Integration Revision Notes — 9709 Paper 1

Reversing differentiation

Syllabus requirement · §1.8

“

understand integration as the reverse process of differentiation, and integrate (ax + b)ⁿ (for any rational n except −1), together with constant multiples, sums and differences

”

The question integration answers

Differentiation takes a curve and gives you its gradient function. Integration runs the arrow the other way: given the gradient, what was the curve?

This is exactly how the syllabus defines integration ("the reverse process of differentiation"), and it is why the rule you are about to meet looks like the power rule done backwards.

Running the arrow backwards also turns out to measure things that seem to have nothing to do with gradients: areas of curved regions, and volumes of solids. Those come later in this note. First, the rule itself.

Building the rule by guessing and checking

There is no need to be told the rule. You can recover it by asking a question you can already answer.

What differentiates to give $x^2$ ?

Try $x^3$ . Differentiating gives $3x^2$ — the right power, but three times too big. So divide the guess by $3$ :

\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{x^3}{3}\right) = \frac{3x^2}{3} = x^2 \ \checkmark

Try the same on $x^5$ . Guess $x^6$ ; differentiating gives $6x^5$ , six times too big; so divide by $6$ and $\dfrac{x^6}{6}$ works.

The pattern in both: the power went up by one, and you divided by the new power. That is the whole rule, and it is forced — differentiation multiplies by the power and drops it, so reversing must raise the power and divide.

Want to integrate	Guess	Differentiating the guess gives	Fix
$x^2$	$x^3$	$3x^2$	divide by $3$ → $\dfrac{x^3}{3}$
$x^5$	$x^6$	$6x^5$	divide by $6$ → $\dfrac{x^6}{6}$
$x^{-3}$	$x^{-2}$	$-2x^{-3}$	divide by $-2$ → $\dfrac{x^{-2}}{-2}$
$x^{n}$	$x^{n+1}$	$(n+1)x^{n}$	divide by $n+1$

The last row is the rule. Note the third: raising a negative power still means adding one, so −3 becomes −2.

The constant that integration cannot recover

There is one thing the backwards arrow can never know. Differentiate each of

x^3 + 3, \qquad x^3, \qquad x^3 - 2

and every one gives $3x^2$ . The constant term vanished, because a constant shifts a curve up or down without changing its gradient anywhere.

So when you integrate $3x^2$ you cannot tell which of those curves you started from. The honest answer names them all at once:

\int 3x^2\,\mathrm{d}x = x^3 + c

where $c$ stands for any constant, called the constant of integration. It is part of the answer: leaving it off states something false.

y = x³ + 3, y = x³ and y = x³ − 2 are the same shape shifted up or down. Each has gradient 3 at x = 1, so from the gradient alone you cannot tell them apart: that is what the + c records.

Reading the notation

The symbol $\int$ is a stretched-out S, for "sum"; the Area under a curve section below explains why.

\int \underbrace{3x^2}_{\text{what to integrate}} \ \underbrace{\mathrm{d}x}_{\text{with respect to } x} \ = \ x^3 + c

The expression being integrated is called the integrand. The $\mathrm{d}x$ is not decoration: it names the variable you are integrating with respect to. On Paper 1 you will nearly always see $\mathrm{d}x$ , but some areas and the volumes about the $y$ -axis at the end of this note use $\mathrm{d}y$ , where the variable is $y$ . The rule is the same in either letter.

An expression like $\int 3x^2\,\mathrm{d}x$ , with no numbers on the $\int$ sign, is an indefinite integral: its answer is a function, with $+\,c$ .

\int x^n\,\mathrm{d}x = \frac{x^{\,n+1}}{n+1} + c \qquad (n \neq -1)

The power rule, reversed

Add one to the power, divide by the new power, and add c. A constant multiple stays in front, and a sum or difference is integrated term by term, exactly as in differentiation.

Why n = −1 is excluded

Put $n = -1$ into the rule and you are dividing by $n + 1 = 0$ . So $\displaystyle\int \frac1x\,\mathrm{d}x$ cannot be found this way; it needs logarithms, which you meet in Paper 2 or Paper 3.

So on Paper 1, if an integral leaves you needing $\int x^{-1}\,\mathrm{d}x$ (or $\int (ax+b)^{-1}\,\mathrm{d}x$ ), you have made an algebra slip somewhere. Go back and check the rewrite: Paper 1 questions never need it.

