Learn7357Paper 1 · PureIntegrationIndefinite integration

Integration · Subtopic 2 of 13

Indefinite Integration

Integration as differentiation in reverse: integrating powers of $x$ , why there is always a $+c$ , and how one point pins down the exact curve.

Cambridge 9709 Paper 1Edexcel YMA01 WMA11Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 40 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Integrate $ax^n$ for rational $n\neq-1$ , and sums of such terms.
Explain the constant of integration $+c$ .
Rewrite roots, products and fractions as powers of $x$ before integrating.
Find the equation of a curve from its gradient function and one point.

Pause and think

Which functions have derivative $2x$ ? Find at least three.

Section 2 of 14

Core idea

Start hereReading the speedometer backwards

Differentiation turns position into speed. Suppose instead you only have the speedometer readings, say $v=2t$ , and want to know where you are. You need to undo differentiation: which function has derivative $2t$ ? $t^2$ does, but so do $t^2+5$ and $t^2-3$ . Undoing differentiation is called integration, and that "plus anything" is its signature.

The big idea

$\displaystyle\int x^n\,dx=\frac{x^{n+1}}{n+1}+c$ for $n\neq-1$ . The $+c$ (constant of integration) is needed because constants vanish when differentiated. A known point fixes $c$ .

1. Reversing the power rule

Finding the rule

$\frac{d}{dx}x^{n+1}=(n+1)x^n$
Power rule.
$\frac{d}{dx}\frac{x^{n+1}}{n+1}=x^n$
Divide both sides by $n+1$ (needs $n\neq-1$ ).
$\int x^n\,dx=\frac{x^{n+1}}{n+1}+c$
So that's an antiderivative of $x^n$ ; add $c$ .

"Add one to the power, divide by the new power."

\int f(x)\,dx

$\int$: the integral sign (a long S, for "sum").
$dx$: says the variable is $x$ .
$c$: the constant of integration.

2. Why +c

Every curve y = x² + c has gradient 2x

c = -3.0

Sliding the curve up or down doesn't change any gradients. That's why integration can only find

y

up to an unknown constant, until you are told a point on the curve.

3. Examples

Worked example $\displaystyle\int\left(6x^2-\frac4{x^3}+\sqrt x\right)dx$

Integrate.

Goal: An antiderivative + $c$ .
Strategy: Powers first, then the rule term by term.

$\int\left(6x^2-4x^{-3}+x^{1/2}\right)dx$
Rewrite.
$=\frac{6x^3}{3}-\frac{4x^{-2}}{-2}+\frac{x^{3/2}}{3/2}+c$
Add one; divide.
$=2x^3+\frac2{x^2}+\tfrac23x^{3/2}+c$
Tidy: dividing by $\frac32$ is multiplying by $\frac23$ .

Answer

2x^3+\dfrac2{x^2}+\frac23x^{3/2}+c

✓ CheckDifferentiate back:

6x^2-4x^{-3}+x^{1/2}

✓.

Worked exampleFinding the curve through a point

$\frac{dy}{dx}=3x^2-4$ and the curve passes through $(2,5)$ . Find $y$ .

Goal: $y$ as a function of $x$ .
Strategy: Integrate; use the point for $c$ .

$y=x^3-4x+c$
Integrate.
$5=8-8+c\Rightarrow c=5$
Substitute $(2,5)$ .

Answer

y=x^3-4x+5

✓ Check

y(2)=8-8+5=5

✓; derivative

3x^2-4

✓.

Common misconceptionIntegrating products term by term

\int(x+2)^2dx

is not

\frac{(x+2)^3}{3}\cdot

(anything) by guesswork; expand:

x^2+4x+4\to\frac{x^3}3+2x^2+4x+c

Edge case $n=-1$

\int x^{-1}dx

would need dividing by

0

. Its integral is

\ln|x|+c

, met later.

Pause and think

$f'(x)=\frac1{\sqrt x}$ , $f(9)=10$ . Find $f(x)$ .

Summary

Add one to the power, divide by it.
Always $+c$ .
Check by differentiating.

Integration: Reverse of differentiation.
Constant of integration: The $+c$ .

HistoryThe long S. Leibniz introduced the symbol

\int

in 1675. It is an elongated S, for the Latin "summa" (sum), because he thought of an integral as adding up infinitely many thin strips, which is exactly how areas are found two lessons from now.

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