Integrating $f(ax+b)$

In a nutshell

It is possible to use the chain rule for differentiation to integrate functions of the form $f(ax+b)$ , where $a$ and $b$ are constants.

Deriving the rule

Consider the function $f(x)$ , and its derivative $f'(x)$ . By definition:

$\int f'(x)\, dx = f(x)+C$

Suppose you want to integrate $f'(ax+b)$ . This can be done by considering the derivative of $f(ax+b)$ .

Differentiating $f(ax+b)$ requires use of the chain rule:

Let $y=f(ax+b)=f(u)$ , where $u(x)=ax+b$ .

The chain rule is described by the formula:

$\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}$

$\begin{aligned} y=f(ax+b) &\Rightarrow \dfrac{dy}{dx}=(f(ax+b))'\\ y=f(u)&\Rightarrow \dfrac{dy}{du}=f'(u)=f'(ax+b)\\u=ax+b &\Rightarrow \dfrac{du}{dx}=a \end{aligned}$

Substitute these expressions into the formula:

$\begin{aligned}(f(ax+b))'&=f'(ax+b)\times a \\ f'(ax+b)&=\dfrac1a (f(ax+b))'\end{aligned}$

Integrate both sides of this expression:

$\begin{aligned}\int f'(ax+b) \, dx =& \int \dfrac1a (f(ax+b))' \,dx\\=& \dfrac1a \int (f(ax+b))' \,dx\end{aligned}$

This provides a formula to integrate functions that have an argument of $ax+b$ :

$\boxed{\int f'(ax+b) \, dx = \dfrac1a f(ax+b)+C}$

In other words, integrate the function normally and divide by the coefficient of $x$ .

Example 1

Compute the following integrals:

i) $\int \cos(2x-1)\,dx$ 

ii) $\int e^{1-x} \, dx$ 

iii) $\int_0^1 (5-\dfrac{1}{2}x)^2\,dx$ 

Part i):

Integrate the function "normally":

$\cos(x)$ integrates to $\sin(x)$ .

So, $\cos(2x-1)$ should integrate to something similar to $\sin(2x-1)$ .

Divide by the coefficient of $x$ , which in this case is $2$ .

So, $\cos(2x-1)$ integrates to $\dfrac{\sin(2x-1)}{2}$ .

$\int \cos(2x-1)\,dx=\underline{\dfrac12 \sin(2x-1)+C}$

Part ii):

$e^x$ integrates to $e^x$ .

So, $e^{1-x}$ should integrate to something similar to $e^{1-x}$ .

The coefficient of $x$ is $-1$ , so divide by this.

$e^{1-x}$ integrates to $\dfrac{e^{1-x}}{-1}$ .

$\int e^{1-x} \, dx=\underline{-e^{1-x}+C}$

Part iii):

$x^2$ integrates to $\dfrac13 x^3$ .

So, $(5-\dfrac{1}{2}x)^2$ should integrate to something similar to $\dfrac 13(5-\dfrac{1}{2}x)^3$ .

The coefficient of $x$ is $-\frac12$ , so divide by this.

$\begin{aligned}\int_0^1 (5-\dfrac{1}{2}x)^2 \, dx &=\left[\dfrac{\dfrac13(5-\dfrac{1}{2}x)^3}{-\dfrac12}\right]_0^1 \\&= \left[-\dfrac23(5-\dfrac{1}{2}x)^3\right]_0^1 \\&=\left(-\dfrac23 (5-\dfrac12 (1))^3 \right) - \left(-\dfrac23 (5-\dfrac12 (0))^3 \right) \\&= \left(-\dfrac23 \left(\dfrac92\right)^3\right) - \left(- \dfrac23 (5)^3\right) \\&=\left( -\dfrac{243}{4}\right) - \left(-\dfrac{250}{3}\right)\\&=\dfrac{271}{12} \end{aligned}$

$\int_0^1 (5-\dfrac{1}{2}x)^2 \, dx=\underline{\dfrac{271}{12}}$

Note: This only works when the argument of the function is a linear expression ( $ax+b$ ).