Integrating (ax + b)^n

AS · P1 · 9 min

The power rule integrates $x^5$ , but what about $(3x - 2)^5$ or $\dfrac{1}{\sqrt{4x + 1}}$ ? Expanding works for small whole-number powers, but it is slow, and it is impossible for roots and negative powers. The syllabus requires you to integrate $(ax + b)^n$ for any rational $n$ except $-1$ , and the method is the chain rule run backwards. It turns up constantly in Paper 1: in finding curves, in areas, and especially in volumes of revolution, where squaring a root of a linear expression leaves exactly this kind of bracket.

Where the rule comes from

Differentiate $(3x - 2)^6$ with the chain rule. The power comes down, the power drops by one, and the derivative of the inside, $3$ , multiplies the result:

\frac{d}{dx}(3x - 2)^6 = 6(3x - 2)^5 \times 3 = 18(3x - 2)^5

So $(3x - 2)^6$ is an integral of $18(3x - 2)^5$ , and dividing by $18$ :

\int (3x - 2)^5\, dx = \frac{(3x - 2)^6}{18} + c

The $18$ is the new power $6$ times the coefficient of $x$ , $3$ . The power rule in reverse gives the $6$ ; the extra factor of $3$ cancels the factor that the chain rule would produce.

Key result

\int (ax + b)^n\, dx = \frac{(ax + b)^{n + 1}}{a(n + 1)} + c \qquad \text{for rational } n \neq -1,\ a \neq 0

In words: add one to the power, divide by the new power, and divide by the coefficient of $x$ .

Why the inside must be linear

The method works because the derivative of $ax + b$ is the constant $a$ , which can be divided out. If the inside is not linear, its derivative involves $x$ , and dividing by an expression in $x$ does not undo the chain rule.

For example, " $\displaystyle\int (x^2 + 1)^3\, dx = \frac{(x^2 + 1)^4}{4 \times 2x}$ " is wrong. Differentiating the right-hand side does not give $(x^2 + 1)^3$ . In Paper 1, a non-linear bracket raised to a whole-number power must be expanded first.

Using the rule

Integrating (ax + b)^n

Write the integrand in the form $k(ax + b)^n$ : move constants out of the denominator, and turn roots and reciprocals into powers. For example $\dfrac{6}{(2x + 1)^3} = 6(2x + 1)^{-3}$ and $\dfrac{1}{\sqrt{5 - 2x}} = (5 - 2x)^{-\frac{1}{2}}$ .
Identify $a$ (the coefficient of $x$ , including its sign) and the new power $n + 1$ .
Write $\dfrac{k(ax + b)^{n+1}}{a(n + 1)}$ and simplify the constant.
Add $+\,c$ for an indefinite integral.
Check by differentiating: you should get back exactly the integrand.

A few results worth knowing by sight:

Integrand	Rewrite	Integral (without $+\,c$ )
$(3x - 2)^5$	already a power	$\dfrac{(3x - 2)^6}{18}$
$\sqrt{4x + 1}$	$(4x + 1)^{\frac{1}{2}}$	$\dfrac{(4x + 1)^{\frac{3}{2}}}{6}$
$\dfrac{1}{\sqrt{5 - 2x}}$	$(5 - 2x)^{-\frac{1}{2}}$	$-\sqrt{5 - 2x}$
$\dfrac{6}{(2x + 1)^3}$	$6(2x + 1)^{-3}$	$-\dfrac{3}{2(2x + 1)^2}$
$\dfrac{4}{(2x - 3)^2}$	$4(2x - 3)^{-2}$	$-\dfrac{2}{2x - 3}$

The third row has $a = -2$ . Forgetting that minus sign is the single most common error with this rule.

Excluded case

The rule fails for $n = -1$ , since it would divide by zero. So $\displaystyle\int \frac{1}{2x + 1}\, dx$ cannot be done in Paper 1. It equals $\tfrac{1}{2}\ln|2x + 1| + c$ , which you will meet in Paper 3. If a P1 question seems to need it, you have probably made an error earlier.

