Indefinite Integration

This lesson covers indefinite integration — the reverse process of differentiation — as required by the Edexcel A-Level Mathematics specification (9MA0). You need to be able to integrate standard functions, apply the power rule for integration, understand the constant of integration, and recognise standard results.

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Integration as the Reverse of Differentiation

Differentiation takes a function and produces its derivative (the rate of change). Integration is the reverse process — it takes a derivative and recovers the original function (up to a constant).

If dy/dx = f(x), then y = ∫ f(x) dx

The symbol ∫ is the integral sign, f(x) is the integrand, and dx tells us we are integrating with respect to x.

Why a Reverse Process?

When we differentiate y = x³, we get dy/dx = 3x². So if we are told that dy/dx = 3x², we can work backwards and say y = x³ + c, where c is a constant. This "working backwards" is integration.

Key Point: Integration and differentiation are inverse operations. If you differentiate a function and then integrate the result, you get back to the original function (plus a constant).

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The Power Rule for Integration

The most fundamental integration rule is the power rule. It reverses the power rule for differentiation.

If dy/dx = xⁿ (where n ≠ -1), then:

∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + c

In words: increase the power by 1, then divide by the new power, and add a constant of integration.

Worked Examples

Example 1: Find ∫ x⁴ dx

• Increase the power: 4 → 5
• Divide by the new power: x⁵ / 5
• Add the constant: x⁵ / 5 + c

Example 2: Find ∫ x⁻³ dx

• Increase the power: -3 → -2
• Divide by the new power: x⁻² / (-2) = -1/(2x²)
• Answer: -1/(2x²) + c

Example 3: Find ∫ √x dx

• Rewrite: √x = x^(1/2)
• Increase the power: 1/2 → 3/2
• Divide by the new power: x^(3/2) / (3/2) = (2/3)x^(3/2)
• Answer: (2/3)x^(3/2) + c

Example 4: Find ∫ 1/x² dx

• Rewrite: 1/x² = x⁻²
• Increase the power: -2 → -1
• Divide by the new power: x⁻¹ / (-1) = -x⁻¹ = -1/x
• Answer: -1/x + c

Exam Tip: Always rewrite roots and fractions as powers of x before integrating. For example, √x = x^(1/2), 1/x³ = x⁻³, and ³√x = x^(1/3).

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The Constant of Integration

When we differentiate y = x³ + 7, we get dy/dx = 3x². But we also get dy/dx = 3x² when we differentiate y = x³ + 100, or y = x³ - 42, or y = x³ + any constant.

This means that when we integrate 3x², we cannot determine the original constant. We write:

∫ 3x² dx = x³ + c

The constant of integration c represents all possible constant values. It is essential to include c in every indefinite integral.

When Can We Find c?

If we are given a boundary condition (a known point on the curve), we can find the specific value of c.

Example: Given dy/dx = 4x - 3 and the curve passes through (2, 5), find y.

• Step 1: Integrate — y = ∫ (4x - 3) dx = 2x² - 3x + c
• Step 2: Substitute (2, 5) — 5 = 2(4) - 6 + c → 5 = 8 - 6 + c → 5 = 2 + c → c = 3
• Step 3: y = 2x² - 3x + 3

Exam Tip: Forgetting the constant of integration is one of the most common mistakes. Always write + c for indefinite integrals. Only omit c when evaluating definite integrals (the constants cancel).

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Integrating Constant Multiples and Sums

Constant Multiple Rule

∫ k × f(x) dx = k × ∫ f(x) dx

You can take a constant factor outside the integral.

Example: ∫ 5x³ dx = 5 × ∫ x³ dx = 5 × (x⁴/4) + c = 5x⁴/4 + c

Sum/Difference Rule

∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx

You can integrate term by term.

