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 ($$ax+b$$).

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