Before you integrate

1
Rewrite every term as $kx^n$ or $k(ax+b)^n$ . $\frac{3}{x^2} = 3x^{-2}, \qquad \sqrt{x} = x^{1/2}, \qquad \frac{2x^3 - x}{x} = 2x^2 - 1$
Exactly the same first step as differentiation — you cannot integrate a fraction or a root as written.
2
Expand any brackets that are not of the form $(ax+b)^n$ . A product like $x(2x-1)$ must be multiplied out first.
Paper 1 has no method for integrating a product directly (Paper 3 has one, integration by parts). Here, expanding is the method.
3
Integrate term by term, adding one to each power and dividing by the new power.
4
Add $+\,c$ — every time, unless the integral is definite.
5
Check by differentiating your answer. It should give back exactly what you started with.
It takes seconds and catches almost every slip this section can produce: a wrong power, a missed division, a lost sign.

Four to start with

Find: (a) $\displaystyle\int \left(6x^2 - 4x + 5\right)\mathrm{d}x$ (b) $\displaystyle\int \left(\frac{3}{x^2} + 5\sqrt{x}\right)\mathrm{d}x$ (c) $\displaystyle\int \frac{2x^3 - x}{x}\,\mathrm{d}x$ (d) $\displaystyle\int x(2x-1)\,\mathrm{d}x$

Show full working

1
(a) Take the terms one at a time. For $6x^2$ : the power $2$ becomes $3$ , then divide by $3$ : $6x^2 \ \longrightarrow \ \frac{6x^3}{3} = 2x^3$
2
For $-4x$ : the power is $1$ , so it becomes $2$ , and divide by $2$ : $-4x \ \longrightarrow \ \frac{-4x^2}{2} = -2x^2$
3
For the constant $5$ : think of it as $5x^0$ . The power becomes $1$ , and dividing by $1$ changes nothing: $5 \ \longrightarrow \ 5x$
A constant always integrates to a linear term. Leaving the 5 as a 5 is a common slip — check it by differentiating: 5x gives 5. ✓
4
Assemble, and add the constant: $\int \left(6x^2 - 4x + 5\right)\mathrm{d}x = 2x^3 - 2x^2 + 5x + c$
5
(b) Rewrite both terms as powers first. $x^2$ on the bottom is a power $-2$ , and the root is a power $\tfrac12$ : $\frac{3}{x^2} + 5\sqrt{x} = 3x^{-2} + 5x^{1/2}$
6
First term: the power $-2$ becomes $-2 + 1 = -1$ , and divide by $-1$ : $3x^{-2} \ \longrightarrow \ \frac{3x^{-1}}{-1} = -3x^{-1} = -\frac{3}{x}$
Dividing by −1 flips the sign. Note the answer has power −1 — that is fine. It is only integrating x⁻¹ that is impossible, not producing it.
7
Second term: the power $\tfrac12$ becomes $\tfrac12 + 1 = \tfrac32$ , and divide by $\tfrac32$ : $5x^{1/2} \ \longrightarrow \ \frac{5x^{3/2}}{3/2}$
8
Dividing by $\tfrac32$ is multiplying by $\tfrac23$ : $\frac{5x^{3/2}}{3/2} = 5 \times \frac23 x^{3/2} = \frac{10}{3}x^{3/2}$
This flip is where fractional-power questions are won and lost. Write “÷ 3/2 means × 2/3” out rather than doing it mentally.
9
So $\int \left(\frac{3}{x^2} + 5\sqrt{x}\right)\mathrm{d}x = -\frac{3}{x} + \frac{10}{3}x^{3/2} + c$
10
(c) The denominator is the single term $x$ , so split the fraction and simplify before integrating: $\frac{2x^3 - x}{x} = \frac{2x^3}{x} - \frac{x}{x} = 2x^2 - 1$
Never integrate a quotient as it stands. Splitting works because the bottom is one term — it would NOT work for 1/(x+1).
11
Now integrate: $2x^2$ gives $\dfrac{2x^3}{3}$ , and $-1$ gives $-x$ : $\int \frac{2x^3 - x}{x}\,\mathrm{d}x = \frac{2}{3}x^3 - x + c$
12
(d) This is a product, and there is no product rule. Expand it: $x(2x-1) = 2x^2 - x$
13
Integrate term by term. $2x^2$ : the power $2$ becomes $3$ , divide by $3$ , giving $\dfrac{2x^3}{3}$ . Then $-x$ : the power $1$ becomes $2$ , divide by $2$ , giving $-\dfrac{x^2}{2}$ . $\int x(2x-1)\,\mathrm{d}x = \frac{2x^3}{3} - \frac{x^2}{2} + c$
14
Check (d) by differentiating: $\dfrac{2 \times 3x^2}{3} - \dfrac{2x}{2} = 2x^2 - x$ ✓ — which is what we expanded at the start.