Example: ∫ (3x² + 2x - 7) dx = x³ + x² - 7x + c

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Standard Results You Must Know

The following standard integrals appear frequently at A-Level:

Function f(x)	Integral ∫ f(x) dx
xⁿ (n ≠ -1)	xⁿ⁺¹ / (n + 1) + c
1/x (i.e. x⁻¹)	ln
eˣ	eˣ + c
e^(kx)	(1/k)e^(kx) + c
sin x	-cos x + c
cos x	sin x + c
sec²x	tan x + c
sin(kx)	-(1/k)cos(kx) + c
cos(kx)	(1/k)sin(kx) + c

Worked Examples of Standard Results

Example 1: ∫ 1/x dx = ln|x| + c

Note the modulus sign — we need |x| because ln is only defined for positive values.

Example 2: ∫ e^(3x) dx = (1/3)e^(3x) + c

Example 3: ∫ cos(2x) dx = (1/2)sin(2x) + c

Example 4: ∫ (3eˣ + 2/x - sin x) dx = 3eˣ + 2 ln|x| + cos x + c

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Integrating Expressions That Need Expanding

Sometimes you need to expand or simplify before integrating.

Example 1: Find ∫ x(x + 3) dx

• Expand: x² + 3x
• Integrate: x³/3 + 3x²/2 + c

Example 2: Find ∫ (2x + 1)² dx (when not using substitution)

• Expand: 4x² + 4x + 1
• Integrate: 4x³/3 + 2x² + x + c

Example 3: Find ∫ (x² + 3)/x dx

• Split: x²/x + 3/x = x + 3/x
• Integrate: x²/2 + 3 ln|x| + c

Exam Tip: You cannot integrate a product or a quotient term by term (i.e., ∫ f(x)g(x) dx ≠ ∫ f(x) dx × ∫ g(x) dx). You must expand or simplify first, or use a technique like substitution or parts.

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Finding the Equation of a Curve

A common exam question gives you a gradient function and a point on the curve, then asks for the equation.

Example: A curve has gradient function dy/dx = 6x² - 4x + 1 and passes through the point (1, 3). Find the equation of the curve.

• Step 1: Integrate — y = ∫ (6x² - 4x + 1) dx = 2x³ - 2x² + x + c
• Step 2: Substitute (1, 3) — 3 = 2(1) - 2(1) + 1 + c → 3 = 1 + c → c = 2
• Step 3: y = 2x³ - 2x² + x + 2

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Common Mistakes to Avoid

1. Forgetting + c — Every indefinite integral must include the constant of integration.
2. Wrong power rule application — Remember to add 1 to the power AND divide by the new power.
3. Trying to integrate products directly — ∫ x × sin x dx ≠ (x²/2)(−cos x). You need integration by parts.
4. Forgetting to rewrite — Always convert roots and fractions to index form before applying the power rule.
5. The case n = −1 — The power rule does not work when n = -1. Instead, ∫ x⁻¹ dx = ln|x| + c.

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Summary

• Integration is the reverse of differentiation — it recovers a function from its derivative.
• The power rule: ∫ xⁿ dx = xⁿ⁺¹/(n + 1) + c, provided n ≠ -1.
• Always include the constant of integration c for indefinite integrals.
• Use boundary conditions (known points) to find the value of c.
• Know the standard results for eˣ, 1/x, sin x, cos x, sec²x.
• Expand brackets and simplify fractions before integrating.
• You can integrate sums/differences term by term and take constant factors outside the integral.

A-Level Deep Dive: Indefinite Integration

Spec mapping

Edexcel 9MA0 specification section 10 — Integration covers the Fundamental Theorem of Calculus; integrate $x^n$xn (excluding $n = -1$n=−1) and related sums, differences and constant multiples; integrate $e^{kx}$ekx, $1/x$1/x, $\sin kx$sinkx, $\cos kx$coskx and related sums (refer to the official specification document for exact wording). Section 10 is examined across both 9MA0-01 (Pure Mathematics 1) and 9MA0-02 (Pure Mathematics 2), and integration techniques propagate into 9MA0-03 Mechanics (kinematics with variable acceleration) and 9MA0-03 Statistics (probability density functions in Year 2). The Edexcel formula booklet lists $\int \sec^2 kx\,dx$∫sec2kxdx and the standard exponential/logarithmic integrals, but the basic power rule and the trigonometric integrals $\int \sin x\,dx$∫sinxdx, $\int \cos x\,dx$∫cosxdx must be memorised.

