Learn9709Paper 1 · Pure Mathematics 1IntegrationIntegrating (ax + b)ⁿ

Integration · Subtopic 2 of 5

Integrating (ax + b)ⁿ

How to integrate a power of a linear bracket in one step by reversing the chain rule: raise the power, divide by the new power, and divide by the coefficient of $x$ .

Cambridge 9709 Paper 1Edexcel YMA01 WMA13Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 35 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Integrate $(ax+b)^n$ for any rational $n\neq-1$ .
Write roots and reciprocals of brackets as powers before integrating.
Check an answer by differentiating with the chain rule.
Find the equation of a curve whose gradient involves $(ax+b)^n$ .

Pause and think

Differentiate $(3x+1)^5$ . Then use your answer to write down $\displaystyle\int(3x+1)^4\,dx$ .

Section 2 of 14

Core idea

Start hereUndoing the chain rule

Differentiating $(3x-1)^5$ with the chain rule gave $5(3x-1)^4\times3$ . So to integrate $(3x-1)^4$ we must undo both the power step and that extra $\times3$ .

The big idea

For a linear inside $ax+b$ : $\displaystyle\int(ax+b)^n\,dx=\frac{(ax+b)^{n+1}}{a(n+1)}+c$ ( $n\neq-1$ ). Integrate as usual, then divide by $a$ .

1. Derivation

Differentiate and integrate are opposite journeys

Each multiplication on the way there is a division on the way back.

The rule

$\frac{d}{dx}(ax+b)^{n+1}=(n+1)(ax+b)^n\cdot a$
Chain rule.
$\frac{d}{dx}\frac{(ax+b)^{n+1}}{a(n+1)}=(ax+b)^n$
Divide by $a(n+1)$ .
$\int(ax+b)^n\,dx=\frac{(ax+b)^{n+1}}{a(n+1)}+c$
Read it backwards.

2. Examples

Worked example $\displaystyle\int(3x-1)^4\,dx$

Integrate.

Goal: Antiderivative.
Strategy: $a=3$ , $n=4$ .

$\frac{(3x-1)^5}{5}$
Raise the power, divide by it.
$\div3:\ \frac{(3x-1)^5}{15}+c$
Divide by $a$ .

Answer

\dfrac{(3x-1)^5}{15}+c

✓ Check

\frac{5(3x-1)^4\cdot3}{15}=(3x-1)^4

✓.

Worked example $\displaystyle\int\frac{6}{\sqrt{2x+5}}\,dx$

Integrate.

Goal: Antiderivative.
Strategy: Power form; $a=2$ , $n=-\frac12$ .

$6(2x+5)^{-1/2}$
Rewrite.
$6\cdot\frac{(2x+5)^{1/2}}{\tfrac12\cdot2}=6\sqrt{2x+5}+c$
$a(n+1)=2\cdot\frac12=1$ .

Answer

6\sqrt{2x+5}+c

✓ Check

\frac{d}{dx}6(2x+5)^{1/2}=6\cdot\frac12(2x+5)^{-1/2}\cdot2=\frac6{\sqrt{2x+5}}

✓.

Worked example $\displaystyle\int_0^4\sqrt{2x+1}\,dx$

Evaluate.

Goal: A number.
Strategy: Integrate, then top minus bottom.

$\left[\frac{(2x+1)^{3/2}}{2\cdot\frac32}\right]_0^4=\left[\frac{(2x+1)^{3/2}}3\right]_0^4$
Rule with $a=2$ , $n=\frac12$ .
$=\frac{9^{3/2}-1^{3/2}}{3}=\frac{27-1}3=\frac{26}3$
$9^{3/2}=27$ .

Answer

\frac{26}3

✓ CheckIntegrand between

1

and

3

over width

4

: area between

4

and

12

;

8.67

✓.

Common misconceptionOnly for linear insides

\int(x^2+1)^3dx\neq\frac{(x^2+1)^4}{4\cdot2x}

. Dividing by an

x

-expression doesn't undo the chain rule; expand instead.

Pause and think

$\int\frac{2}{(1-4x)^3}dx$ ?

Summary

Divide by the new power and by $a$ .
Linear insides only.

Linear expression: $ax+b$ .

HistoryCalculus for hire. In 1691 Johann Bernoulli taught calculus to the French Marquis de l'Hôpital, and later agreed to send him new discoveries in exchange for a salary. L'Hôpital published the first calculus textbook in 1696, including a rule Bernoulli had found. It is still called l'Hôpital's rule.

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