Integrating Simple Linear Partial Fractions

Why Linear Partial Fractions Produce Logarithms

A partial fraction decomposition can turn a rational function into a sum of simpler terms. When one of those terms has a constant numerator and a linear denominator, its antiderivative is a logarithm. The key detail is that the coefficient of the logarithm must account for the slope of the denominator.

For example, \(1/x\) has antiderivative \(\ln|x|\) wherever \(x\ne 0\). But if the denominator is \(3x-2\), differentiating \(\ln|3x-2|\) produces an extra factor of \(3\). Correcting for that factor is the central rule of this tutorial.

The Antiderivative Rule

For constants \(A,a,b\) with \(a\ne 0\), the rule is:

Here \(C\) is an arbitrary constant on the interval where the antiderivative is considered. The restriction \(ax+b\ne 0\) excludes the point where the denominator vanishes. Since \(a\ne 0\), that point is \(x=-b/a\).

The absolute value makes the logarithm real-valued on both sides of the excluded point. For any nonzero real number \(u\), \(\ln|u|\) is defined whether \(u\) is positive or negative. An antiderivative is considered on an interval that does not cross the point where the denominator is zero; the constant of integration may be chosen separately on different such intervals.

Why the Rule Works

The rule follows from the derivative of the logarithm and the chain rule. If \(u(x)\ne 0\), then \(\frac{d}{dx}\ln|u(x)|=\frac{u'(x)}{u(x)}\). To see why the absolute value causes no change to this derivative, consider the two possible signs of \(u\). If \(u>0\), then \(\ln|u|=\ln u\), whose derivative is \(u'/u\). If \(u<0\), then \(\ln|u|=\ln(-u)\), whose derivative is \((-u)'/(-u)=u'/u\). Thus the same derivative formula holds wherever \(u\ne 0\).

Apply this fact to \(u(x)=ax+b\), whose derivative is \(a\). Then:

This verifies that the proposed logarithm is an antiderivative on any interval where \(ax+b\ne 0\). The factor \(1/a\) is necessary because differentiating the logarithm introduces the factor \(a\) from the linear expression inside it.

Worked Examples

Using the Rule on a Sum

Once a rational function has been expressed as a sum of simple linear partial fractions, integrate the terms one at a time. If the decomposition is \(\sum_{i=1}^{n}\frac{A_i}{a_i x+b_i}\), with each \(a_i\ne 0\), then linearity of integration gives:

This formula applies on intervals where every denominator in the sum is nonzero. The decomposition itself does not change the excluded points: they are the real zeros of the original denominator, after any valid cancellation has been taken into account. Each logarithm can be checked independently by differentiation, and the derivative of the sum is the sum of those derivatives.

The rule concerns terms whose numerators are constants and whose denominators are linear. It does not by itself integrate terms with linear numerators over quadratic denominators or constant numerators over irreducible quadratic denominators. Those require different antiderivative forms.

Common Pitfall: Forgetting the Slope Factor

A frequent error is to treat \(\ln|ax+b|\) as though its derivative were \(1/(ax+b)\). The chain rule shows that this misses a factor of \(a\). For example, the following calculation is wrong:

Differentiating the proposed expression reveals the error:

which is three times the required integrand. To correct the coefficient, divide by the slope \(3\):

Differentiating this corrected expression gives \(\frac{7}{3}\cdot\frac{3}{3x-2}=\frac{7}{3x-2}\), as required. A useful habit is to differentiate every proposed logarithmic antiderivative, rather than relying on the appearance of the formula alone.

Absolute Values and Excluded Points

Omitting the absolute value can make a real antiderivative undefined on part of the domain. For instance, \(\ln(3x-2)\) is real only when \(x>2/3\). But the integrand \(7/(3x-2)\) is also defined for \(x<2/3\). The expression \(\ln|3x-2|\) works on both sides, with the singular point \(x=2/3\) excluded.

That does not mean an antiderivative can be continued through a denominator zero. The integrand is undefined there, and the logarithmic expression is also singular there. Instead, consider each interval of the domain separately. In the first example, these intervals are \((-\infty,2/3)\) and \((2/3,\infty)\). On either interval, differentiation verifies the same formula; the arbitrary constant may be different on the two intervals.