Why a Constant Factor Matters
Substitution is a way to recognize the reverse chain rule. If an integrand contains a function of an inner expression and also contains that expression's derivative, the change of variable can turn the integral into a familiar antiderivative.
Sometimes the factor present in the integrand is not exactly the inner derivative. It may be a constant multiple of that derivative. The substitution still works, but the constant multiple must be carried through the calculation. Losing it—or using its reciprocal in the wrong direction—changes the answer.
The Constant-Factor Rule
Suppose the integrand contains \(F(g(x))\), where \(g\) is the inner function, and suppose the remaining factor is a constant multiple \(k g'(x)\). Let \(H\) be an antiderivative of \(F\), so \(H'(u)=F(u)\) on the relevant range. Then the substitution \(u=g(x)\) gives:
Here \(k\) is a constant, and \(C\) is an arbitrary constant of integration. The equation applies on an interval where the functions and antiderivatives involved are defined and the substitution is valid. The key bookkeeping step is that \(du=g'(x)\,dx\), so the factor \(k\) remains outside the integral.
A useful equivalent form begins with an available factor \(a(x)\) satisfying \(a(x)=k g'(x)\). The integrand then has the form \(F(g(x))a(x)\), and the result is \(kH(g(x))+C\). If instead \(a(x)=g'(x)/k\), the multiplier is \(1/k\), not \(k\).
Why the Rule Works
The rule follows directly from differentiation. Suppose \(H'=F\). By the chain rule, the derivative of \(H(g(x))\) is \(F(g(x))g'(x)\). Multiplying this expression by \(k\) gives precisely the integrand with factor \(k g'(x)\).
Therefore \(kH(g(x))\) is an antiderivative of \(kF(g(x))g'(x)\). All antiderivatives on the same interval differ by a constant, which accounts for \(+C\). This differentiation argument both explains the scaling and gives a reliable way to check the final result.
Worked Example: A Power Antiderivative with a Larger Available Factor
Worked Example: A Power Antiderivative with a Larger Available Factor
Evaluate \(\displaystyle \int 6x(x^2+5)^4\,dx\).
The inner function is \(g(x)=x^2+5\), so its derivative is \(g'(x)=2x\). The factor \(6x\) in the integrand is three times that derivative:
Set \(u=x^2+5\). Then \(du=2x\,dx\), and the integral becomes:
The factor is \(3\), not \(1/3\), because the available factor is three times \(du\). Check by differentiating:
This is the original integrand, so the antiderivative is verified.
Worked Example: A Smaller Available Factor
Worked Example: A Smaller Available Factor
Evaluate \(\displaystyle \int x(4+x^2)^3\,dx\).
Choose \(u=4+x^2\). Then \(du=2x\,dx\), which means \(x\,dx=\tfrac12\,du\). The available factor is half the inner derivative, so the substitution contributes a factor of \(\tfrac12\):
Check the result:
The derivative recovers the original integrand. Notice that the multiplier is \(\tfrac12\), since \(x\,dx=\tfrac12\,du\).
Worked Example: A Logarithmic Antiderivative
Worked Example: A Logarithmic Antiderivative
Evaluate \(\displaystyle \int \frac{4x}{x^2+9}\,dx\).
Let \(u=x^2+9\). Then \(du=2x\,dx\), and \(4x\,dx=2\,du\). Thus the available numerator factor is twice the inner derivative:
The logarithmic antiderivative uses \(\int \frac{1}{u}\,du=\ln|u|+C\), for \(u\ne 0\). Here \(x^2+9>0\) for every real \(x\), so \(\ln|x^2+9|=\ln(x^2+9)\). Differentiate to check:
The result differentiates to the stated integrand.
A Reliable Multiplier Check
Before integrating in \(u\), write the relationship between the factor in the integral and \(du\) explicitly. For example, if \(du=2x\,dx\) but the integral contains \(6x\,dx\), write \(6x\,dx=3\,du\). This line makes the multiplier visible before any antiderivative rule is used.
Write \(u=g(x)\) and calculate \(du=g'(x)\,dx\).
Express the available factor as a constant times \(g'(x)\,dx\), or express it as a constant times \(du\).
Keep that multiplier outside the integral in \(u\), and then use the appropriate antiderivative rule.
Replace \(u\) by \(g(x)\), include \(+C\), and differentiate the result to check it.
When This Adjustment Does Not Apply
This rule requires a constant ratio between the available factor and the inner derivative. If the factor is not a constant multiple of \(g'(x)\), you cannot use this particular adjustment to make the substitution match. A changing ratio is not an outside constant and cannot simply be moved outside the integral.
For instance, if \(g'(x)=2x\) but the available factor is \(x+1\), the ratio \((x+1)/(2x)\) is not constant on an interval. The constant-factor rule does not turn \(x+1\) into a constant multiple of \(2x\). That does not by itself prove that no method can evaluate an integral; it means only that this specific multiplier adjustment is not justified.
Common Mistake: Using the Reciprocal in the Wrong Direction
Consider the integral \(\int 6x(x^2+5)^4\,dx\). Since \(d(x^2+5)=2x\,dx\), a common incorrect calculation is:
This incorrectly uses the reciprocal of the multiplier. In fact, \(6x\,dx=3\,du\), not \(\tfrac13\,du\). The correction is:
Differentiating the incorrect result exposes the error:
which is not \(6x(x^2+5)^4\). Differentiating the corrected result gives \(6x(x^2+5)^4\), as required. A derivative check is especially useful when the multiplier is easy to reverse accidentally.
Why the Constant Is Not Optional
The constant factor changes the size of the derivative, so it changes the size of the antiderivative needed to recover the integrand. If \(G'(x)\) equals the desired integrand divided by \(k\), then \(kG(x)\), rather than \(G(x)\), has the desired derivative:
This is the same scaling principle behind the substitution rule. It is also why an answer that looks structurally correct can still be wrong by a constant factor. The constant of integration \(C\) does not fix such an error: adding \(C\) changes the value of an antiderivative by a constant, but it does not change its derivative.
Check Your Understanding
For each question, pay attention to the relationship between the available factor and the inner derivative.
- In \(\int 8x(x^2+3)^2\,dx\), what constant relates \(8x\,dx\) to \(d(x^2+3)\)?
- Evaluate \(\int 2x(1+x^2)^5\,dx\), and verify the result by differentiation.
- For \(\int \frac{6x}{x^2+4}\,dx\), what multiplier appears after setting \(u=x^2+4\)?
- If the inner derivative is \(4x\) but the available factor is \(x\,dx\), what constant multiple of \(du\) is available?
- Why does the constant-factor rule not apply directly if the ratio of the available factor to the inner derivative varies with \(x\)?
