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Integration Strategy · Tutorial 5 of 998

Recognizing Compositions by the Chain Rule

Learn to spot composite-function patterns in integrands and check whether the inner function’s derivative is present, possibly up to a constant factor.

Beginner 9 min read

What You'll Learn

  • Identify the inner and outer functions in a composite expression.
  • Match an integrand to the chain rule in reverse.
  • Recognize when the inner derivative is present up to a constant factor.
  • Use differentiation to verify a proposed antiderivative.
  • Distinguish a genuine composition pattern from a superficially similar expression.

Why Composition Patterns Matter

An integrand can look complicated because one function is nested inside another. Recognizing that structure can make the antiderivative visible without expanding or rearranging the expression extensively.

The key question is whether the integrand contains not just a function of an inner expression, but also the derivative of that inner expression. This is the chain rule read backward.

Central idea: Look for an outer-function derivative evaluated at an inner function, multiplied by the derivative of that inner function. When those pieces fit, the antiderivative is the outer function evaluated at the inner function.

Composition and the Chain Rule

If \(g\) maps an interval into the domain of \(F\), the composition \(F\circ g\) is the function defined by \((F\circ g)(x)=F(g(x))\). The function \(g\) is the inner function; \(F\) is the outer function.

The chain rule states that if \(F\) and \(g\) are differentiable, then

$$ \frac{d}{dx}F(g(x))=F'(g(x))g'(x). $$

This formula explains the pattern we seek in an integrand: an expression of the form \(F'(g(x))g'(x)\). The factor \(F'(g(x))\) is the derivative of the outer function, evaluated at the inner function. The factor \(g'(x)\) accounts for the rate at which that inner function changes.

Recognition rule: If \(F\) and \(g\) are differentiable on the relevant intervals, then $$ \int F'(g(x))g'(x)\,dx=F(g(x))+C. $$ Here \(C\) is an arbitrary constant. The equality applies on an interval where the functions and derivatives involved are defined.

This is a recognition rule, not a license to ignore a missing factor. If the integrand contains \(F'(g(x))\) but not \(g'(x)\), the displayed pattern does not apply as written. Sometimes the needed factor is present up to a constant multiple; that case is still manageable, as the examples show.

Why the Recognition Rule Works

The justification is short: differentiate the proposed antiderivative. If the result is the original integrand, the proposal is an antiderivative by definition.

1
Choose the proposed form.
For an integrand \(F'(g(x))g'(x)\), propose \(F(g(x))\).
2
Differentiate it.
By the chain rule, \(\frac{d}{dx}F(g(x))=F'(g(x))g'(x)\).
3
Compare with the integrand.
The derivative equals the given integrand, so \(F(g(x))+C\) is an antiderivative.

This argument also explains why the inner derivative is essential: it is part of the derivative of the proposed composition.

Worked Examples: Matching the Inner Derivative

Worked Example: A Trigonometric Function of a Quadratic

Recognize the pattern in \(\int 6x\cos(3x^2+1)\,dx\).

The expression inside the cosine is \(g(x)=3x^2+1\), so \(g'(x)=6x\). The outer function whose derivative is cosine is \(F(u)=\sin u\), since \(F'(u)=\cos u\). Thus the integrand is \(F'(g(x))g'(x)\).

$$ \int 6x\cos(3x^2+1)\,dx =\sin(3x^2+1)+C. $$

Check:

$$ \frac{d}{dx}\sin(3x^2+1) =\cos(3x^2+1)\cdot\frac{d}{dx}(3x^2+1) =\cos(3x^2+1)\cdot 6x =6x\cos(3x^2+1). $$

The derivative reproduces the integrand exactly.

Worked Example: A Logarithm with a Positive Inner Function

Find \(\int \frac{2x}{x^2+4}\,dx\) by recognizing a composition pattern.

Take \(g(x)=x^2+4\), giving \(g'(x)=2x\). The outer function is \(F(u)=\ln u\), whose derivative is \(F'(u)=1/u\) for \(u>0\). Since \(x^2+4>0\) for every real \(x\), the logarithm is defined throughout the real line. The integrand is \(F'(g(x))g'(x)\).

$$ \int \frac{2x}{x^2+4}\,dx =\ln(x^2+4)+C. $$

Check:

$$ \frac{d}{dx}\ln(x^2+4) =\frac{1}{x^2+4}\cdot 2x =\frac{2x}{x^2+4}. $$

The domain condition is satisfied, and differentiation confirms the answer.

Worked Example: An Exponential with a Constant Inner Derivative

Evaluate \(\int e^{5x-2}\,dx\).

