Why Substitution Reverses the Chain Rule
The Chain Rule explains how to differentiate a composition. If an outer function is evaluated at an inner function, differentiating the result produces the derivative of the outer function, evaluated at the inner function, multiplied by the derivative of the inner function.
Indefinite substitution uses this same relationship in reverse. When an integrand contains a composition together with the derivative of its inner function, the integrand has the form of a Chain Rule derivative. The antiderivative can then be written as a composition.
The key is to distinguish the function inside the composition from an antiderivative of the outside function. Let \(f\) be the outside function, and let \(F\) be an antiderivative of \(f\). Thus \(F'=f\). The substitution rule gives an antiderivative of \(f(g(x))g'(x)\) by evaluating \(F\) at \(g(x)\).
The formula describes a family of antiderivatives, not a single function: differentiating \(F(g(x))+C\) gives the integrand for every constant \(C\). As usual, on an interval, any two antiderivatives of the same function differ by a constant.
Deriving the Rule from the Chain Rule
The result follows directly from the Chain Rule. To see why, differentiate the proposed antiderivative \(F(g(x))\). The outer function is \(F\), and the inner function is \(g\), so the Chain Rule gives the following calculation.
Consider the composite function \(F(g(x))\), where \(F'=f\).
By the Chain Rule, its derivative is \(F'(g(x))g'(x)\).
Since \(F'=f\), the derivative becomes \(f(g(x))g'(x)\), exactly the integrand.
Therefore \(F(g(x))\) is an antiderivative of \(f(g(x))g'(x)\). Adding any constant does not change its derivative, which proves the displayed integration formula. The argument requires differentiability of \(g\), and differentiability of \(F\) on a domain containing the values of \(g\). If \(F\) is an antiderivative of \(f\), then the equality \(F'=f\) supplies the needed outer derivative.
This proof also explains the matching-factor requirement. The factor \(g'(x)\) is not an optional detail: it is part of the derivative of a composition. Without it, differentiating \(F(g(x))\) would not, in general, produce the integrand.
Worked Examples
Worked Example: A Polynomial Power Composition
Evaluate \(\int (3x^2+1)^4(6x)\,dx\).
The inner expression is \(g(x)=3x^2+1\). Its derivative is
The remaining outside function is \(f(t)=t^4\), where \(t\) is an input variable. An antiderivative of \(f\) is \(F(t)=\frac{t^5}{5}\), since
The integrand is precisely \(f(g(x))g'(x)\). Apply the rule:
Check by differentiating the answer:
The derivative agrees with the integrand, so the antiderivative is correct.
Worked Example: A Polynomial Outside Function
Evaluate \(\int (2x^2+1)(2x)\,dx\).
Rewrite the first factor in terms of \(g(x)=x^2-1\):
The derivative of the inner function is
Thus the integrand is \((2g(x)+3)g'(x)\). For the outside function \(f(t)=2t+3\), an antiderivative is \(F(t)=t^2+3t\), because
The substitution rule now gives
This answer can also be simplified:
Check by differentiating the simplified answer:
The last equality follows by multiplying: \((2x^2+1)(2x)=4x^3+2x\). The check matches the original integrand.
Worked Example: An Exponential Composition
Evaluate \(\int e^{x^2-2x}(2x-2)\,dx\).
Take \(g(x)=x^2-2x\). Differentiating gives
The outside function is \(f(t)=e^t\), and an antiderivative is \(F(t)=e^t\), because \(\frac{d}{dt}e^t=e^t\). The integrand therefore has the required form \(f(g(x))g'(x)\). Applying the rule,
Check by differentiation:
This is the original integrand, so the answer is verified.
What the Formula Does—and Does Not—Say
The substitution rule is a conditional recognition rule. It applies when the integrand can be written as an outside function evaluated at an inner function, multiplied by the derivative of that inner function. In the formula, \(f\) is the outside function, \(g\) is the inner function, and \(F\) is an antiderivative of \(f\).
The condition concerns the entire product. For example, in the first worked example, the inner function \(3x^2+1\) has derivative \(6x\), and that exact factor appears in the integrand. The rule then turns the outside power \(t^4\) into its antiderivative \(t^5/5\), evaluated at \(3x^2+1\).
The formula does not claim that every integral containing a composition can be evaluated this way. If the inner derivative is absent, or if the remaining factors do not make the integrand fit \(f(g(x))g'(x)\), this direct application is not justified. Other methods may be relevant, but they are not consequences of this formula alone.
A Common Missing-Factor Error
A frequent mistake is to integrate a composed power as though the derivative of its inner expression did not matter. For instance, the following proposed calculation is incorrect:
Differentiate the proposed answer to test it:
The derivative contains an extra factor \(6x\), so it is not the original integrand \((3x^2+1)^4\). The calculation incorrectly used the substitution rule without the matching inner derivative. The rule does correctly give
This correction does not evaluate the integral with the missing factor; it only identifies why the proposed expression is not its antiderivative. Always differentiate a proposed answer when checking whether the inner derivative has been accounted for.
Why This Derivation Matters
The substitution rule is not a separate trick that happens to produce convenient formulas. It is a direct consequence of a general differentiation rule. Recognizing this connection gives a reliable test: differentiate the proposed composite antiderivative and compare the result with the integrand.
This also clarifies the role of each part of the formula. The outside function \(f\) is integrated to obtain \(F\); the inner function \(g\) remains in place as the input to \(F\); and \(g'(x)\) is the factor created by differentiating that composition. When all three roles match, the formula follows. When one does not match, the Chain Rule does not justify the proposed answer.
Check Your Understanding
For each question, pay attention to the relationship between the inner function and its derivative.
- State the indefinite substitution rule using \(f\), \(g\), and an antiderivative \(F\) of \(f\).
- In the integrand \((x^2+5)^3(2x)\), identify the inner function, its derivative, and the outside function.
- Evaluate \(\int (x^2+4)^2(2x)\,dx\), then verify your answer by differentiation.
- Why does \(\frac{(3x^2+1)^5}{5}+C\) fail to be an antiderivative of \((3x^2+1)^4\)?
- What conditions on \(F\) and \(g\) are used to justify the substitution formula?
