, dx$ — product of polynomial and exponential, unrelated $\Rightarrow$ by parts ($u = x$).
(b) $\int \dfrac{2x+1}{x^2 + x + 1}\, dx$ — numerator is the derivative of the denominator $\Rightarrow$ $f'/f$ log rule.
(c) $\int \sqrt{4 - x^2}\, dx$ — radical of $a^2 - x^2$ form $\Rightarrow$ trig sub $x = 2\sin\theta$.
(d) $\int \dfrac{1}{x^2 - 1}\, dx$ — proper rational with factorable denominator $(x-1)(x+1)$ $\Rightarrow$ partial fractions.

Why the order matters. Recognition and substitution are cheap. Partial fractions and trig sub require setup. Spending 10 seconds asking "is this just $f'/f$?" before launching into partial fractions saves time and avoids algebra errors.

Decision order: recognition $\to$ substitution / $f'/f$ $\to$ by parts $\to$ partial fractions $\to$ trig sub $\to$ $t$-sub · Trig sub triggers: $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, $\sqrt{x^2 - a^2}$ · $t = \tan(x/2)$ for rational expressions in $\sin x, \cos x$ · Always read the shape before choosing the tool

Pause — copy the decision order (recognition → $u$-sub/$f'/f$ → IBP → partial fractions → trig sub → $t$-sub), the three trig-sub triggers, and the $t$-sub trigger (rational in $\sin x, \cos x$) into your book.

Quick check: Which technique is best suited to $\displaystyle \int \frac{x + 3}{(x - 1)(x + 2)}\, dx$?

Close-call patterns — when two techniques both apply

core concept

We just saw the integration decision tree: recognition → substitution/$f'/f$ → IBP → partial fractions → trig sub → $t$-sub, with trig-sub triggers ($\sqrt{a^2-x^2}$, $\sqrt{a^2+x^2}$, $\sqrt{x^2-a^2}$) and $t$-sub for rational trig expressions. That raises a question: what happens when two techniques both look valid — which wins? This card answers it → $f'/f$ beats partial fractions when the numerator is already the derivative; standard arctan/arcsin forms beat trig sub when the integrand already matches; substitution beats IBP when an inner function and its derivative are both present.

Sometimes the integrand fits more than one pattern. Pick the cheaper one.

Rational with $f'/f$ form vs partial fractions. $\int \frac{2x}{x^2 - 1}\, dx$ fits $f'/f$ (answer: $\ln|x^2 - 1| + C$). Avoid partial-fractions setup — it would give the same answer with more work.
Substitution vs by parts. $\int x \sqrt{x^2 + 1}\, dx$ fits substitution ($u = x^2 + 1$), not by parts. By parts would work but takes longer.
Trig sub vs standard form. $\int \frac{1}{1 + x^2}\, dx$ is the standard form $\arctan x + C$. Don't substitute $x = \tan\theta$ — recognise it.
Partial fractions vs trig sub. $\int \frac{1}{x^2 + 4}\, dx$ has an unfactorable denominator over $\mathbb{R}$; partial fractions fail. Recognise the $\frac{1}{a^2 + x^2}$ standard form: $\tfrac{1}{2}\arctan(x/2) + C$.

$$\int \frac{1}{a^2 + x^2}\, dx = \tfrac{1}{a}\arctan\!\left(\tfrac{x}{a}\right) + C \qquad \int \frac{1}{\sqrt{a^2 - x^2}}\, dx = \arcsin\!\left(\tfrac{x}{a}\right) + C$$

Common mistake. Reaching for partial fractions whenever you see a rational integrand. Always check the $f'/f$ pattern first, and check whether the denominator factorises over $\mathbb{R}$ at all.

$f'/f$ rule beats partial fractions when the numerator is the derivative of the denominator · Standard forms $\arctan, \arcsin$ beat trig sub when the integrand already matches · Substitution beats by parts when an inner function and its derivative are both present · The cheapest valid technique is always the right answer in an exam

Pause — copy the three close-call resolution rules ($f'/f$ over partial fractions; standard form over trig sub; $u$-sub over IBP) and the exam principle "cheapest valid technique" into your book.

