Integration Techniques

This lesson covers three key A-Level integration methods: integration by substitution, integration by parts, and integration using partial fractions.

Integration by Substitution

Substitution reverses the chain rule. Replace a complicated expression with a simpler variable u.

Method

Choose a substitution u = g(x).
Find du/dx and hence dx in terms of du.
Replace all x-terms with u-terms.
Integrate with respect to u.
Substitute back to x (for indefinite integrals).

Worked Example 1

Find ∫2x(x² + 1)⁴ dx.

Let u = x² + 1, so du = 2x dx.

∫2x(x² + 1)⁴ dx = ∫u⁴ du = u⁵/5 + c

= (x² + 1)⁵/5 + c

Worked Example 2

Find ∫x√(3x − 1) dx.

Let u = 3x − 1, so x = (u + 1)/3 and dx = du/3.

∫[(u + 1)/3] √u × du/3 = (1/9) ∫(u + 1)u^(1/2) du = (1/9) ∫(u^(3/2) + u^(1/2)) du

= (1/9)[2u^(5/2)/5 + 2u^(3/2)/3] + c

= 2(3x − 1)^(5/2)/45 + 2(3x − 1)^(3/2)/27 + c

Definite Integrals with Substitution

Change the limits as well: when x = a → u = g(a); when x = b → u = g(b).

Worked Example 3

Evaluate ∫₀² x(x² + 1)³ dx.

Let u = x² + 1, du = 2x dx → x dx = du/2.

When x = 0, u = 1; when x = 2, u = 5.

∫₁⁵ u³ × (1/2) du = (1/2)[u⁴/4]₁⁵ = (1/8)(625 − 1)

= 78

Integration by Parts

Integration by parts reverses the product rule:

∫u (dv/dx) dx = uv − ∫v (du/dx) dx

Choosing u and dv/dx

Use the LIATE priority: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential — let u be the function that appears earliest in this list.

Worked Example 4

Find ∫x eˣ dx.

Let u = x (algebraic), dv/dx = eˣ.
du/dx = 1, v = eˣ.

∫x eˣ dx = xeˣ − ∫eˣ dx = xeˣ − eˣ + c

= eˣ(x − 1) + c

Worked Example 5

Find ∫x² sin x dx.

Let u = x², dv/dx = sin x → du/dx = 2x, v = −cos x.

= −x² cos x + ∫2x cos x dx

Now apply by parts again to ∫2x cos x dx:
u = 2x, dv/dx = cos x → du/dx = 2, v = sin x.

= −x² cos x + [2x sin x − ∫2 sin x dx]
= −x² cos x + 2x sin x + 2 cos x + c

= (2 − x²) cos x + 2x sin x + c

Worked Example 6

Find ∫ln x dx.

Let u = ln x, dv/dx = 1 → du/dx = 1/x, v = x.

∫ln x dx = x ln x − ∫x × (1/x) dx = x ln x − ∫1 dx

= x ln x − x + c

Integration Using Partial Fractions

Decompose a rational function into partial fractions, then integrate each term separately.

Key Integral Results

Fraction	Integral
A/(x − a)	A ln
A/(x − a)²	−A/(x − a) + c

Worked Example 7

Find ∫(5x − 1)/[(x + 1)(x − 2)] dx.

Decompose: (5x − 1)/[(x + 1)(x − 2)] = A/(x + 1) + B/(x − 2)

5x − 1 = A(x − 2) + B(x + 1)

x = 2: 9 = 3B → B = 3
x = −1: −6 = −3A → A = 2

∫[2/(x + 1) + 3/(x − 2)] dx

= 2 ln|x + 1| + 3 ln|x − 2| + c

Worked Example 8

Find ∫(3x + 5)/[(x + 1)²] dx.

Decompose: (3x + 5)/(x + 1)² = A/(x + 1) + B/(x + 1)²

3x + 5 = A(x + 1) + B

x = −1: 2 = B
Compare x coefficients: 3 = A

∫[3/(x + 1) + 2/(x + 1)²] dx

= 3 ln|x + 1| − 2/(x + 1) + c

Exam Tips

For substitution, choose u to be the “awkward” part — often the expression inside a root or power.
Always change limits when using substitution with definite integrals.
For by parts, the LIATE rule guides which function to choose as u.
If by parts leads to the original integral, denote it as I and solve algebraically (e.g. ∫eˣ sin x dx).
Always decompose into partial fractions before attempting to integrate a rational function.
Check your integration by differentiating the result.

