Mathematics 9709 · Pure Mathematics 1

Integrating (ax + b)ⁿ

When the thing raised to a power is a linear bracket like $(2x-3)$ , you integrate the whole bracket. You do not expand it.
It is the normal power rule plus one extra step. Most students lose marks on that step.

Add one to the power. Divide by the new power. Then divide by $a$ (the number in front of $x$ ): $\int(ax+b)^n\,dx=\dfrac{(ax+b)^{n+1}}{a(n+1)}+c$ .

The rule, and the extra divide by a

It reads just like the basic power rule (§8.1): add one to the power and divide by the new power.
The one difference: you also divide by $a$ , the number in front of $x$ :

$\int(ax+b)^n\,dx=\dfrac{(ax+b)^{n+1}}{a(n+1)}+c\qquad(n\neq-1)$

The bracket stays the same. The gold step (divide by a) is the one people forget.

This only works when the inside is linear. For something like $(x^2+1)^6$ the rule does not work. You cannot fix it by dividing by a number.

Why does that extra a appear?

Check it backwards. Differentiate $(ax+b)^{n+1}$ . The chain rule gives an extra factor of $a$ . (The derivative of the inside, $ax+b$ , is $a$ .) So when we integrate, we divide by $a$ first to cancel it. Integration is the chain rule done backwards.

This also shows why a non-linear inside does not work. For $(x^2+1)^6$ the inside differentiates to $2x$ , not a number. So no single number undoes it. Dividing by a number only fixes a linear inside.

Roots and fractions: rewrite as a power first

The rule needs a power. So turn any root or fraction into a power first, then use the rule.
You never expand the bracket. Four steps, in order:

Rewrite as a power: a root becomes a fractional index, a fraction $\tfrac{1}{(\,\cdot\,)^k}$ becomes a negative one.
Add one to the power, then divide by that new power — the normal rule.
Divide by $a$ , the number in front of $x$ inside the bracket. This is the extra step.
Add $+c$ (or put in the limits, for a definite integral).

Worked example

Find $\int(2x-3)^4\,dx$

Add one to the power (it is now $5$ ), divide by that $5$ , and divide by $a=2$ :

$\int(2x-3)^4\,dx=\dfrac{(2x-3)^5}{2(5)}+c$

Answer

\tfrac{1}{10}(2x-3)^5+c

Worked example

Find $\int \dfrac{5}{\sqrt{2x+7}}\,dx$

Rewrite the root as a power: $\dfrac{5}{\sqrt{2x+7}}=5(2x+7)^{-\frac12}$ . The new power is $\tfrac12$ , and $a=2$ :

$=\dfrac{5(2x+7)^{\frac12}}{2\left(\frac12\right)}+c=5(2x+7)^{\frac12}+c$

Answer

5\sqrt{2x+7}+c

Where the marks go

Common mistake

Do not forget to divide by $a$ . This is the mark most students lose in this topic. The answer looks right, but it is too big or too small by a number.

Common mistake

When

a

is negative and the power is negative, be careful with the signs. For

(1-4x)^{-6}

you have

a=-4

and

n+1=-5

. So you divide by

(-4)(-5)=20

, which is positive. You get

\dfrac{(1-4x)^{-5}}{20}+c

. (9709-style)

“Differentiate this, hence find that”

For a non-linear bracket like $(x^2+2)^4$ there is no rule to integrate it directly.
So the exam helps you. A two-part question first asks you to differentiate some $F(x)$ . Then it says “hence find” an integral.
Hence tells you what to do: take your part-(a) answer and run it backwards. That is the integral you need.

$\text{if }\dfrac{d}{dx}\big[F(x)\big]=f(x),\text{ then }\int f(x)\,dx=F(x)+c$

Worked example

(a) Differentiate $(x^2+2)^4$ . (b) Hence find $\int x(x^2+2)^3\,dx$ .

(a) Chain rule: bring the power down, drop it by one, then times the derivative of the inside $(2x)$ :

$\dfrac{d}{dx}(x^2+2)^4=4(x^2+2)^3\cdot 2x=8x(x^2+2)^3$

(b) That answer is $8$ times the thing we want to integrate. Integrating undoes the differentiating, so we just divide by $8$ :

$\int 8x(x^2+2)^3\,dx=(x^2+2)^4+c\;\Rightarrow\;\int x(x^2+2)^3\,dx=\tfrac18(x^2+2)^4+c$

Answer

\tfrac18(x^2+2)^4+c

Common mistake

When the thing you integrate is a number times your part-(a) answer, you must match them. Here part (a) gave

8x(x^2+2)^3

but the integral has

x(x^2+2)^3

. So divide by

8

. Do not forget the

+c

. (9709-style)

Now you try

A curve passes through $\left(2,5\tfrac{2}{3}\right)$ and has $\tfrac{dy}{dx}=\dfrac{8}{(3x+2)^2}$ . Find the equation of the curve. (9709/12 Oct/Nov 2021 Q4)

Your turn— tap to reveal the worked answer (9709/12 Oct/Nov 2021 Q4)

Rewrite as a power: $\tfrac{dy}{dx}=8(3x+2)^{-2}$ .

Integrate. New power is $-1$ , and divide by $a=3$ : $y=\dfrac{8(3x+2)^{-1}}{3(-1)}+c=-\dfrac{8}{3(3x+2)}+c$ .

