Integrating Logarithms Of Affine Expressions

Integrating a Logarithm with an Affine Argument

The natural logarithm of a linear expression, \(\ln(ax+b)\), appears constantly in applications. The integral \(\int \ln x\,dx\) is already a classic use of integration by parts, because \(\ln x\) alone has no obvious antiderivative. The affine version adds one extra step: the derivative of the inside function contributes a factor \(a\). In this tutorial \(a\ne0\) and \(b\) are constants, and the integral is considered on an interval where \(ax+b>0\), since the logarithm requires a positive argument.

Derivation by Parts

Let \(u=\ln(ax+b)\) and \(dv=dx\). The chain rule gives \(du=\frac{a}{ax+b}\,dx\), and \(v=x\). Then

The new integral is a rational function whose numerator and denominator have the same degree. Rewrite the numerator by adding and subtracting \(b\): \(ax=(ax+b)-b\). Then

Integrating term by term, \(\int 1\,dx=x\) and \(\int\frac{b}{ax+b}\,dx=\frac{b}{a}\ln(ax+b)\), the latter because the argument is positive and the derivative of \(ax+b\) is \(a\). So \(\int\frac{ax}{ax+b}\,dx=x-\frac{b}{a}\ln(ax+b)\), and

The two logarithmic terms combine because \(x+\frac ba=\frac{ax+b}{a}\). Notice the pleasant form of the answer: the factor in front of the logarithm is the original affine expression divided by \(a\), exactly as in the basic result \(\int\ln x\,dx=x\ln x-x\).

A Shortcut by Substitution

The same result follows from a substitution. Let \(w=ax+b\), so \(dw=a\,dx\). Then \(\int\ln(ax+b)\,dx=\frac1a\int\ln w\,dw=\frac1a\bigl(w\ln w-w\bigr)+C=\frac{ax+b}{a}\ln(ax+b)-\frac{ax+b}{a}+C\). The last term equals \(-x-\frac ba\), and the constant \(-\frac ba\) can be absorbed into \(C\). The two answers differ only by a constant, which is exactly what two correct antiderivatives may do. This agreement is a useful consistency check on both methods.

Worked Examples

Common Mistake: Dropping the Factor from the Chain Rule

A frequent error is to differentiate \(\ln(ax+b)\) as \(\frac{1}{ax+b}\), forgetting the inner derivative \(a\). That produces \(du=\frac{1}{ax+b}\,dx\) and the incorrect leftover integral \(\int\frac{x}{ax+b}\,dx\). The resulting antiderivative differs from the correct one by a factor of \(a\) in the second term. A derivative check detects the problem: the proposed answer will not differentiate back to \(\ln(ax+b)\) unless the chain-rule factor is present.

A second common slip is to treat the negative-slope case carelessly. With \(a=-1\), the fraction \(\frac{ax+b}{a}\) becomes \(-(ax+b)\), and forgetting that minus sign changes the answer. Writing the result in terms of the original expression, as above, makes the sign easy to track.

Domain and Interpretation

The antiderivative is valid only where \(ax+b>0\). If an interval contains the point \(x=-b/a\), where the argument is zero, the integrand has a vertical asymptote there. The integral is improper at that point and converges, because \(\ln\) is integrable near zero. If the argument is negative, one must use \(\ln|ax+b|\), and the antiderivative takes the same form with absolute values: \(\frac{ax+b}{a}\ln|ax+b|-x+C\).

A Reliable Procedure

A fixed sequence of steps makes these problems routine, regardless of the signs of \(a\) and \(b\).

Relation to the Basic Logarithm Integral

Setting \(a=1\) and \(b=0\) in the formula gives \(\int\ln x\,dx=x\ln x-x+C\), the basic result for the logarithm. Every affine version is a stretched and shifted copy of this one. The shift \(b\) moves the point where the logarithm vanishes, while the slope \(a\) both stretches the horizontal scale and divides the antiderivative by \(a\), because \(\int f(ax+b)\,dx=\frac1aF(ax+b)\) whenever \(F'=f\). Recognizing this scaling rule lets you write the answer without any parts at all once the basic integral is known.

This is also why the answer is naturally written in terms of the affine expression. The form \(\frac{ax+b}{a}\ln(ax+b)-x\) displays the same structure as \(w\ln w-w\), with \(w=ax+b\), up to a constant. When you are asked to express the result using the original argument, keep \(ax+b\) intact rather than expanding it.

Quick Reference

An Improper Endpoint

The antiderivative behaves well at the point where the argument vanishes. Since \(w\ln w\to0\) as \(w\to0^+\), the limit of \(\frac{ax+b}{a}\ln(ax+b)\) at \(x=-b/a\) is zero, so the improper integral converges. For example,

The value is negative because \(\ln(x+1)<0\) on \((-1,0)\), where the argument lies between \(0\) and \(1\). The integrand has a vertical asymptote at \(x=-1\), yet the area is finite. This is a common surprise: logarithmic singularities are integrable, in contrast with a pole such as \(1/(x+1)\).

A Final Habit: Check the Domain First

Before integrating \(\ln(ax+b)\), confirm that \(ax+b>0\) on the whole interval. If the interval contains the point \(x=-\frac ba\), the integrand is undefined there and the integral cannot be evaluated by a simple substitution of endpoints. This check takes a few seconds, and it prevents the most common error with logarithms of affine expressions: using the antiderivative on an interval where the integrand does not exist.

Where This Fits in the Course

This tutorial belongs to the module “Integration by Parts I”. The tutorial just before it is “Power-Weighted Logarithm Integrals,” and the one that follows is “Integrating The Logarithm Of A Quadratic.” Here the logarithm's argument is a general linear expression, so the chain rule contributes an extra factor.