Answer

(a) $2x^3 - 2x^2 + 5x + c$ (b) $-\dfrac{3}{x} + \dfrac{10}{3}x^{3/2} + c$ (c) $\dfrac{2}{3}x^3 - x + c$ (d) $\dfrac{2x^3}{3} - \dfrac{x^2}{2} + c$

Every one of these began with a rewrite, not an integration. Roots to fractional powers, denominators to negative powers, brackets expanded, quotients split — get that line down before you touch the rule.

Integrating $(ax+b)^n$

The syllabus asks for one more form: a linear expression raised to a power, such as $(2x+1)^5$ or $\dfrac{4}{(3x-2)^2}$ .

You could expand $(2x+1)^5$ — but there is a rule, and it is worth deriving rather than memorising, because the derivation tells you where its odd extra factor comes from.

Work forwards first. What happens if you differentiate $(ax+b)^{n+1}$ ? By the chain rule, you bring the power down, reduce it, and multiply by the derivative of the inside, which is $a$ :

\frac{\mathrm{d}}{\mathrm{d}x}\left(ax+b\right)^{n+1} = (n+1)\left(ax+b\right)^{n} \times a

So differentiating produced two unwanted factors: the $(n+1)$ , and the $a$ . To go backwards you must divide by both.

\int (ax+b)^n\,\mathrm{d}x = \frac{(ax+b)^{\,n+1}}{a\,(n+1)} + c \qquad (n \neq -1)

Raise the power, divide by the new power — and divide by a as well

The extra ÷a is the chain rule's factor being undone. Forgetting it is the standard error.

Only for a LINEAR inside

This rule works because the derivative of $ax+b$ is the constant $a$ , which can simply be divided out. It does not extend to anything else:

\int \left(x^2+1\right)^{5}\mathrm{d}x \ \neq \ \frac{\left(x^2+1\right)^{6}}{2x \times 6}

because $2x$ is not a constant and cannot be pulled through the integral. Paper 1 never asks you to integrate a non-linear bracket raised to a power — if you meet one, either it expands, or you have misread the question.

The (ax + b)ⁿ rule, on invented numbers

Find: (a) $\displaystyle\int (2x+1)^5\,\mathrm{d}x$ (b) $\displaystyle\int \frac{4}{(3x-2)^2}\,\mathrm{d}x$ (c) $\displaystyle\int 6\sqrt{4x+1}\,\mathrm{d}x$

Show full working

1
(a) Here $a = 2$ , $b = 1$ , $n = 5$ . Raise the power by one, to $6$ : $(2x+1)^6$
2
Divide by the new power $6$ , and by $a = 2$ : $\int (2x+1)^5\,\mathrm{d}x = \frac{(2x+1)^6}{2 \times 6} + c = \frac{(2x+1)^6}{12} + c$
Write the denominator as 2 × 6 before simplifying it to 12 — that way you can see both divisions happened.
3
Check by differentiating: $\dfrac{6(2x+1)^5 \times 2}{12} = (2x+1)^5$ ✓
4
(b) Rewrite with a negative power — the bracket goes upstairs: $\frac{4}{(3x-2)^2} = 4(3x-2)^{-2}$
5
Here $a = 3$ and $n = -2$ . The new power is $-2 + 1 = -1$ : $4(3x-2)^{-1}$
6
Divide by the new power $(-1)$ and by $a = 3$ : $\int 4(3x-2)^{-2}\,\mathrm{d}x = \frac{4(3x-2)^{-1}}{3 \times (-1)} + c = -\frac{4}{3}(3x-2)^{-1} + c$
The new power −1 is negative, so this division flips the sign. Both the 3 and the −1 must appear.
7
Tidy into fraction form: $= -\frac{4}{3(3x-2)} + c$
8
(c) Rewrite the root as a power $\tfrac12$ , keeping the $6$ in front: $6\sqrt{4x+1} = 6(4x+1)^{1/2}$
9
New power is $\tfrac12 + 1 = \tfrac32$ , and $a = 4$ : $\int 6(4x+1)^{1/2}\,\mathrm{d}x = \frac{6(4x+1)^{3/2}}{4 \times \tfrac32} + c$
10
Work the denominator out: $4 \times \tfrac32 = 6$ . $= \frac{6(4x+1)^{3/2}}{6} + c = (4x+1)^{3/2} + c$
The numbers cancelling completely is a hint the question was designed that way. If yours does not cancel, that is fine too — just leave it as a tidy fraction.