Use the point. At $x=2$ , $3x+2=8$ , so the term is $-\tfrac{8}{24}=-\tfrac13$ . Then $5\tfrac23=-\tfrac13+c\Rightarrow c=6$ .

Answer

y=6-\dfrac{8}{3(3x+2)}

PreviousReversing differentiation & the constant NextDefinite integrals

Integrating (ax + b)ⁿ

When the thing raised to a power is a linear bracket like $(2x-3)$ , you integrate the whole bracket. You do not expand it.
It is the normal power rule plus one extra step. Most students lose marks on that step.

Add one to the power. Divide by the new power. Then divide by $a$ (the number in front of $x$ ): $\int(ax+b)^n\,dx=\dfrac{(ax+b)^{n+1}}{a(n+1)}+c$ .

The rule, and the extra divide by a

It reads just like the basic power rule (§8.1): add one to the power and divide by the new power.
The one difference: you also divide by $a$ , the number in front of $x$ :

$\int(ax+b)^n\,dx=\dfrac{(ax+b)^{n+1}}{a(n+1)}+c\qquad(n\neq-1)$

The bracket stays the same. The gold step (divide by a) is the one people forget.

This only works when the inside is linear. For something like $(x^2+1)^6$ the rule does not work. You cannot fix it by dividing by a number.

Why does that extra a appear?

Roots and fractions: rewrite as a power first

The rule needs a power. So turn any root or fraction into a power first, then use the rule.
You never expand the bracket. Four steps, in order:

Rewrite as a power: a root becomes a fractional index, a fraction $\tfrac{1}{(\,\cdot\,)^k}$ becomes a negative one.
Add one to the power, then divide by that new power — the normal rule.
Divide by $a$ , the number in front of $x$ inside the bracket. This is the extra step.
Add $+c$ (or put in the limits, for a definite integral).

Worked example

Find $\int(2x-3)^4\,dx$

Add one to the power (it is now $5$ ), divide by that $5$ , and divide by $a=2$ :

$\int(2x-3)^4\,dx=\dfrac{(2x-3)^5}{2(5)}+c$

Answer

\tfrac{1}{10}(2x-3)^5+c

Worked example

Find $\int \dfrac{5}{\sqrt{2x+7}}\,dx$

Rewrite the root as a power: $\dfrac{5}{\sqrt{2x+7}}=5(2x+7)^{-\frac12}$ . The new power is $\tfrac12$ , and $a=2$ :

$=\dfrac{5(2x+7)^{\frac12}}{2\left(\frac12\right)}+c=5(2x+7)^{\frac12}+c$

Answer

5\sqrt{2x+7}+c

Where the marks go

Common mistake

Do not forget to divide by $a$ . This is the mark most students lose in this topic. The answer looks right, but it is too big or too small by a number.

Common mistake

When

a

is negative and the power is negative, be careful with the signs. For

(1-4x)^{-6}

you have

a=-4

and

n+1=-5

. So you divide by

(-4)(-5)=20

, which is positive. You get

\dfrac{(1-4x)^{-5}}{20}+c

. (9709-style)

“Differentiate this, hence find that”

For a non-linear bracket like $(x^2+2)^4$ there is no rule to integrate it directly.
So the exam helps you. A two-part question first asks you to differentiate some $F(x)$ . Then it says “hence find” an integral.
Hence tells you what to do: take your part-(a) answer and run it backwards. That is the integral you need.

$\text{if }\dfrac{d}{dx}\big[F(x)\big]=f(x),\text{ then }\int f(x)\,dx=F(x)+c$

Worked example

(a) Differentiate $(x^2+2)^4$ . (b) Hence find $\int x(x^2+2)^3\,dx$ .

(a) Chain rule: bring the power down, drop it by one, then times the derivative of the inside $(2x)$ :

$\dfrac{d}{dx}(x^2+2)^4=4(x^2+2)^3\cdot 2x=8x(x^2+2)^3$

(b) That answer is $8$ times the thing we want to integrate. Integrating undoes the differentiating, so we just divide by $8$ :

$\int 8x(x^2+2)^3\,dx=(x^2+2)^4+c\;\Rightarrow\;\int x(x^2+2)^3\,dx=\tfrac18(x^2+2)^4+c$

Answer

\tfrac18(x^2+2)^4+c

Common mistake

When the thing you integrate is a number times your part-(a) answer, you must match them. Here part (a) gave

8x(x^2+2)^3

but the integral has

x(x^2+2)^3

. So divide by

8

. Do not forget the

+c

. (9709-style)

Now you try

A curve passes through $\left(2,5\tfrac{2}{3}\right)$ and has $\tfrac{dy}{dx}=\dfrac{8}{(3x+2)^2}$ . Find the equation of the curve. (9709/12 Oct/Nov 2021 Q4)

Your turn— tap to reveal the worked answer (9709/12 Oct/Nov 2021 Q4)

Rewrite as a power: $\tfrac{dy}{dx}=8(3x+2)^{-2}$ .

Integrate. New power is $-1$ , and divide by $a=3$ : $y=\dfrac{8(3x+2)^{-1}}{3(-1)}+c=-\dfrac{8}{3(3x+2)}+c$ .

Use the point. At $x=2$ , $3x+2=8$ , so the term is $-\tfrac{8}{24}=-\tfrac13$ . Then $5\tfrac23=-\tfrac13+c\Rightarrow c=6$ .

Answer

y=6-\dfrac{8}{3(3x+2)}