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Worked example with full mark scheme

Question (8 marks): Find $\displaystyle\int \left(6x^2 - \dfrac{4}{x} + 3\sin x + 2e^x - \sec^2 x\right) dx$∫(6x2−x4​+3sinx+2ex−sec2x)dx, giving your answer in fully simplified form and stating any restriction on the domain.

Solution with mark scheme:

Step 1 — split the integral into standard pieces.

Integration is linear, so:

$\int \left(6x^2 - \dfrac{4}{x} + 3\sin x + 2e^x - \sec^2 x\right) dx = 6\int x^2\,dx - 4\int \dfrac{1}{x}\,dx + 3\int \sin x\,dx + 2\int e^x\,dx - \int \sec^2 x\,dx$∫(6x2−x4​+3sinx+2ex−sec2x)dx=6∫x2dx−4∫x1​dx+3∫sinxdx+2∫exdx−∫sec2xdx

M1 — splitting using linearity and pulling constants outside the integral. Candidates who try to integrate the whole sum without separating it tend to mishandle one or more terms.

Step 2 — apply the power rule to $6x^2$6x2.

$6 \int x^2\,dx = 6 \cdot \dfrac{x^3}{3} = 2x^3$6∫x2dx=6⋅3x3​=2x3

M1 — correct application of $\int x^n\,dx = \dfrac{x^{n+1}}{n+1}$∫xndx=n+1xn+1​ with $n = 2$n=2.

Step 3 — handle the $1/x$1/x term using the special case.

The power rule fails at $n = -1$n=−1 because the exponent denominator becomes zero. Instead use the standard result $\int \dfrac{1}{x}\,dx = \ln|x|$∫x1​dx=ln∣x∣:

$-4 \int \dfrac{1}{x}\,dx = -4 \ln|x|$−4∫x1​dx=−4ln∣x∣

M1 — recognising the $n = -1$n=−1 exception and using $\ln|x|$ln∣x∣ (not $\ln x$lnx). The modulus is essential because the original integrand $1/x$1/x is defined for both $x > 0$x>0 and $x < 0$x<0.

Step 4 — integrate the trig and exponential terms.

$3 \int \sin x\,dx = -3\cos x \qquad 2 \int e^x\,dx = 2e^x \qquad \int \sec^2 x\,dx = \tan x$3∫sinxdx=−3cosx2∫exdx=2ex∫sec2xdx=tanx

M1 — $\int \sin x\,dx = -\cos x$∫sinxdx=−cosx (note the sign — this is the classic slip). A1 — $\int e^x\,dx = e^x$∫exdx=ex and $\int \sec^2 x\,dx = \tan x$∫sec2xdx=tanx both correct.

Step 5 — combine and add the constant of integration.

$\int \left(6x^2 - \dfrac{4}{x} + 3\sin x + 2e^x - \sec^2 x\right) dx = 2x^3 - 4\ln|x| - 3\cos x + 2e^x - \tan x + C$∫(6x2−x4​+3sinx+2ex−sec2x)dx=2x3−4ln∣x∣−3cosx+2ex−tanx+C

A1 — fully assembled answer with all terms correct.

A1 — explicit $+ C$+C included.

A1 — domain restriction stated: $x \neq 0$x=0 (from the $1/x$1/x term) and $x \neq \tfrac{\pi}{2} + k\pi$x=2π​+kπ for integer $k$k (from $\sec^2 x$sec2x).

Total: 8 marks (M4 A4).

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Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): Given that $\dfrac{dy}{dx} = \dfrac{(x + 2)^2}{\sqrt{x}}$dxdy​=x​(x+2)2​ for $x > 0$x>0, and that $y = 5$y=5 when $x = 1$x=1, find $y$y in terms of $x$x.