Answer

(a) $\dfrac{(2x+1)^6}{12} + c$ (b) $-\dfrac{4}{3(3x-2)} + c$ (c) $(4x+1)^{3/2} + c$

Say it as three moves: "power up one, divide by the new power, divide by $a$ ." Missing the third is the commonest error with this rule.

Common mistakes

Omitting the $+\,c$
$\displaystyle\int 3x^2\,\mathrm{d}x = x^3 + c$
The constant was destroyed by differentiating, so it must be restored. Without it, later parts asking for “the equation of the curve” cannot be done at all.
$\displaystyle\int (2x+1)^5\,\mathrm{d}x = \frac{(2x+1)^6}{6}$
$= \dfrac{(2x+1)^6}{12}$
The ÷a is missing. Differentiating the wrong answer gives 2(2x+1)⁵ — twice what you wanted.
$\displaystyle\int x^{-2}\,\mathrm{d}x = \frac{x^{-1}}{-2}$
$= \dfrac{x^{-1}}{-1} = -x^{-1}$
Divide by the NEW power, not the old one. Raising −2 by one gives −1.
Integrating $x(2x-1)$ as $\dfrac{x^2}{2} \times$ something
Expand to $2x^2 - x$ first
You cannot integrate the two factors separately and multiply. Every Paper 1 product either expands or is already in (ax+b)ⁿ form.
$\displaystyle\int \frac{1}{x}\,\mathrm{d}x = \frac{x^0}{0}$
Not integrable by this rule — recheck your algebra
n = −1 is excluded because the rule divides by zero. Paper 1 never requires it, so its appearance signals an earlier slip.

In the exam

54 parts · 233 marks · in 36 of 37 papers, 2021–2025

Integrating powers is almost never a question on its own. It is the first step inside nearly every constant-of-integration, area and volume question in the rest of this note, and mark schemes usually give a separate mark for each correctly integrated term.

What is tested is plain: rewrite into powers, integrate term by term, and handle $(ax+b)^n$ with its extra $\div a$ . The challenge is doing it accurately, several times, inside a longer question.

Your turn

Rewrite before integrating on every one, and differentiate your answer to check. Question 3 uses the (ax + b)ⁿ rule, with both of its divisions in play.