Using a given derivative

Sometimes a question gives you a derivative and asks you to use it, with the word "hence", to integrate something that is not of the form $(ax + b)^n$ . This is still integration as the reverse of differentiation: if $\dfrac{d}{dx}F(x) = f(x)$ , then $\displaystyle\int f(x)\, dx = F(x) + c$ , and constant multiples adjust as needed. Example 6 shows the idea. The general technique for spotting such integrals without help is in Paper 3.

Worked examples

A whole-number power

Find $\displaystyle\int (3x - 2)^5\, dx$ , and check your answer by differentiation.

Solution

Here $a = 3$ and the new power is $6$ :

\int (3x - 2)^5\, dx = \frac{(3x - 2)^6}{3 \times 6} + c = \frac{(3x - 2)^6}{18} + c

Check: $\dfrac{d}{dx}\left[\dfrac{(3x - 2)^6}{18}\right] = \dfrac{6(3x - 2)^5 \times 3}{18} = (3x - 2)^5$ .

A negative power with a constant

Find $\displaystyle\int \frac{6}{(2x + 1)^3}\, dx$ .

Solution

Rewrite: $6(2x + 1)^{-3}$ . Here $a = 2$ and the new power is $-2$ :

\int 6(2x + 1)^{-3}\, dx = \frac{6(2x + 1)^{-2}}{2 \times (-2)} + c = -\frac{3}{2}(2x + 1)^{-2} + c = -\frac{3}{2(2x + 1)^2} + c

Roots, including a negative coefficient

Find (a) $\displaystyle\int \sqrt{4x + 1}\, dx$ and (b) $\displaystyle\int \frac{1}{\sqrt{5 - 2x}}\, dx$ .

Solution

(a) $(4x + 1)^{\frac{1}{2}}$ , with $a = 4$ and new power $\tfrac{3}{2}$ :

\frac{(4x + 1)^{\frac{3}{2}}}{4 \times \tfrac{3}{2}} + c = \frac{(4x + 1)^{\frac{3}{2}}}{6} + c

(b) $(5 - 2x)^{-\frac{1}{2}}$ , with $a = -2$ and new power $\tfrac{1}{2}$ :

\frac{(5 - 2x)^{\frac{1}{2}}}{(-2) \times \tfrac{1}{2}} + c = -(5 - 2x)^{\frac{1}{2}} + c = -\sqrt{5 - 2x} + c

Check (b): $\dfrac{d}{dx}\left[-(5 - 2x)^{\frac{1}{2}}\right] = -\tfrac{1}{2}(5 - 2x)^{-\frac{1}{2}} \times (-2) = (5 - 2x)^{-\frac{1}{2}}$ .

Finding the equation of a curve

A curve is such that $\dfrac{dy}{dx} = \dfrac{6}{\sqrt{4x + 1}}$ , and the curve passes through the point $(2, 5)$ . Find the equation of the curve.

Solution

y = \int 6(4x + 1)^{-\frac{1}{2}}\, dx = \frac{6(4x + 1)^{\frac{1}{2}}}{4 \times \tfrac{1}{2}} + c = 3\sqrt{4x + 1} + c

At $(2, 5)$ : $5 = 3\sqrt{9} + c = 9 + c$ , so $c = -4$ .

The curve is $y = 3\sqrt{4x + 1} - 4$ .

y = 3 sqrt(4x + 1) - 4 (2, 5) (0, -1)

A definite integral

Evaluate $\displaystyle\int_1^5 \frac{1}{\sqrt{2x - 1}}\, dx$ .

Solution

\int (2x - 1)^{-\frac{1}{2}}\, dx = \frac{(2x - 1)^{\frac{1}{2}}}{2 \times \tfrac{1}{2}} = \sqrt{2x - 1}

\int_1^5 \frac{1}{\sqrt{2x - 1}}\, dx = \Big[\sqrt{2x - 1}\Big]_1^5 = \sqrt{9} - \sqrt{1} = 2

Definite integrals and what they measure are covered in definite integrals and areas.

Using a given derivative

(a) Find $\dfrac{d}{dx}\left[(x^2 + 3)^{\frac{3}{2}}\right]$ .

(b) Hence find $\displaystyle\int x\sqrt{x^2 + 3}\, dx$ .