Mark scheme decomposition by AO:

• M1 (AO1.1a) — expanding the numerator: $(x + 2)^2 = x^2 + 4x + 4$(x+2)2=x2+4x+4.
• M1 (AO1.1b) — splitting and rewriting in index form: $\dfrac{x^2 + 4x + 4}{x^{1/2}} = x^{3/2} + 4x^{1/2} + 4x^{-1/2}$x1/2x2+4x+4​=x3/2+4x1/2+4x−1/2.
• M1 (AO1.1b) — applying the power rule term by term to obtain $\tfrac{2}{5}x^{5/2} + \tfrac{8}{3}x^{3/2} + 8x^{1/2} + C$52​x5/2+38​x3/2+8x1/2+C.
• A1 (AO1.1b) — correct integrated expression including $+ C$+C.
• M1 (AO3.1a) — substituting $x = 1, y = 5$x=1,y=5 to determine $C$C: $5 = \tfrac{2}{5} + \tfrac{8}{3} + 8 + C$5=52​+38​+8+C, giving $C = 5 - \tfrac{2}{5} - \tfrac{8}{3} - 8 = -\tfrac{91}{15}$C=5−52​−38​−8=−1591​.
• A1 (AO2.5) — final answer $y = \tfrac{2}{5}x^{5/2} + \tfrac{8}{3}x^{3/2} + 8x^{1/2} - \tfrac{91}{15}$y=52​x5/2+38​x3/2+8x1/2−1591​ in fully simplified exact form.

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. This is the canonical Edexcel "find the particular solution" structure: rewrite, integrate, use the boundary condition to pin $C$C.

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Synoptic links

1. Section 9 — Differentiation: integration is the inverse process. Verifying $\int 3x^2\,dx = x^3 + C$∫3x2dx=x3+C is identical to checking that $\dfrac{d}{dx}(x^3 + C) = 3x^2$dxd​(x3+C)=3x2. Every integration result can — and should — be differentiated back to confirm.

2. Section 6 — Logarithms and exponentials: the special case $\int \dfrac{1}{x}\,dx = \ln|x|$∫x1​dx=ln∣x∣ is the defining property of the natural logarithm. The constant $e$e is precisely the number for which $\int_1^e \dfrac{1}{x}\,dx = 1$∫1e​x1​dx=1.

3. Section 5 — Trigonometry: the trig integrals $\int \sin x\,dx = -\cos x$∫sinxdx=−cosx and $\int \cos x\,dx = \sin x$∫cosxdx=sinx assume $x$x is in radians. In degree mode the chain rule introduces a $\pi/180$π/180 factor, which is why all A-Level calculus is done in radians.

4. Section 2 — Algebra and functions: integrating expressions like $(2x + 1)/\sqrt{x}$(2x+1)/x​ requires confident manipulation of fractional and negative indices. The integration is straightforward; the algebra is where marks are lost.

5. Mechanics 9MA0-03: if $a(t)$a(t) is acceleration, then velocity is $v(t) = \int a(t)\,dt$v(t)=∫a(t)dt and displacement is $s(t) = \int v(t)\,dt$s(t)=∫v(t)dt. The constants of integration are fixed by initial conditions $v(0)$v(0) and $s(0)$s(0) — exactly the boundary-condition technique above.

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Mark-scheme literacy

Indefinite integration questions on 9MA0 split AO marks toward AO1 with a strong AO2 thread:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	60–70%	Applying the power rule, recalling standard integrals, manipulating expressions into integrable form
AO2 (reasoning / interpretation)	20–30%	Recognising the $n = -1$n=−1 special case, using $\ln\lvert x\rvert$ln∣x∣ correctly, presenting the constant of integration, simplifying to requested form
AO3 (problem-solving)	5–15%	Determining the constant from a boundary condition; combining integration with mechanics or growth/decay contexts

Examiner-rewarded phrasing: "since this is an indefinite integral, we add $+ C$+C"; "the power rule fails at $n = -1$n=−1, so we use $\int \dfrac{1}{x}\,dx = \ln\lvert x\rvert$∫x1​dx=ln∣x∣"; "rewriting $\sqrt{x}$x​ as $x^{1/2}$x1/2 before integrating". Phrases that lose marks: omitting $+ C$+C entirely (an automatic loss of an A1); writing $\ln x$lnx without modulus signs when the domain admits negative $x$x; integrating $\sin x$sinx to $\cos x$cosx with the wrong sign.