Here \(g(x)=5x-2\), so \(g'(x)=5\). The outer function is \(F(u)=e^u\), for which \(F'(u)=e^u\). The integrand includes \(e^{g(x)}\), but it does not include the factor \(g'(x)=5\). Since \(e^{g(x)}=\frac{1}{5}e^{g(x)}g'(x)\), a factor of \(1/5\) is needed in the antiderivative.

$$ \int e^{5x-2}\,dx =\frac{1}{5}e^{5x-2}+C. $$

Check:

$$ \frac{d}{dx}\left(\frac{1}{5}e^{5x-2}\right) =\frac{1}{5}e^{5x-2}\cdot 5 =e^{5x-2}. $$

The constant factor precisely compensates for the inner derivative.

Worked Example: A Power of a Cubic Expression

Recognize and evaluate \(\int 3x^2(1+x^3)^4\,dx\).

Choose \(g(x)=1+x^3\), so \(g'(x)=3x^2\). The outer expression is the fourth power. A suitable outer function is \(F(u)=u^5/5\), since \(F'(u)=u^4\). Therefore \(F'(g(x))g'(x)\) is exactly \((1+x^3)^4(3x^2)\), the given integrand.

$$ \int 3x^2(1+x^3)^4\,dx =\frac{(1+x^3)^5}{5}+C. $$

Check:

$$ \frac{d}{dx}\left(\frac{(1+x^3)^5}{5}\right) =\frac{1}{5}\cdot 5(1+x^3)^4\cdot 3x^2 =3x^2(1+x^3)^4. $$

The derivative matches every factor in the integrand.

How to Read a Candidate Pattern

A reliable first pass is to identify a plausible inner expression and differentiate it. Then ask whether that derivative is represented among the remaining factors. Finally, identify an outer function whose derivative matches the function of the inner expression.

The match need not appear in a particular written order. For example, multiplication is commutative, so \(g'(x)F'(g(x))\) is the same product as \(F'(g(x))g'(x)\). But the factors must still be present, perhaps with a constant adjustment. A visual resemblance alone is not enough.

Nested expressions can have more than one layer. For instance, in \(\sin((x^2+1)^3)\), the outermost function is sine, and its input is \((x^2+1)^3\). That input is itself a composition. For the full derivative, the factors from each layer of the chain rule must be accounted for. This layered view helps prevent overlooking part of the inner derivative.

A Common Mistake: Forgetting the Inner Derivative

A frequent error is to take an antiderivative of the visible outer-function pattern while ignoring how quickly its input changes. For example, the calculation \(\int \cos(4x)\,dx=\sin(4x)+C\) is incorrect. Differentiating the proposed answer gives

$$ \frac{d}{dx}\sin(4x)=4\cos(4x), $$

which is four times the given integrand, not the integrand itself. Since the inner derivative is \(4\), the corrected calculation is

$$ \int \cos(4x)\,dx=\frac{1}{4}\sin(4x)+C. $$

Indeed, \(\frac{d}{dx}\left(\frac{1}{4}\sin(4x)\right)=\frac{1}{4}\cdot 4\cos(4x)=\cos(4x)\). Differentiating a proposed answer is a direct way to expose a missing factor.

When the Pattern Is Not Present

Do not force a composition match when the inner derivative is absent and cannot be supplied by a constant factor. For example, \(x\cos(x^2)\) contains the cosine of \(x^2\), but the derivative of the inner function \(x^2\) is \(2x\), not \(x\). In this case, the integrand is one-half of \(\cos(x^2)\cdot 2x\), so the factor is present up to a constant, and the match does work.

By contrast, \(\cos(x^2)\) alone does not contain the factor \(2x\), nor can a constant multiple turn \(\cos(x^2)\) into \(2x\cos(x^2)\) for every \(x\). The direct pattern described here therefore does not apply as written. Recognizing this limitation is part of recognizing the method accurately; a similar-looking outer function does not guarantee a chain-rule match.

Why This Recognition Is Useful

Composition patterns occur throughout integration because derivatives of nested functions naturally contain both an outer derivative and an inner derivative. Spotting those paired pieces lets you see an antiderivative without being distracted by the complexity of the inner expression.

For a proposed match, keep three questions in view:

  • What is a plausible inner function \(g(x)\)?
  • What outer function \(F\) has a derivative matching the remaining expression?
  • Does the integrand contain \(g'(x)\), possibly after accounting for a constant factor?

If the answer to all three is yes, the candidate is \(F(g(x))+C\), with any necessary constant adjustment. Verify it by differentiation before accepting it.

Check Your Understanding

For each question, identify the proposed inner function and use differentiation to assess the pattern.

  1. In \(\int 10x\,e^{5x^2}\,dx\), what are the inner and outer functions, and what antiderivative do they suggest?
  2. Why does \(\int \cos(7x)\,dx\) require a constant factor in front of \(\sin(7x)\)?
  3. For \(\int \frac{4x^3}{x^4+2}\,dx\), identify the inner function and the outer derivative pattern.
  4. Does \(x^2\sin(x^3)\) contain the derivative of the inner function \(x^3\) up to a constant multiple? Explain.
  5. What calculation verifies that a proposed composition antiderivative is correct?

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