A-Level Deep Dive: Integration Techniques

Spec mapping

AQA 7357 Pure Mathematics specification, Section H — Integration (Year 2 content) covers using substitution; integrate by parts; integrate using partial fractions that are linear in the denominator; recognise standard forms ( $\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C$ , $\int \frac{1}{a^2 + x^2}\,dx$ via inverse trig where listed) (refer to the official specification document for exact wording). Section H sits at the apex of Pure: Year 1 integration (power rule, basic trig, exponentials) is the prerequisite, and the Year 2 techniques unlock differential equations (Section H sub-strand on first-order separable ODEs) and any modelling question that produces a non-elementary integrand. The AQA formula booklet provides the standard integrals $\int x^n\,dx$ , $\int e^{kx}\,dx$ , $\int \frac{1}{x}\,dx$ , basic trig integrals, and the integration-by-parts formula $\int u\,dv = uv - \int v\,du$ — but does not list reduction patterns or partial-fraction decompositions; those must be derived in the answer.

The technique is examined across all three AQA papers: Paper 1 (Pure 1) tests substitution and basic standard integrals; Paper 2 (Pure 2) is where the harder by-parts and partial-fraction work appears; Paper 3 (Mechanics + Statistics) uses integration to recover displacement from velocity ( $s = \int v\,dt$ ) and to compute expectations of continuous random variables ( $E(X) = \int x f(x)\,dx$ ).

Worked example with full mark scheme

Question (8 marks): Find $\int x^2 e^x \, dx$ .

Solution with mark scheme:

Step 1 — recognise the structure. The integrand is a polynomial times an exponential. Substitution will not collapse it (no inner function whose derivative appears as a factor). Partial fractions does not apply. The remaining tool is integration by parts, which trades a "harder" $\int u\,dv$ for an "easier" $uv - \int v\,du$ . Choose $u$ as the polynomial (so that differentiating it lowers the degree) and $dv$ as the exponential (so that integrating it leaves the form unchanged).

B1 — correct identification of integration by parts as the appropriate technique, with reasoning recorded.

Step 2 — first application of by-parts.

Let $u = x^2$ and $dv = e^x\,dx$ . Then $du = 2x\,dx$ and $v = e^x$ .

$\int x^2 e^x\,dx = x^2 e^x - \int 2x \cdot e^x \, dx = x^2 e^x - 2\int x e^x\,dx$

M1 — correct choice of $u$ and $dv$ with derivatives and antiderivative computed.

A1 — correct first-stage expression $x^2 e^x - 2\int x e^x\,dx$ .

Step 3 — second application of by-parts on the residual integral $\int x e^x\,dx$ .

Let $u = x$ and $dv = e^x\,dx$ . Then $du = dx$ and $v = e^x$ .

$\int x e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x + C_1$

M1 — correct second application of by-parts.

A1 — correct evaluation $x e^x - e^x$ (constant absorbed at the end).

Step 4 — combine.

$\int x^2 e^x\,dx = x^2 e^x - 2(x e^x - e^x) + C = x^2 e^x - 2x e^x + 2e^x + C$

M1 — correct substitution back into the first-stage expression.

A1 — correct factored or expanded form, e.g. $e^x(x^2 - 2x + 2) + C$ .

A1 — $+\,C$ explicitly included (an A1 mark on every indefinite integral; AQA examiners deduct it routinely when omitted).

Total: 8 marks (B1 M3 A4).

The factored form $e^x(x^2 - 2x + 2) + C$ is preferred because it can be verified instantly: differentiate using the product rule and confirm $x^2 e^x$ falls out. Examiners reward candidates who write "Check: $\frac{d}{dx}[e^x(x^2 - 2x + 2)] = e^x(x^2 - 2x + 2) + e^x(2x - 2) = x^2 e^x$ " because it demonstrates AO3 self-checking discipline.

Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks): Use the substitution $u = 1 + x^2$ to evaluate $\int_0^1 \dfrac{x}{(1 + x^2)^3}\,dx$ , giving your answer as an exact fraction.

Mark scheme decomposition by AO:

M1 (AO1.1a) — computing $\dfrac{du}{dx} = 2x$ so $x\,dx = \tfrac{1}{2}\,du$ .
M1 (AO1.1b) — transforming the integrand: $\dfrac{x\,dx}{(1 + x^2)^3} = \dfrac{1}{2u^3}\,du$ .
B1 (AO2.5) — transforming the limits: $x = 0 \Rightarrow u = 1$ , $x = 1 \Rightarrow u = 2$ .
M1 (AO1.1b) — antidifferentiating: $\int \tfrac{1}{2}u^{-3}\,du = -\tfrac{1}{4}u^{-2}$ .
A1 (AO1.1b) — correct evaluation at limits: $\left[-\tfrac{1}{4u^2}\right]_1^2 = -\tfrac{1}{16} + \tfrac{1}{4} = \tfrac{3}{16}$ .
A1 (AO2.5) — answer presented as the exact fraction $\dfrac{3}{16}$ (not the decimal 0.1875).