Did you get this? True or false: $\displaystyle \int \frac{2x}{x^2 - 1}\, dx$ is best handled by partial fractions rather than the $f'(x)/f(x)$ log rule.

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

PROBLEM 1 · BY PARTS

Evaluate $\displaystyle \int x e^{2x}\, dx$.

Identify the technique. The integrand is a product of a polynomial ($x$) and an exponential ($e^{2x}$) — unrelated types. LIATE: Algebraic beats Exponential, so $u = x$.

With no substitution available (the derivative of $2x$ is just a constant, no inner $x$ to absorb), by parts is the natural tool. LIATE pinpoints $u$.

PROBLEM 2 · PARTIAL FRACTIONS

Evaluate $\displaystyle \int \frac{1}{x^2 - 1}\, dx$.

Identify the technique. The integrand is a proper rational with factorable denominator $x^2 - 1 = (x-1)(x+1)$, and the numerator is not the derivative of the denominator (which would be $2x$). Use partial fractions.

Check $f'/f$ first: derivative of $x^2 - 1$ is $2x$, but the numerator is $1$. No match. Move to partial fractions.

PROBLEM 3 · TRIG SUBSTITUTION

Evaluate $\displaystyle \int \frac{1}{\sqrt{9 - x^2}}\, dx$.

Identify the technique. The radical has form $\sqrt{a^2 - x^2}$ with $a = 3$. (Also recognisable as a standard $\arcsin$ form — both routes give the same answer.) Use $x = 3\sin\theta$.

Trig sub neutralises $\sqrt{a^2 - x^2}$ via $1 - \sin^2\theta = \cos^2\theta$. Alternatively, quote the standard form $\arcsin(x/a)$ directly.

Fill the gap: For $\int \sqrt{a^2 + x^2}\, dx$ the standard trig substitution is $x =$ $\theta$, and for $\int \sqrt{a^2 - x^2}\, dx$ the substitution is $x =$ $\theta$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Defaulting to partial fractions for every rational

Not every rational integrand calls for partial fractions. If the numerator is the derivative of the denominator (e.g., $\int \frac{2x}{x^2 + 3}\, dx$), the log rule gives the answer immediately. If the denominator is irreducible (e.g., $x^2 + 4$), partial fractions over $\mathbb{R}$ are not available — use the $\arctan$ standard form.

Trap 02

Choosing the wrong $u$ in by-parts

For $\int x \ln x\, dx$, the choice $u = x$, $dv = \ln x\, dx$ requires you to already know $\int \ln x\, dx$. LIATE picks $u = \ln x$ — Logarithm beats Algebraic — so differentiation removes the log and leaves a clean power.

Trap 03

Reaching for $t$-substitution too soon

$t = \tan(x/2)$ is the heaviest tool in the kit. Reserve it for rational expressions in $\sin x, \cos x$ that resist everything else (e.g., $\int \frac{1}{1 + \sin x}\, dx$). Many trig integrals yield to simpler substitution first.

Did you get this? True or false: for $\displaystyle \int \frac{1}{x^2 + 4}\, dx$ the right move is partial fractions over $\mathbb{R}$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Identify the technique you would use for $\displaystyle \int x^2 \ln x\, dx$, then evaluate.

Identify the technique you would use for $\displaystyle \int \frac{1}{(x-2)(x+3)}\, dx$, then evaluate.

Identify the technique you would use for $\displaystyle \int x \sqrt{x^2 + 4}\, dx$, then evaluate.

Identify the technique you would use for $\displaystyle \int \frac{1}{\sqrt{1 - 4x^2}}\, dx$, then evaluate.

Identify the technique you would use for $\displaystyle \int \frac{1}{1 + \cos x}\, dx$, then evaluate.

Odd one out: Three of these integrals are most efficiently handled by integration by parts. Which one is NOT?

