Powers Of Affine Expressions

Integrating a Power of a Linear Expression

A power such as \((3x-2)^4\) is a composite function: a linear expression is raised to a power. Its antiderivative is not found by simply increasing the exponent; the slope of the inner expression must also be accounted for. The reverse chain rule gives a systematic way to do this.

This tutorial considers integrands of the form \((ax+b)^n\), where \(a\) and \(b\) are constants and \(n\) is an integer. The method works for positive exponents, for negative exponents wherever the integrand is defined, and for the exponent \(n=-1\), which requires a separate logarithmic formula.

The Formula and Its Conditions

For an integer \(n\ne -1\), the power rule and the chain rule give the following antiderivative:

When \(n=-1\), the integrand is the reciprocal of the affine expression. Its antiderivative is logarithmic:

Here \(C\) is an arbitrary constant. If \(n\) is a negative integer, \((ax+b)^n\) is defined only where \(ax+b\ne 0\). In that case, the formula applies on any interval that does not contain the point \(-b/a\). The logarithmic formula has the same restriction. For positive integer \(n\), the integrand and the displayed antiderivative are defined for every real \(x\).

If \(a=0\), the expression is constant rather than a nonconstant affine expression. When \(b^n\) is defined, the integral is simply \(b^n x+C\); the formulas involving division by \(a\) do not apply. For a negative integer exponent, this constant case also requires \(b\ne 0\).

Why the Factor \(a\) Appears

Set \(u=ax+b\). Then \(du=a\,dx\), so \(dx=\frac{1}{a}\,du\), provided \(a\ne0\). The factor \(\frac{1}{a}\) is essential: it is the adjustment for the rate at which the inner expression changes with \(x\).

For \(n\ne-1\), the power rule gives \(\int u^n\,du=\frac{u^{n+1}}{n+1}+C\). Substituting \(u=ax+b\) yields the stated formula. For \(n=-1\), the corresponding integral is \(\int \frac{1}{u}\,du=\ln|u|+C\), giving the logarithmic formula after the same factor \(\frac{1}{a}\).

The reason these formulas work can also be checked directly by differentiation. This check verifies both the power and logarithmic cases and makes the chain factor explicit.

More explicitly, for \(n\ne-1\), the chain rule gives

This calculation is valid wherever the powers involved are defined. For the logarithmic case, \(\frac{d}{dx}\ln|ax+b|=\frac{a}{ax+b}\) wherever \(ax+b\ne0\), so the factor \(\frac{1}{a}\) again leaves precisely the original integrand.

Worked Examples

A Common Mistake: Losing the Slope Factor

A frequent error is to increase the exponent and divide by the new exponent, but forget to divide by the slope \(a\). For example, the incorrect calculation

does not pass the derivative check:

The derivative is four times the integrand, so the proposed antiderivative is incorrect. The correction is to include the factor \(\frac{1}{a}=\frac14\):

Indeed, differentiating gives \(\frac{3(4x+1)^2\cdot4}{12}=(4x+1)^2\). A derivative check is especially useful here because it exposes a missing chain factor immediately.

Choosing the Correct Case

Before applying the formula, identify the exponent and check the domain. The distinction between \(n=-1\) and other integer exponents matters because the usual power rule for antiderivatives divides by \(n+1\), which is zero when \(n=-1\).

When a negative power is involved, one antiderivative formula describes a family of antiderivatives on each interval where the integrand is defined. The constants on the two sides of the excluded point need not be the same, because those intervals are separate domains.

Summary

For a nonconstant affine expression \(ax+b\), substitution gives \(du=a\,dx\), so every formula includes the factor \(\frac{1}{a}\). For integer powers other than \(-1\), increase the exponent by one and divide by both the new exponent and \(a\). For exponent \(-1\), use the logarithm \(\frac{1}{a}\ln|ax+b|\). Negative powers and the logarithmic case are defined only away from the zero of the affine expression. Differentiating the proposed antiderivative checks the result and confirms that no factor has been lost.