Solution

(a) By the chain rule,

\frac{d}{dx}(x^2 + 3)^{\frac{3}{2}} = \tfrac{3}{2}(x^2 + 3)^{\frac{1}{2}} \times 2x = 3x\sqrt{x^2 + 3}

(b) Part (a) says $(x^2 + 3)^{\frac{3}{2}}$ is an integral of $3x\sqrt{x^2 + 3}$ . The integrand wanted is one third of this, so

\int x\sqrt{x^2 + 3}\, dx = \tfrac{1}{3}(x^2 + 3)^{\frac{3}{2}} + c

The bracket here is not linear, so the $(ax + b)^n$ rule could not be used directly. The "hence" supplies the result.

Integration and stationary points

A curve is defined for $x > -2$ and has $\dfrac{dy}{dx} = 4 - \dfrac{9}{(x + 2)^2}$ . The curve passes through $(1, 7)$ .

(a) Find the equation of the curve.

(b) Find the coordinates of the stationary point and determine its nature.

Solution

(a)

y = \int \left(4 - 9(x + 2)^{-2}\right) dx = 4x - \frac{9(x + 2)^{-1}}{1 \times (-1)} + c = 4x + \frac{9}{x + 2} + c

At $(1, 7)$ : $7 = 4 + 3 + c$ , so $c = 0$ and $y = 4x + \dfrac{9}{x + 2}$ .

(b) $\dfrac{dy}{dx} = 0$ gives $(x + 2)^2 = \dfrac{9}{4}$ , so $x + 2 = \pm\dfrac{3}{2}$ . Since $x > -2$ , $x + 2 > 0$ , so $x + 2 = \dfrac{3}{2}$ and $x = -\dfrac{1}{2}$ .

y = -2 + \frac{9}{3/2} = -2 + 6 = 4

$\dfrac{d^2y}{dx^2} = 18(x + 2)^{-3} = \dfrac{18}{(3/2)^3} = \dfrac{16}{3} > 0$ , so $\left(-\tfrac{1}{2}, 4\right)$ is a minimum point.

y = 4x + 9/(x + 2) x = -2 (-0.5, 4) (1, 7)

Watch out

Forgetting to divide by $a$ . $\displaystyle\int (3x - 2)^5\, dx \neq \dfrac{(3x - 2)^6}{6} + c$ . Differentiate that and you get $3(3x - 2)^5$ , three times too big.

Multiplying by $a$ instead of dividing. That is what differentiation does. Integration divides.

Losing the sign of $a$ . In $(5 - 2x)^n$ the coefficient of $x$ is $-2$ . The answer changes sign.

Applying the rule to a non-linear bracket. $(x^2 + 1)^3$ must be expanded: $\displaystyle\int (x^6 + 3x^4 + 3x^2 + 1)\, dx$ . Never divide by $2x$ .

Leaving a constant in the denominator. $\dfrac{2}{3(x - 4)^5}$ is $\tfrac{2}{3}(x - 4)^{-5}$ ; the $3$ is not part of the bracket.

Using $n = -1$ . $\displaystyle\int \frac{1}{ax + b}\, dx$ is not $\dfrac{(ax + b)^0}{0}$ . It is outside Paper 1.

Exam tip

Write the rewritten power form first, such as $6(4x + 1)^{-\frac{1}{2}}$ . Most mark schemes give a mark for the correct power (the bracket raised to $n + 1$ ) and a separate mark for the correct coefficient.
The coefficient mark is where most candidates lose out. Write $\dfrac{1}{a(n + 1)}$ explicitly, then simplify.
Do not expand $(3x - 2)^6$ in your answer; leaving the bracket is expected and saves time.
"Hence" means use the previous result. Starting from scratch with a method not in the syllabus may not earn the marks.
Check by differentiating. For this rule a check catches both the power and the coefficient in one go.

Summary

$\displaystyle\int (ax + b)^n\, dx = \frac{(ax + b)^{n + 1}}{a(n + 1)} + c$ for rational $n \neq -1$ .
Add one to the power, divide by the new power, divide by the coefficient of $x$ (with its sign).
Rewrite roots and reciprocals as powers, and move constants out of the bracket, first.
The inside must be linear. Non-linear brackets to whole-number powers are expanded.
$n = -1$ gives a logarithm, which is Paper 3.
"Hence" questions use a given derivative to integrate non-linear forms.
Always check by differentiating.