A specific Edexcel pattern: when a question says "find the general solution" or "find an expression for $y$y", the constant $C$C must appear. When the question says "given that $y = a$y=a when $x = b$x=b, find $y$y", the constant must be evaluated — leaving $C$C in the final answer is then a losing answer.

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Grade-band model answers

3-mark question

Question: Find $\displaystyle\int (4x^3 - 6x + 5)\,dx$∫(4x3−6x+5)dx.

Grade C response (~150 words):

Integrate term by term using the power rule $\int x^n\,dx = \dfrac{x^{n+1}}{n+1}$∫xndx=n+1xn+1​:

$\int 4x^3\,dx = x^4$∫4x3dx=x4. $\int -6x\,dx = -3x^2$∫−6xdx=−3x2. $\int 5\,dx = 5x$∫5dx=5x.

So $\int (4x^3 - 6x + 5)\,dx = x^4 - 3x^2 + 5x + C$∫(4x3−6x+5)dx=x4−3x2+5x+C.

Examiner commentary: Full marks (3/3). Each term is integrated correctly, the coefficients are right, and the constant of integration is included. The work is brief but complete. This is the model Grade C response — procedural competence with no flourishes — and it secures every mark on offer.

Grade A response (~210 words):*

Integration is linear, so I split the sum:

$\int (4x^3 - 6x + 5)\,dx = 4\int x^3\,dx - 6\int x\,dx + 5\int 1\,dx$∫(4x3−6x+5)dx=4∫x3dx−6∫xdx+5∫1dx

Applying $\int x^n\,dx = \dfrac{x^{n+1}}{n+1}$∫xndx=n+1xn+1​ for $n \neq -1$n=−1:

$= 4 \cdot \dfrac{x^4}{4} - 6 \cdot \dfrac{x^2}{2} + 5x + C = x^4 - 3x^2 + 5x + C$=4⋅4x4​−6⋅2x2​+5x+C=x4−3x2+5x+C

The constant $C$C is required because differentiating any constant gives zero, so the antiderivative is determined only up to an additive constant.

Examiner commentary: Full marks (3/3). The candidate articulates why $+ C$+C is necessary rather than simply tacking it on, names linearity explicitly, and references the $n \neq -1$n=−1 restriction even though it is not directly invoked here. None of this is required for the marks, but the metacognitive habit is what carries through to harder questions where understanding why the rules are as they are determines whether AO2/AO3 marks can be earned.

6-mark question

Question: Find $\displaystyle\int \left(\dfrac{3}{x} + 4\cos x - 2e^x\right) dx$∫(x3​+4cosx−2ex)dx, stating any restrictions on the domain.

Grade B response (~220 words):

Integrate each term:

$\int \dfrac{3}{x}\,dx = 3\ln x$∫x3​dx=3lnx. $\int 4\cos x\,dx = 4\sin x$∫4cosxdx=4sinx. $\int -2e^x\,dx = -2e^x$∫−2exdx=−2ex.

So the answer is $3\ln x + 4\sin x - 2e^x + C$3lnx+4sinx−2ex+C, with $x > 0$x>0.

Examiner commentary: Method correct throughout, but $\ln x$lnx should be $\ln\lvert x\rvert$ln∣x∣ to handle both positive and negative $x$x. The candidate has restricted to $x > 0$x>0 unnecessarily — the original integrand $3/x$3/x is defined for all $x \neq 0$x=0, so the antiderivative is also valid there. Loses one accuracy mark for the modulus omission. Score: 5/6. The trig and exponential terms are correct; the constant of integration is present. The slip is exactly the kind of thing distinguishing Grade B from Grade A — the maths is right, the domain reasoning isn't fully sharp.

