When the Inner Derivative Almost Matches
Substitution works most directly when an integrand contains an inner function together with its derivative. Sometimes the derivative is present, but only after a constant adjustment. For instance, the derivative of \(x^2+4\) is \(2x\), while an integrand might contain \(x\). The factor \(x\) is half of \(2x\), so the substitution still applies, provided that the factor of one-half is carried through the calculation.
The key is to distinguish the function of the inner expression from the factor that changes the variable. If that changing factor is a constant multiple of the inner derivative, the constant remains in the integral. Depending on how the integrand is written, it may appear as a multiplier or as a divisor.
The Scaling Rule
Let \(g\) be differentiable on an interval \(I\), and suppose \(h\) has an antiderivative \(H\) on the values \(g(x)\) takes. This means \(H'(u)=h(u)\) for those values of \(u\). The chain rule gives a precise rule for an integrand that contains a scaled copy of \(g'(x)\).
Here \(C\) is an arbitrary constant. In the common near-match form, the integrand contains \(q(x)h(g(x))\), where \(q(x)=c\,g'(x)\). If \(c\ne 0\), the same statement can be written as
These forms express the same idea. If the factor present is \(c\,g'(x)\), keep \(c\) as a multiplier. If the factor present is \(g'(x)/c\), keep \(1/c\) as a multiplier. In particular, do not discard a constant merely because the remaining expression has the right substitution pattern.
Why the Rule Works
The justification is a direct application of the chain rule. Differentiate the proposed antiderivative \(cH(g(x))\). Since \(H'(u)=h(u)\), differentiating the composition gives:
This is exactly the integrand \(c\,g'(x)h(g(x))\). Therefore \(cH(g(x))\) is an antiderivative, and the indefinite integral includes an arbitrary constant \(C\). The constant factor is not a separate trick: it is necessary for the derivative of the proposed answer to reproduce the original integrand.
The equivalent differential notation makes the same scaling visible. If \(u=g(x)\), then \(du=g'(x)\,dx\). Thus, if the original factor is \(g'(x)/c\), the differential portion becomes \(du/c\), not \(du\). The factor \(1/c\) must remain when the integral is rewritten in terms of \(u\).
A Reliable Procedure
When the derivative match is off by a constant, use the following sequence. The main task is to identify and preserve the scale, rather than to manipulate the entire integrand at once.
Identify the expression \(g(x)\) that appears inside another function.
Compute \(g'(x)\) and compare it with the factor multiplying the composition.
Rewrite the factor as \(c\,g'(x)\), or equivalently as \(g'(x)/k\) for a nonzero constant \(k\).
Integrate the remaining function of \(u=g(x)\), keeping \(c\) or \(1/k\) outside the integral.
Replace \(u\) by \(g(x)\), include \(C\), and differentiate the result to compare with the original integrand.
Worked Examples
Worked Example: The Integrand Has Half the Inner Derivative
Evaluate \(\int x(x^2+4)\,dx\). The inner expression is \(g(x)=x^2+4\), so \(g'(x)=2x\). The factor \(x\) in the integrand is half of this derivative:
Set \(u=x^2+4\). Then \(du=2x\,dx\), so \(x\,dx=\frac{1}{2}\,du\). Substitute both the inner expression and its scaled differential:
Check by differentiating the answer. With \(g(x)=x^2+4\),
The derivative matches the original integrand.
Worked Example: One Third of the Inner Derivative
Evaluate \(\int (x^2+1)\bigl((x^3+3x)^2+1\bigr)\,dx\). Choose \(u=x^3+3x\). Its derivative is
The factor in the integral is therefore \((x^2+1)\,dx=\frac{1}{3}\,du\). The remaining expression becomes \(u^2+1\), so
To check, differentiate the result in its compact form. The derivative of \(u^3/9+u/3\) with respect to \(u\) is \(u^2/3+1/3=(u^2+1)/3\), while \(du/dx=3(x^2+1)\). Thus
This is the original integrand after substituting \(u=x^3+3x\), so the result checks.
Worked Example: Three Times the Inner Derivative
Evaluate \(\int 6x\bigl((x^2+1)^2-2(x^2+1)\bigr)\,dx\). Let \(u=x^2+1\). Then \(du=2x\,dx\), and \(6x\,dx=3\,du\). The factor is three times the inner derivative, so the integral becomes
Differentiate the answer. The derivative of \(u^3-3u^2\) with respect to \(u\) is \(3u^2-6u\), and \(du/dx=2x\). Therefore
Replacing \(u\) by \(x^2+1\) gives exactly the original integrand.
The Missing-Scale Pitfall
A common error is to notice the correct inner expression and then replace its derivative factor as if it matched exactly. In the first example, for instance, it would be incorrect to write:
This calculation treats \(x\,dx\) as \(du\). But \(du=2x\,dx\), so the correct relation is \(x\,dx=du/2\). The missing factor changes the answer. Differentiating the incorrect result makes the error clear:
which is twice the required integrand. The correction is to include the missing half:
A useful habit is to check the differential relation explicitly before integrating. Writing \(du=g'(x)\,dx\) shows exactly which constant is present; it prevents the visually similar factors from being mistaken for equal ones.
What the Rule Does and Does Not Cover
The scaling rule applies when the integrand can be written as a constant times \(g'(x)h(g(x))\), with \(h\) having an antiderivative on the relevant values. The constant is independent of \(x\). If the factor multiplying \(h(g(x))\) is not a constant multiple of \(g'(x)\), this particular rule does not justify the substitution in this form. A resemblance between part of an integrand and an inner derivative is not enough; the factors must match after accounting for a constant.
For example, if \(g'(x)=4x\) and the integrand contains \(x\), then \(x=g'(x)/4\), so a factor of \(1/4\) must appear after substitution. If instead the integrand contains \(12x\), then \(12x=3g'(x)\), so the substitution carries a factor of \(3\). The direction of the adjustment depends on the actual factor in the integrand.
This method matters because many useful compositions do not display their inner derivative with coefficient exactly one. Keeping track of a constant makes the reverse chain rule applicable without changing the integrand or guessing an answer. Differentiation then provides a direct test: a correct antiderivative must return the original integrand, including its scale.
Check Your Understanding
For each question, identify the constant scale and keep it in the calculation.
- If \(u=x^2+5\), express \(x\,dx\) in terms of \(du\).
- For \(\int 4x\bigl((x^2+2)^2+1\bigr)\,dx\), what is the inner function and what constant multiplies its derivative?
- In \(\int (x^2+1)\bigl((x^3+3x)^2+1\bigr)\,dx\), why does the substitution introduce a factor of \(1/3\)?
- Differentiate \(\frac{1}{4}(x^2+4)^2\). What does the result tell you about the missing-scale error?
- State the condition on the factor multiplying \(h(g(x))\) that allows this constant-scaling rule to apply.