Revisit your thinking

Earlier you classified four integrals: $\int x e^x\, dx$, $\int \frac{2x+1}{x^2 + x + 1}\, dx$, $\int \sqrt{4 - x^2}\, dx$, $\int \frac{1}{x^2 - 1}\, dx$.

The answers: (a) by parts ($u = x$); (b) log rule, since the numerator is exactly the derivative of $x^2 + x + 1$; (c) trig sub $x = 2\sin\theta$; (d) partial fractions on $(x-1)(x+1)$. The shape of the integrand told you the technique. Becoming fast at this classification — within seconds, not minutes — is what separates a confident MEX-C1 candidate from one who guesses.

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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 $\displaystyle \int x \cos x\, dx$, stating the technique used. (2 marks)

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ApplyBand 43 marks

Q2. Evaluate $\displaystyle \int \frac{3x + 5}{(x + 1)(x - 2)}\, dx$, stating the technique used. (3 marks)

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AnalyseBand 53 marks

Q3. For each integral, state the most appropriate technique and apply it: (a) $\int \dfrac{x}{\sqrt{1 - x^2}}\, dx$; (b) $\int \dfrac{1}{4 + x^2}\, dx$; (c) $\int \dfrac{\sin x}{1 + \cos x}\, dx$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. By parts, $u = \ln x$, $dv = x^2\, dx$: $\int x^2 \ln x\, dx = \tfrac{x^3}{3}\ln x - \tfrac{x^3}{9} + C$.

2. Partial fractions: $\tfrac{1}{(x-2)(x+3)} = \tfrac{1/5}{x-2} - \tfrac{1/5}{x+3}$. Integral $= \tfrac{1}{5}\ln\!\left|\dfrac{x-2}{x+3}\right| + C$.

3. Substitution $u = x^2 + 4$, $du = 2x\, dx$: $\int x\sqrt{x^2 + 4}\, dx = \tfrac{1}{3}(x^2 + 4)^{3/2} + C$.

4. Standard form (or $u = 2x$): $\int \frac{1}{\sqrt{1 - 4x^2}}\, dx = \tfrac{1}{2}\arcsin(2x) + C$.

5. $t$-sub: $1 + \cos x = \tfrac{2}{1+t^2}$, $dx = \tfrac{2}{1+t^2}\, dt$. Integrand becomes $1$, so $\int 1\, dt = t + C = \tan(x/2) + C$.

Q1 (2 marks): By parts with $u = x$, $dv = \cos x\, dx$ $\Rightarrow$ $du = dx$, $v = \sin x$ [1]. $\int x \cos x\, dx = x \sin x - \int \sin x\, dx = x \sin x + \cos x + C$ [1].

Q2 (3 marks): Partial fractions: $\tfrac{3x + 5}{(x+1)(x-2)} = \tfrac{A}{x+1} + \tfrac{B}{x-2}$ [1]. Cover-up: at $x = -1$, $A = \tfrac{-3 + 5}{-3} = -\tfrac{2}{3}$; at $x = 2$, $B = \tfrac{6 + 5}{3} = \tfrac{11}{3}$ [1]. Integral $= -\tfrac{2}{3}\ln|x+1| + \tfrac{11}{3}\ln|x-2| + C$ [1].

Q3 (3 marks): (a) Substitution $u = 1 - x^2$, $du = -2x\, dx$: $\int \tfrac{x}{\sqrt{1 - x^2}}\, dx = -\sqrt{1 - x^2} + C$ [1]. (b) Standard form $\int \tfrac{1}{a^2 + x^2}\, dx = \tfrac{1}{a}\arctan(x/a)$: with $a = 2$, answer $= \tfrac{1}{2}\arctan(x/2) + C$ [1]. (c) $f'/f$ rule with $f(x) = 1 + \cos x$, $f'(x) = -\sin x$: $\int \tfrac{\sin x}{1 + \cos x}\, dx = -\ln|1 + \cos x| + C$ [1].

Boss battle · The Technique Selector

earn bronze · silver · gold

Five timed questions on choosing the right integration technique. 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 quick 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 2