Recognizing a Reversed Chain Rule
An integral may look complicated because one function is nested inside another. Sometimes, however, the integrand contains exactly the factor produced by differentiating that composition. In that case, the chain rule can be read backward in a single step.
The key is to recognize both parts of the pattern: an outer function evaluated at an inner function, and the derivative of that inner function as a factor in the integrand. When the match is exact, there is no need to carry out a separate substitution calculation.
This is a recognition method, not a new differentiation rule. Its reliability comes from checking that the proposed answer differentiates to the original integrand. That check is especially useful when constants appear in the inner derivative.
The One-Step Rule
Let \(I\) and \(J\) be intervals. Suppose \(g\) is differentiable on \(I\), its values lie in \(J\), and \(F\) is differentiable on \(J\). Write \(f=F'\). Then for every \(x\) in \(I\), the chain rule gives the identity below.
This identity immediately gives the antiderivative rule. If the integrand has the form on the right, an antiderivative is obtained by evaluating the outer antiderivative \(F\) at the inner function \(g(x)\).
Here \(C\) is an arbitrary constant. On an interval, any two antiderivatives of the same function differ by a constant. The rule assumes that the stated derivatives exist on the relevant intervals; it does not require the integrand to have a particular appearance beyond the exact factorization shown.
The notation \(F'=f\) means that \(F\) is an antiderivative of \(f\). The inner function \(g\) and the factor \(g'\) must match: \(g'\) is the derivative of the same expression that appears as the input of \(f\). The outer antiderivative \(F\) must also be correct. Recognizing only a familiar inner expression is not enough.
Why the Rule Works
The complete justification is a direct application of the chain rule. The chain rule says that differentiating an outer function evaluated at an inner function multiplies the derivative of the outer function, evaluated at the inner function, by the derivative of the inner function.
Choose \(F\) so that \(F'(u)=f(u)\) for inputs \(u\) in the relevant interval.
The candidate antiderivative is \(F(g(x))\).
By the chain rule, its derivative is \(F'(g(x))g'(x)=f(g(x))g'(x)\), exactly the integrand.
This proof also explains the practical check: differentiate the proposed answer and compare the result with the original integrand, including every constant and sign.
Worked Examples
Worked Example: A Trigonometric Outer Function
Evaluate \(\int (3x^2+2)\cos(x^3+2x)\,dx\).
First identify the inner function and its derivative:
The remaining factor is \(\cos(g(x))\). An outer antiderivative of \(\cos(u)\) is \(\sin(u)\), since \(\frac{d}{du}\sin(u)=\cos(u)\). The inner derivative is present exactly, so the one-step rule gives:
Check by differentiating the result:
This is the original integrand, so the antiderivative is verified.
Worked Example: A Logarithm with a Matching Derivative
Evaluate \(\int \frac{4x}{2x^2+7}\,dx\).
The denominator is the inner function, and its derivative is the numerator:
For positive \(u\), an outer antiderivative of \(1/u\) is \(\ln(u)\). Since \(2x^2+7\geq 7>0\) for every real \(x\), the logarithm is defined throughout the domain. Thus:
To check, differentiate the proposed antiderivative:
The derivative agrees with the integrand, including the numerator factor.
Worked Example: Correcting for a Constant Factor
Evaluate \(\int x\cos(x^2+1)\,dx\).
Set \(g(x)=x^2+1\), so \(g'(x)=2x\). The integrand contains \(x\), not \(2x\). Since \(x=\frac12(2x)\), the integrand is one half of the exact chain-rule pattern:
An outer antiderivative of \(\cos(u)\) is \(\sin(u)\). Keeping the factor \(\frac12\) gives:
Verify the result by differentiating:
The factor of \(2\) from differentiating the inner function cancels the factor of \(\frac12\), leaving exactly the original integrand.
Worked Example: A Constant Inside an Exponential
Evaluate \(\int e^{5x+1}\,dx\).
Here the inner function is \(g(x)=5x+1\), with \(g'(x)=5\). The integrand does not include that factor of \(5\). Write it as one fifth of the matching pattern:
An outer antiderivative of \(e^u\) is \(e^u\). Therefore:
Differentiate to check the coefficient:
The result differentiates to the original integrand.
Handling Constant Factors Reliably
A constant factor can be handled by factoring it out of the integral or by comparing the inner derivative with the factor that is present. Suppose \(F'=f\) and \(c\) is a constant. The chain rule shows that:
Consequently, if the integrand is \(c f(g(x))g'(x)\), an antiderivative is \(cF(g(x))+C\). Equivalently, if the inner derivative is a constant multiple of the factor that appears, rewrite the integrand to expose the exact derivative and keep the compensating constant.
For instance, in the exponential example, the derivative of \(5x+1\) is \(5\), while the integrand has no factor \(5\). The required antiderivative therefore carries the factor \(\frac15\). In the trigonometric example, the inner derivative was \(2x\), but the integrand had \(x\); the needed factor was again \(\frac12\).
Do not change a constant merely because an inner function looks familiar. Compare the derivative of that exact inner function with the factor actually present in the integrand. A factor that is absent must be accounted for, and a factor already present must not be inserted again.
A Common Mistake: Forgetting the Outer Antiderivative
A frequent error is to preserve the inner expression and the derivative factor but choose the wrong outer antiderivative. For example, it is incorrect to claim:
Differentiating the proposed answer reveals the error:
which is not the original integrand \(3x^2\cos(x^3)\). The outer function is \(\cos(u)\), whose antiderivative is \(\sin(u)\), not \(\cos(u)\). The corrected calculation is:
and the check is:
This illustrates why a verification by differentiation is more than a formality: it detects errors in the outer antiderivative, the inner derivative, and constant factors.
When the One-Step Match Is Not Enough
The pattern is useful only when the integrand really has the required structure. Before applying it, check these points:
- The expression inside the outer function is identified as \(g(x)\).
- The integrand includes \(g'(x)\), either exactly or up to a constant factor.
- The outer antiderivative \(F\) satisfies \(F'=f\) on the values taken by \(g(x)\).
- The proposed result differentiates back to the entire integrand.
If the derivative factor does not match, the one-step rule cannot be applied in that form. A different evaluation strategy may be needed. In particular, a superficial resemblance between part of an integrand and a familiar composite function does not establish the chain-rule pattern.
Reversing the chain rule matters because it turns many composite integrals into quick, verifiable calculations. It also provides a useful way to judge a proposed antiderivative: the answer is correct precisely when differentiation reconstructs the integrand on the interval under consideration.
Check Your Understanding
For each question, identify the inner function and use differentiation to check any proposed antiderivative.
- Evaluate \(\int 2x\sin(x^2)\,dx\).
- Evaluate \(\int \frac{6x}{3x^2+2}\,dx\), and state why the logarithm is defined for every real \(x\).
- Find an antiderivative of \(e^{4x}\). What constant factor is needed, and why?
- Explain why \(\cos(2x^2)\) is not an antiderivative of \(4x\sin(2x^2)\).
- State the chain-rule identity that justifies the one-step antiderivative rule.