1
Find (a) $\displaystyle\int \left(4x^3 - 6x + 2\right)\mathrm{d}x$ (b) $\displaystyle\int \left(\frac{5}{x^3} - 2\sqrt{x}\right)\mathrm{d}x$
Stuck? Show hint
In (b), $\dfrac{5}{x^3} = 5x^{-3}$ and $2\sqrt{x} = 2x^{1/2}$ . Adding one to $-3$ gives $-2$ .
Show solution
1. 1
  (a) Term by term. $4x^3$ : power $3 \to 4$ , divide by $4$ : $\frac{4x^4}{4} = x^4$
2. 2
  $-6x$ : power $1 \to 2$ , divide by $2$ : $\frac{-6x^2}{2} = -3x^2$
3. 3
  $+2$ : a constant, so it becomes $2x$ .
4. 4
  $\int \left(4x^3 - 6x + 2\right)\mathrm{d}x = x^4 - 3x^2 + 2x + c$
5. 5
  (b) Rewrite: $\frac{5}{x^3} - 2\sqrt{x} = 5x^{-3} - 2x^{1/2}$
6. 6
  First term: power $-3 \to -2$ , divide by $-2$ : $\frac{5x^{-2}}{-2} = -\frac52 x^{-2} = -\frac{5}{2x^2}$
  Dividing by the negative new power flips the sign.
7. 7
  Second term: power $\tfrac12 \to \tfrac32$ , divide by $\tfrac32$ (that is, multiply by $\tfrac23$ ): $\frac{-2x^{3/2}}{3/2} = -2 \times \frac23 x^{3/2} = -\frac43 x^{3/2}$
8. 8
  $\int \left(\frac{5}{x^3} - 2\sqrt{x}\right)\mathrm{d}x = -\frac{5}{2x^2} - \frac43 x^{3/2} + c$
Answer
(a) $x^4 - 3x^2 + 2x + c$ (b) $-\dfrac{5}{2x^2} - \dfrac43 x^{3/2} + c$
2
Find $\displaystyle\int \frac{x^3 - 4x^2}{2x}\,\mathrm{d}x$ .
Stuck? Show hint
The denominator is a single term, so split the fraction and simplify each piece before integrating.
Show solution
1. 1
  Split the fraction over the single denominator: $\frac{x^3 - 4x^2}{2x} = \frac{x^3}{2x} - \frac{4x^2}{2x}$
2. 2
  Simplify each piece with the index laws: $\frac{x^3}{2x} = \frac12 x^{2}, \qquad \frac{4x^2}{2x} = 2x$
3. 3
  So the integrand is $\tfrac12 x^2 - 2x$ .
  Now it is a sum of powers. Attempting the original quotient directly has no available method.
4. 4
  Integrate the first term: power $2 \to 3$ , divide by $3$ : $\frac{\tfrac12 x^3}{3} = \frac{x^3}{6}$
5. 5
  Second term: power $1 \to 2$ , divide by $2$ : $\frac{-2x^2}{2} = -x^2$
6. 6
  $\int \frac{x^3 - 4x^2}{2x}\,\mathrm{d}x = \frac{x^3}{6} - x^2 + c$
7. 7
  Check: differentiating gives $\dfrac{3x^2}{6} - 2x = \dfrac{x^2}{2} - 2x$ ✓
Answer
$\dfrac{x^3}{6} - x^2 + c$
3
Find (a) $\displaystyle\int 12(2x-5)^2\,\mathrm{d}x$ (b) $\displaystyle\int \frac{9}{(5x+4)^{1/2}}\,\mathrm{d}x$
Stuck? Show hint
In (b), write it as $9(5x+4)^{-1/2}$ . The new power is $-\tfrac12 + 1 = \tfrac12$ .
Show solution
1. 1
  (a) Here $a = 2$ , $n = 2$ . Raise the power to $3$ : $12(2x-5)^3$
2. 2
  Divide by the new power $3$ and by $a = 2$ : $\int 12(2x-5)^2\,\mathrm{d}x = \frac{12(2x-5)^3}{3 \times 2} + c$
3. 3
  $3 \times 2 = 6$ , and $\dfrac{12}{6} = 2$ : $= 2(2x-5)^3 + c$
  Check by differentiating: 2 × 3(2x − 5)² × 2 = 12(2x − 5)² ✓
4. 4
  (b) Rewrite with a negative power: $\frac{9}{(5x+4)^{1/2}} = 9(5x+4)^{-1/2}$
5. 5
  New power: $-\tfrac12 + 1 = \tfrac12$ . Here $a = 5$ : $\frac{9(5x+4)^{1/2}}{5 \times \tfrac12} + c$
6. 6
  Work out the denominator: $5 \times \tfrac12 = \tfrac52$ . Dividing by $\tfrac52$ means multiplying by $\tfrac25$ : $= 9 \times \frac25 (5x+4)^{1/2} + c = \frac{18}{5}(5x+4)^{1/2} + c$
  The new power is positive here (+1/2), so there is no sign change this time. You will meet this exact integral again in an area question later in the note.
Answer
(a) $2(2x-5)^3 + c$ (b) $\dfrac{18}{5}(5x+4)^{1/2} + c$

Practise integratingReal past-paper questions · Integration as reverse differentiation; integrating (ax + b)^n

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Can you do all of these?

Explain, using $x^3+3$ and $x^3-2$ , why $\int 3x^2\,\mathrm{d}x$ needs a $+\,c$
Integrate $\dfrac{3}{x^2}$ , $5\sqrt{x}$ , $x(2x-1)$ and $(3x-1)^4$ , remembering the $+\,c$
Find the equation of a curve from $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ and one point on it, including when it is given as $\mathrm{f}'(x)$ and $\mathrm{f}(a)$
Find a curve from $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$ and a stationary point, using $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$ for the first constant and the point for the second
Evaluate $\displaystyle\int_1^4 \left(2x - x^{-1/2}\right)\mathrm{d}x$ , writing the square-bracket line first
Evaluate $\displaystyle\int_1^{\infty} x^{-2}\,\mathrm{d}x$ and $\displaystyle\int_0^1 x^{-1/2}\,\mathrm{d}x$ , saying what happens at the awkward end
Find the limits for an area question by solving two equations simultaneously
Find the area between a curve and a line as one integral of the difference, splitting the region if the top boundary changes
Explain why a region below the $x$ -axis gives a negative integral, and how to report its area
Find an area against the $y$ -axis with $\int x\,\mathrm{d}y$ , using $y$ -values as limits
Explain why $\int y^2\,\mathrm{d}x$ is not $\left(\int y\,\mathrm{d}x\right)^2$
Rearrange a curve for $x$ and find a volume of revolution about the $y$ -axis
Find the volume of a hollow solid for a region bounded below by $y = k$ rather than the axis