Practice questions

Question

Find $\displaystyle\int (5x + 2)^4\, dx$ .
Find $\displaystyle\int \frac{4}{(2x - 3)^2}\, dx$ .
Find $\displaystyle\int \sqrt{6x + 3}\, dx$ .
Find $\displaystyle\int \frac{1}{\sqrt{1 - 3x}}\, dx$ .
Evaluate $\displaystyle\int_0^4 \sqrt{2x + 1}\, dx$ .
A curve has $\dfrac{dy}{dx} = \dfrac{12}{(3x + 1)^2}$ and passes through $(1, 2)$ . Find its equation.
Find $\displaystyle\int \left((2x + 1)^2 + \frac{1}{(2x + 1)^2}\right) dx$ .
Explain why $\displaystyle\int (x^2 + 1)^3\, dx$ is not $\dfrac{(x^2 + 1)^4}{8x} + c$ , and find the correct integral.
A curve has $\dfrac{dy}{dx} = k\sqrt{2x + 1}$ , where $k$ is a constant, and passes through the points $(0, 1)$ and $(4, 27)$ . Find $k$ and the equation of the curve.

Answers

$\dfrac{(5x + 2)^5}{5 \times 5} + c = \dfrac{(5x + 2)^5}{25} + c$ .
$4(2x - 3)^{-2}$ integrates to $\dfrac{4(2x - 3)^{-1}}{2 \times (-1)} + c = -\dfrac{2}{2x - 3} + c$ .
$\dfrac{(6x + 3)^{\frac{3}{2}}}{6 \times \tfrac{3}{2}} + c = \dfrac{(6x + 3)^{\frac{3}{2}}}{9} + c$ .
$(1 - 3x)^{-\frac{1}{2}}$ integrates to $\dfrac{(1 - 3x)^{\frac{1}{2}}}{(-3) \times \tfrac{1}{2}} + c = -\tfrac{2}{3}\sqrt{1 - 3x} + c$ .
$\displaystyle\int (2x + 1)^{\frac{1}{2}}\, dx = \dfrac{(2x + 1)^{\frac{3}{2}}}{3}$ . So $\left[\dfrac{(2x + 1)^{\frac{3}{2}}}{3}\right]_0^4 = \dfrac{27}{3} - \dfrac{1}{3} = \dfrac{26}{3}$ .
$y = \dfrac{12(3x + 1)^{-1}}{3 \times (-1)} + c = -\dfrac{4}{3x + 1} + c$ . At $(1, 2)$ : $2 = -1 + c$ , $c = 3$ . So $y = 3 - \dfrac{4}{3x + 1}$ .
$\dfrac{(2x + 1)^3}{2 \times 3} + \dfrac{(2x + 1)^{-1}}{2 \times (-1)} + c = \dfrac{(2x + 1)^3}{6} - \dfrac{1}{2(2x + 1)} + c$ .
Differentiating $\dfrac{(x^2 + 1)^4}{8x}$ does not give $(x^2 + 1)^3$ : the inside $x^2 + 1$ is not linear, so its derivative $2x$ varies and cannot be divided out. Expand instead: $(x^2 + 1)^3 = x^6 + 3x^4 + 3x^2 + 1$ , so the integral is $\dfrac{x^7}{7} + \dfrac{3x^5}{5} + x^3 + x + c$ .
$y = \dfrac{k(2x + 1)^{\frac{3}{2}}}{2 \times \tfrac{3}{2}} + c = \dfrac{k}{3}(2x + 1)^{\frac{3}{2}} + c$ . At $(0, 1)$ : $1 = \dfrac{k}{3} + c$ . At $(4, 27)$ : $27 = \dfrac{k}{3}(27) + c = 9k + c$ . Subtracting: $26 = 9k - \dfrac{k}{3} = \dfrac{26k}{3}$ , so $k = 3$ and $c = 0$ . The curve is $y = (2x + 1)^{\frac{3}{2}}$ .

Where the rule comes from

Why the inside must be linear

Using the rule

Using a given derivative

Worked examples

Practice questions

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