Mathematics/Lessons/Integration/Integrating f(ax + b)
Pure · Integration · Year 2

Integrating f(ax + b)

Whenever the variable is replaced by a linear expression $ax+b$, the standard result still applies — divide by $a$ and carry on.

8 min read 4 worked examples

Stretching a standard result

Differentiating $f(ax+b)$ with the chain rule multiplies the derivative by $a$ — the coefficient of $x$ inside the bracket. Integrating reverses this: replacing $x$ with the linear expression $ax+b$ in a standard result introduces an extra factor of $\dfrac1a$, to compensate.

The rule

If $\displaystyle\int f(x)\,dx = F(x)+c$, then $\displaystyle\int f(ax+b)\,dx = \dfrac1a F(ax+b)+c$.

Applied to the standard results from the previous lesson, this gives a short list of linear-composite versions:

In practice

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

$\displaystyle\int e^{ax+b}\,dx = \dfrac1a e^{ax+b}+c$

$\displaystyle\int \dfrac1{ax+b}\,dx = \dfrac1a\ln|ax+b|+c$

$\displaystyle\int \sin(ax+b)\,dx = -\dfrac1a\cos(ax+b)+c$

$\displaystyle\int \cos(ax+b)\,dx = \dfrac1a\sin(ax+b)+c$

$\displaystyle\int \sec^2(ax+b)\,dx = \dfrac1a\tan(ax+b)+c$

Worked example 1

A linear expression raised to a power
Find $\displaystyle\int (2x+5)^4\,dx$.
MethodIntegrate as if x, then divide by the coefficient of x
1
Integrating $(2x+5)^4$ as though it were $x^4$ would give $\dfrac{(2x+5)^5}{5}$.
Apply the power rule to the bracket as a whole.
2
Here $a=2$, so divide by this extra factor: $\dfrac{(2x+5)^5}{5\times2}=\dfrac{(2x+5)^5}{10}$.
Correct for the factor of 2 the chain rule would otherwise introduce on differentiating.
3
$\displaystyle\int (2x+5)^4\,dx = \dfrac{(2x+5)^5}{10}+c$.
State the full answer with the constant of integration.
▪ solved
Quick check

Differentiating $\dfrac{(2x+5)^5}{10}$ with the chain rule gives $\dfrac{5(2x+5)^4\times2}{10}=(2x+5)^4$ — confirming the answer.

Worked example 2

An exponential with a linear exponent
Find $\displaystyle\int e^{3x-1}\,dx$.
MethodApply the exponential result with a = 3
1
Here $a=3$, $b=-1$.
Identify the linear expression inside the exponent.
2
$\displaystyle\int e^{3x-1}\,dx = \dfrac13e^{3x-1}+c$.
Apply the standard exponential result, dividing by a = 3.
▪ solved

Worked example 3

A definite integral with cos(ax + b)
Find the exact value of $\displaystyle\int_0^{\pi/6} \cos(2x)\,dx$.
MethodIntegrate, then evaluate between the limits
1
Here $a=2$, $b=0$, so $\displaystyle\int \cos(2x)\,dx = \dfrac12\sin(2x)+c$.
Apply the standard cos(ax+b) result.
2
At $x=\dfrac\pi6$: $\dfrac12\sin\dfrac\pi3 = \dfrac12\cdot\dfrac{\sqrt3}2=\dfrac{\sqrt3}4$.
Substitute the upper limit, using an exact trig value.
3
At $x=0$: $\dfrac12\sin0=0$.
Substitute the lower limit.
4
$\displaystyle\int_0^{\pi/6}\cos(2x)\,dx = \dfrac{\sqrt3}4-0=\dfrac{\sqrt3}4$.
Subtract to give the exact value.
▪ solved

Worked example 4

A definite integral with 1/(ax + b) · Challenge
Find the exact value of $\displaystyle\int_1^3 \dfrac{5}{2x-1}\,dx$.
MethodApply the 1/(ax+b) result, then evaluate between the limits
1
Here $a=2$, $b=-1$, so $\displaystyle\int \dfrac5{2x-1}\,dx = 5\times\dfrac12\ln|2x-1| = \dfrac52\ln|2x-1|+c$.
Apply the standard 1/(ax+b) result, carrying the constant 5 through.
2
At $x=3$: $\dfrac52\ln|5| = \dfrac52\ln5$.
Substitute the upper limit.
3
At $x=1$: $\dfrac52\ln|1| = \dfrac52\ln1 = 0$.
Substitute the lower limit; ln 1 = 0.
4
$\displaystyle\int_1^3\dfrac5{2x-1}\,dx = \dfrac52\ln5-0=\dfrac52\ln5$.
Subtract to give the exact value.
▪ solved
Check the interval first

For $\dfrac1{2x-1}$ to be defined throughout $[1,3]$, the denominator $2x-1$ must never equal zero on that interval — here $2x-1$ ranges from 1 to 5, safely positive throughout, so no modulus subtlety arises.

Common mistake

Multiplying by $a$ instead of dividing by it — or forgetting the $\dfrac1a$ factor altogether and integrating as though $ax+b$ were simply $x$. The factor of $\dfrac1a$ is what makes the chain-rule differentiation of the answer come back out correctly; it's easy to check by differentiating.

Check yourself

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

Method
1
Integrating as if x gives $\dfrac{(3x-2)^4}{4}$.
Apply the power rule to the bracket.
2
Here $a=3$, so divide by 3: $\dfrac{(3x-2)^4}{4\times3}=\dfrac{(3x-2)^4}{12}$.
Correct for the coefficient of x inside the bracket.
3
$\displaystyle\int (3x-2)^3\,dx = \dfrac{(3x-2)^4}{12}+c$.
State the full answer.
▪ solved
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Integrating Standard Functions