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:
∫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:
dxdy=dudy×dxdu
y=f(ax+b)y=f(u)u=ax+b⇒dxdy=(f(ax+b))′⇒dudy=f′(u)=f′(ax+b)⇒dxdu=a
Substitute these expressions into the formula:
(f(ax+b))′f′(ax+b)=f′(ax+b)×a=a1(f(ax+b))′
Integrate both sides of this expression:
∫f′(ax+b)dx==∫a1(f(ax+b))′dxa1∫(f(ax+b))′dx
This provides a formula to integrate functions that have an argument of ax+b:
∫f′(ax+b)dx=a1f(ax+b)+C
In other words, integrate the function normally and divide by the coefficient of x.
Example 1
Compute the following integrals:
i) ∫cos(2x−1)dx
ii) ∫e1−xdx
iii) ∫01(5−21x)2dx
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 2sin(2x−1).
∫cos(2x−1)dx=21sin(2x−1)+C
Part ii):
ex integrates to ex.
So, e1−x should integrate to something similar to e1−x.
The coefficient of x is −1, so divide by this.
e1−x integrates to −1e1−x.
∫e1−xdx=−e1−x+C
Part iii):
x2 integrates to 31x3.
So, (5−21x)2 should integrate to something similar to 31(5−21x)3.
The coefficient of x is −21, so divide by this.
∫01(5−21x)2dx=−2131(5−21x)301=[−32(5−21x)3]01=(−32(5−21(1))3)−(−32(5−21(0))3)=(−32(29)3)−(−32(5)3)=(−4243)−(−3250)=12271
∫01(5−21x)2dx=12271
Note: This only works when the argument of the function is a linear expression (ax+b).