Grade A response (~245 words):*

Integration is linear; integrate term by term using standard results:

$\int \dfrac{3}{x}\,dx = 3\ln\lvert x\rvert$∫x3​dx=3ln∣x∣ $\int 4\cos x\,dx = 4\sin x$∫4cosxdx=4sinx $\int -2e^x\,dx = -2e^x$∫−2exdx=−2ex

Combining:

$\int \left(\dfrac{3}{x} + 4\cos x - 2e^x\right) dx = 3\ln\lvert x\rvert + 4\sin x - 2e^x + C$∫(x3​+4cosx−2ex)dx=3ln∣x∣+4sinx−2ex+C

The modulus is required on $\ln\lvert x\rvert$ln∣x∣ because $1/x$1/x is defined for all $x \neq 0$x=0. Domain restriction: $x \neq 0$x=0.

Examiner commentary: Full marks (6/6). The modulus on $\ln\lvert x\rvert$ln∣x∣ is correctly placed and explicitly justified. The domain restriction $x \neq 0$x=0 is stated, not silently assumed. The trig term $\int 4\cos x\,dx = 4\sin x$∫4cosxdx=4sinx is correct (no sign slip), and the exponential is unchanged by integration as expected. The candidate's discipline around domain reasoning is the marker of A* work — the same algebra, but with the meta-mathematical justification that the examiner is checking for.

9-mark question

Question: Given that $\dfrac{dy}{dx} = \dfrac{2x^3 - 5}{x^2}$dxdy​=x22x3−5​ for $x > 0$x>0, and $y = 4$y=4 when $x = 1$x=1, find $y$y as a function of $x$x.

Grade A response (~240 words):*

Rewrite the integrand by splitting and using negative indices:

$\dfrac{2x^3 - 5}{x^2} = 2x - 5x^{-2}$x22x3−5​=2x−5x−2

Integrate:

$y = \int (2x - 5x^{-2})\,dx = x^2 - 5 \cdot \dfrac{x^{-1}}{-1} + C = x^2 + \dfrac{5}{x} + C$y=∫(2x−5x−2)dx=x2−5⋅−1x−1​+C=x2+x5​+C

Apply the boundary condition $y = 4$y=4 when $x = 1$x=1:

$4 = 1 + 5 + C \implies C = -2$4=1+5+C⟹C=−2

So $y = x^2 + \dfrac{5}{x} - 2$y=x2+x5​−2.

Examiner commentary: Full marks (9/9). The candidate splits the fraction cleanly, converts to negative-index form before applying the power rule, handles the awkward $x^{-2}$x−2 integral correctly (the answer is $-x^{-1}$−x−1, not $x^{-1}$x−1 or $-x^{-3}/3$−x−3/3), includes $+ C$+C, and uses the boundary condition to pin $C = -2$C=−2 exactly. The final answer expresses $y$y as requested, with the $5/x$5/x term left in fractional form rather than $5x^{-1}$5x−1 — both are valid, but the fraction matches conventional presentation.

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A-Level-depth misconceptions

The errors that distinguish A from A* on indefinite integration:

1. Forgetting the constant of integration. Indefinite integrals are families of antiderivatives differing by a constant. Omitting $+ C$+C loses an A1 every time, no matter how complicated the rest of the integration is. It is the most common single mark loss on Section 10.

2. The $n = -1$n=−1 special case. $\int x^n\,dx = \dfrac{x^{n+1}}{n+1}$∫xndx=n+1xn+1​ fails at $n = -1$n=−1 because the denominator vanishes. The correct result is $\int x^{-1}\,dx = \ln\lvert x\rvert + C$∫x−1dx=ln∣x∣+C. Candidates who blindly apply the power rule at $n = -1$n=−1 produce $x^0/0$x0/0, which is undefined.

