Introduction
An antiderivative problem often asks us to recognize a derivative that has already been formed. The chain rule tells us how to differentiate a composite function: differentiate the outer function at the inner input, then multiply by the derivative of that inner input. Reversing that process gives a useful method for finding antiderivatives.
The central pattern is a function evaluated at an inner expression, multiplied by the derivative of that expression. When the pattern is present, we can temporarily give the inner expression a new name, integrate with respect to that name, and then substitute the original expression back.
Recognizing the pattern
In the expression \(F(g(x))g'(x)\), the function \(g\) is the inner function. The factor \(F(g(x))\) is another function evaluated at the inner expression, and \(g'(x)\) is the derivative of that inner function. The matching factor \(g'(x)\) is essential: it is what the chain rule produces when differentiating a composite.
To use the method, we need an antiderivative \(A\) of the outer function \(F\), so that \(A'(u)=F(u)\). The variable \(u\) is only a temporary name for the input to \(F\). After integrating in terms of \(u\), replace \(u\) with \(g(x)\).
In this tutorial, the method is applied when the integrand has the stated composite-and-derivative form and the outer function has a known antiderivative. The chain rule verifies the result directly. If the inner-derivative factor is missing or does not match, this pattern cannot be applied as written; that does not by itself prove that the integrand has no antiderivative.
Worked Example: A trigonometric function of a composite input
Worked Example: A trigonometric function of a composite input
Find an antiderivative of \(\int 2x\cos(x^2+1)\,dx\).
First identify the inner function and its derivative:
The remaining function is cosine, whose antiderivative with respect to its input \(u\) is \(\sin u\), since \(\frac{d}{du}\sin u=\cos u\). The integrand is therefore \(\cos(g(x))g'(x)\).
Temporarily write \(u=g(x)=x^2+1\). In terms of this input, the antiderivative is \(\sin u\). Substitute \(u=x^2+1\) back:
Check by differentiation. The chain rule gives
This is the original integrand, so the antiderivative is correct.
Why the method works
The justification is exactly the chain rule. Suppose \(g\) is differentiable on an interval \(I\), and \(A\) is differentiable on an interval containing the values \(g(x)\) for \(x\in I\). Suppose also that \(A'(u)=F(u)\) for those inputs. Then the composite function \(A(g(x))\) is differentiable on \(I\), and its derivative is as follows.
Proof. Apply the chain rule to the composite \(A\circ g\). Its outer function is \(A\), and its inner function is \(g\). Therefore, for every \(x\in I\),
The last expression is the stated integrand. By the definition of antiderivative, \(A(g(x))\) is an antiderivative of that integrand. This proves the result.
This argument also explains why the temporary variable is useful. It isolates the input being passed to the outer function. We find an antiderivative in terms of that input, then restore the original expression. The temporary variable does not change the function or the answer.
Worked Example: A polynomial outer function
Worked Example: A polynomial outer function
Find an antiderivative of \(\int (3x^2-2)(x^3-2x+5)^4\,dx\).
Set \(g(x)=x^3-2x+5\). Differentiate it:
Thus the integrand is \(g'(x)\) multiplied by \(g(x)^4\). The outer function is \(F(u)=u^4\), and an antiderivative of it is \(A(u)=\frac{u^5}{5}\), because
Replace the temporary input \(u\) by \(x^3-2x+5\):
Check the result by differentiating the proposed antiderivative:
The derivative matches the original integrand exactly.
A practical sequence of steps
For a candidate integrand, compare the expression inside the composition with the factors multiplying it. The order below makes the check explicit and reduces the chance of losing the inner derivative.
Identify a plausible \(g(x)\) that is being used as the input to another function.
Compute \(g'(x)\) and compare it with the other factors in the integrand.
If the outer function is \(F(u)\), find \(A(u)\) satisfying \(A'(u)=F(u)\).
Write \(A(g(x))+C\), with the original variable \(x\) restored.
Apply the chain rule to the proposed antiderivative and compare the result with the integrand.
Worked Example: A matching derivative found by expansion
Worked Example: A matching derivative found by expansion
Find an antiderivative of \(\int (2x+4)(x^2+4x+7)^3\,dx\).
The inner expression is \(g(x)=x^2+4x+7\). Its derivative is
This matches the factor outside the power, so the integrand has the required form. For the outer function \(F(u)=u^3\), an antiderivative is \(A(u)=\frac{u^4}{4}\), since
Substitute the inner expression into \(A\):
Differentiate to check:
The derivative is precisely the original integrand, confirming the answer.
A common mistake: forgetting the inner derivative
When integrating a function of \(g(x)\), it is tempting to integrate only the visible outer function and omit the factor that the chain rule requires. That produces an answer whose derivative is usually incorrect.
There is a related pattern-recognition mistake: seeing an inner expression does not guarantee that its derivative is present. For example, \(\cos(x^2)\) contains the inner function \(x^2\), but its derivative is \(2x\), and the integrand as written has no such factor. The reverse-chain-rule method does not apply directly in that form. One should not insert a missing factor without also accounting for its effect on the expression.
When this viewpoint is useful
Many complicated-looking integrands are simple once their structure is recognized. A nested expression can conceal the derivative of a familiar outer antiderivative. Looking for the inner derivative is therefore not merely a bookkeeping trick: it is a way to identify how the integrand could have been produced by differentiation.
The final differentiation check is especially valuable. It tests both the outer antiderivative and the inner derivative factor. If the result differs from the original integrand, compare the two expressions carefully before accepting the answer. A missing factor, a sign error, or an incorrect outer antiderivative will be exposed by this check.
Remember that the constant \(C\) is included because adding any constant to an antiderivative does not change its derivative. The reverse-chain-rule step constructs one antiderivative, and \(+C\) records the full family on the interval under consideration.
Check Your Understanding
For each question, identify the inner expression and use differentiation to justify your conclusion.
- Find an antiderivative of \(\int 6x\sin(3x^2+2)\,dx\), and verify it.
- For \(\int (4x^3)(x^4+1)^6\,dx\), identify the inner function, its derivative, and the outer antiderivative.
- Why does the factor \(g'(x)\) appear in an integrand of the form \(F(g(x))g'(x)\)?
- Does the reverse-chain-rule pattern apply directly to \(\int \sin(x^2)\,dx\) as written? Explain what factor you would look for.
- State the chain-rule calculation that verifies \(A(g(x))\) as an antiderivative when \(A'(u)=F(u)\).