Total: 6 marks split AO1 = 4, AO2 = 2. The AO2 marks reward the limit transformation (a step that distinguishes substitution from a "u-only" symbolic exercise) and the exact-form discipline.

Synoptic links

Connects to:

Year 1 basic integration (Section H earlier sub-strand): the power rule $\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$ for $n \neq -1$ is the engine inside every substitution. Once a substitution reduces the integrand to $u^n$ form, Year 1 machinery completes the job.
Differentiation — chain rule (Section G): substitution is the chain rule run in reverse. Whenever you see $\int f(g(x)) g'(x)\,dx$ , substituting $u = g(x)$ converts it to $\int f(u)\,du$ . Recognising the inner function and its derivative as a factor in the integrand is the synoptic skill.
Differentiation — product rule (Section G): integration by parts is the product rule run in reverse. From $\frac{d}{dx}(uv) = u\,dv/dx + v\,du/dx$ , integrating both sides gives $uv = \int u\,dv + \int v\,du$ , hence $\int u\,dv = uv - \int v\,du$ . Without fluency in the product rule, by-parts is opaque procedure.
Algebra — partial fractions (Section B): integrating rational functions like $\dfrac{1}{(x-1)(x+2)}$ requires decomposing into $\dfrac{A}{x-1} + \dfrac{B}{x+2}$ first, then integrating each piece as $\ln$ . The algebraic decomposition is a Section B technique repurposed as a Section H tool.
First-order separable ODEs (Section H later sub-strand): equations like $\dfrac{dy}{dx} = \dfrac{y}{x(x+1)}$ separate into $\int \dfrac{1}{y}\,dy = \int \dfrac{1}{x(x+1)}\,dx$ , where the right-hand side demands partial fractions. Every separable ODE eventually invokes one of the four techniques in this Deep Dive.

Mark-scheme literacy

Integration-techniques questions on AQA 7357 distribute AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	55–70%	Choosing $u$ and $dv$ for by-parts, computing derivatives and antiderivatives, applying the substitution formula
AO2 (reasoning / interpretation)	20–35%	Justifying technique choice, transforming limits in definite integrals, recognising standard forms, presenting exact answers
AO3 (problem-solving)	5–15%	Multi-step problems combining techniques (e.g. substitution then by-parts), modelling integrals, ODE setup

Examiner-rewarded phrasing: "let $u = \dots$ so that $du/dx = \dots$ , hence $dx = \dots$ "; "transforming the limits: when $x = a$ , $u = \dots$ "; "applying integration by parts with $u = \dots$ and $dv = \dots$ chosen so the residual integral is simpler"; " $+\,C$ " on every indefinite integral. Phrases that lose marks: omitting the constant of integration, leaving $\int$ symbols in the final answer, mixing $x$ and $u$ variables in a single line, and presenting decimals where exact form is demanded.

A specific AQA pattern: when a question writes "use the substitution $u = \dots$ " the substitution is prescribed — using a different one (even if it works) loses the M1 for "correct method". When the question writes "find" without prescribing, any valid technique scores.

Grade-band model answers

3-mark question

Question: Find $\int (2x + 1)^5 \, dx$ .

Grade C response (~150 words):

Let $u = 2x + 1$ , so $\dfrac{du}{dx} = 2$ and $dx = \dfrac{du}{2}$ .

The integral becomes $\int u^5 \cdot \dfrac{1}{2}\,du = \dfrac{1}{2} \cdot \dfrac{u^6}{6} + C = \dfrac{u^6}{12} + C$ .

Substituting back: $\dfrac{(2x + 1)^6}{12} + C$ .

Examiner commentary: Full marks (3/3). Substitution is set up cleanly, the constant absorbed into the rescaling $dx = du/2$ , the back-substitution explicit, and $+\,C$ included. This is the standard procedural answer for a single-step substitution. Some candidates skip the substitution and apply the "reverse chain rule" mentally — that is also acceptable but riskier under time pressure because the divide-by-the-derivative-of-the-inner step is easy to forget.

Grade A response (~210 words):*

Integration Techniques

Integration Techniques

Integration by Substitution

Method

Worked Example 1

Worked Example 2

Definite Integrals with Substitution

Worked Example 3

Integration by Parts

Choosing u and dv/dx

Worked Example 4

Worked Example 5

Worked Example 6

Integration Using Partial Fractions

Key Integral Results

Worked Example 7

Worked Example 8

Exam Tips

A-Level Deep Dive: Integration Techniques

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 2 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

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