3. Sign error in $\int \sin x\,dx$∫sinxdx. $\dfrac{d}{dx}\cos x = -\sin x$dxd​cosx=−sinx, so $\int \sin x\,dx = -\cos x + C$∫sinxdx=−cosx+C. Many candidates write $+\cos x$+cosx. The check is to differentiate the answer: $\dfrac{d}{dx}(-\cos x) = \sin x$dxd​(−cosx)=sinx, confirming.

4. Modulus omission on $\ln\lvert x\rvert$ln∣x∣. Writing $\ln x$lnx silently restricts the domain to $x > 0$x>0. If the original $1/x$1/x is defined for negative $x$x too, the antiderivative must be $\ln\lvert x\rvert$ln∣x∣ to remain valid across both branches.

5. Treating integration as multiplication. $\int (f \cdot g)\,dx \neq \left(\int f\,dx\right)\left(\int g\,dx\right)$∫(f⋅g)dx=(∫fdx)(∫gdx). Integration is linear (it respects sums and constant multiples) but not multiplicative. Products require integration by parts (Year 2) or substitution.

6. Confusing $\int e^x\,dx$∫exdx with $\int e^{kx}\,dx$∫ekxdx. $\int e^x\,dx = e^x + C$∫exdx=ex+C, but $\int e^{kx}\,dx = \dfrac{e^{kx}}{k} + C$∫ekxdx=kekx​+C. The factor $1/k$1/k comes from reversing the chain rule and is the most-missed coefficient in Section 10.

7. Integrating $\sec^2 x$sec2x wrong. $\int \sec^2 x\,dx = \tan x + C$∫sec2xdx=tanx+C, recalled by recognising that $\dfrac{d}{dx}\tan x = \sec^2 x$dxd​tanx=sec2x. Candidates sometimes write $\sec x$secx or $\tfrac{1}{3}\sec^3 x$31​sec3x — both are wrong. The Edexcel formula booklet does list this integral, so checking it during the exam is sensible.

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Common errors and mark-loss patterns

• Constant absorption. When integrating expressions like $\int 3 \cdot e^x\,dx$∫3⋅exdx, candidates correctly write $3e^x$3ex but then absorb $+ C$+C into the $3$3, producing $3e^x$3ex alone. The constant of integration is additive, not multiplicative — it is always written separately as $+ C$+C.
• Mid-solution sign drift. When an integrand has both $+\sin x$+sinx and $-\cos x$−cosx terms, the integrated answer mixes $-\cos x$−cosx and $-\sin x$−sinx, and candidates often muddle which sign goes where. Cure: integrate each trig term separately, on its own line, then assemble.
• Premature simplification. Trying to simplify before integrating an awkward-looking expression like $\dfrac{x^2 + 1}{x}$xx2+1​ frequently produces an algebraic mistake. Better: split first as $x + \dfrac{1}{x}$x+x1​, then integrate to $\tfrac{1}{2}x^2 + \ln\lvert x\rvert + C$21​x2+ln∣x∣+C. The split reveals the $1/x$1/x term, which would otherwise be missed.
• Final-form mismatch. Questions asking for the answer "in the form $f(x) + C$f(x)+C" or "in fully simplified form" reject answers like $\dfrac{2x^4}{4}$42x4​ — that is not simplified. Reduce coefficients fully before submitting.

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Going further — university and academic signposting

• Riemann integration (Year 1 Analysis): the rigorous definition of $\int_a^b f(x)\,dx$∫ab​f(x)dx as the limit of upper and lower sums of partitions. The Fundamental Theorem of Calculus — that integration and differentiation are inverse operations — is proved here, not assumed.
• Lebesgue integration (Year 2/3): generalises Riemann's theory to functions that fail to be Riemann-integrable (e.g. the Dirichlet function, which is $1$1 on rationals and $0$0 on irrationals). Modern probability theory, Fourier analysis, and partial differential equations all rest on the Lebesgue integral.
• Antiderivatives in computer algebra systems: the Risch algorithm (1968) decides in finite steps whether an elementary function has an elementary antiderivative. It is the engine behind the symbolic integration in Mathematica, Maple and SageMath. Some functions — like $e^{-x^2}$e−x2, important in statistics — provably have no elementary antiderivative; they are integrated only numerically or via special functions like the error function $\mathrm{erf}(x)$erf(x).
• Differential forms (Year 3/4): integration generalises to higher dimensions through differential forms and Stokes's theorem, of which the Fundamental Theorem of Calculus is the one-dimensional case. This is the language of modern physics — Maxwell's equations in differential-form notation are four lines.

Oxbridge interview prompt: "Why is integration harder than differentiation? Can you give an example of a function whose antiderivative cannot be expressed in closed form, and explain why this matters for applied mathematics?"

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Additional A* practice: rewriting before integrating

A common A* trap on 9MA0 is a fraction that looks like it needs the quotient rule but is really just a sum of power-rule terms in disguise. The technique is to split the fraction first, then integrate term by term.

Worked example: Find $\displaystyle\int \dfrac{2x + 1}{\sqrt{x}}\,dx$∫x​2x+1​dx for $x > 0$x>0.

Rewrite the integrand by splitting and converting roots to fractional indices:

$\dfrac{2x + 1}{\sqrt{x}} = \dfrac{2x}{x^{1/2}} + \dfrac{1}{x^{1/2}} = 2x^{1/2} + x^{-1/2}$x​2x+1​=x1/22x​+x1/21​=2x1/2+x−1/2

(using $\dfrac{x^1}{x^{1/2}} = x^{1/2}$x1/2x1​=x1/2 and $\dfrac{1}{x^{1/2}} = x^{-1/2}$x1/21​=x−1/2).

Now integrate term by term:

$\int (2x^{1/2} + x^{-1/2})\,dx = 2 \cdot \dfrac{x^{3/2}}{3/2} + \dfrac{x^{1/2}}{1/2} + C = \dfrac{4}{3}x^{3/2} + 2x^{1/2} + C$∫(2x1/2+x−1/2)dx=2⋅3/2x3/2​+1/2x1/2​+C=34​x3/2+2x1/2+C

Optionally factor: $\tfrac{2}{3}\sqrt{x}(2x + 3) + C$32​x​(2x+3)+C.

Why A candidates spot this immediately:* the structure "polynomial divided by a single power of $x$x" is the signature of a hidden sum of power-rule terms. Whenever the denominator is a monomial ($x^k$xk or $\sqrt{x}$x​), split first. The same trick handles $\dfrac{x^3 - 4x + 7}{x^2}$x2x3−4x+7​ (split into $x - 4x^{-1} + 7x^{-2}$x−4x−1+7x−2, then integrate) and $\dfrac{(x + 1)^2}{x^{3/2}}$x3/2(x+1)2​ (expand the numerator first, then split). Recognising the pattern across these contexts is exactly the synoptic skill Edexcel rewards.

A subtlety: the power rule $\int x^n\,dx = \dfrac{x^{n+1}}{n+1}$∫xndx=n+1xn+1​ requires $n + 1 \neq 0$n+1=0, i.e. $n \neq -1$n=−1. After splitting, check that no term has exponent $-1$−1. If one does, treat it separately as $\int x^{-1}\,dx = \ln\lvert x\rvert + C$∫x−1dx=ln∣x∣+C.

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Edexcel A-Level alignment footer

This content is aligned with the Pearson Edexcel GCE A Level Mathematics (9MA0) specification, Paper 1 — Pure Mathematics, Section 10: Integration. For the most accurate and up-to-date information, please refer to the official Pearson Edexcel specification document.

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Visual summary

graph TD
    A["Integrand"] --> B{"Term type?"}
    B -->|"x^n, n ≠ -1"| C["Power rule:<br/>x^(n+1)/(n+1)"]
    B -->|"1/x"| D["ln|x|<br/>(special case)"]
    B -->|"e^x, sin x,<br/>cos x, sec²x"| E["Standard<br/>integrals"]
    C --> F["Add + C<br/>and verify by<br/>differentiating"]
    D --> F
    E --> F

    style D fill:#27ae60,color:#fff
    style F fill:#3498db,color:#fff