Logarithmic Terms From Linear Quadratic Numerators

Why a Derivative-Matched Numerator Produces a Logarithm

Partial fractions can produce terms whose denominators are irreducible quadratics and whose numerators are linear. A particularly direct case occurs when the numerator is a constant multiple of the denominator’s derivative. The chain rule then turns the integral into a logarithm.

The key is to compare the numerator with the derivative of the entire quadratic denominator—not merely to notice that both contain \(x\). A missing numerical factor changes the antiderivative, so the comparison must be explicit.

What “Irreducible Quadratic” Means Here

In partial fractions over the real numbers, an irreducible quadratic is a quadratic polynomial that cannot be factored into real linear factors. For \(q(x)=ax^2+bx+c\), with \(a\ne 0\), this is equivalent to its discriminant being negative:

Such a quadratic has no real zeros, so its reciprocal is defined for every real \(x\). It also has a constant sign: if it changed sign, continuity would force it to equal zero somewhere. The absolute value in \(\ln|q(x)|\) therefore makes the logarithm valid whether \(q\) is positive or negative. In many partial-fraction examples the quadratic is positive, but the formula does not require that.

The relevant derivative is

Thus the numerator must have exactly the same \(x\)-term and constant-term ratio as \(2ax+b\), up to one overall constant multiplier. Finding that multiplier is the main algebraic step.

Deriving the Logarithmic Antiderivative

The formula follows directly from the chain rule and the derivative of the natural logarithm. If \(q(x)\ne 0\), then \(\frac{d}{dx}\ln|q(x)|=\frac{q'(x)}{q(x)}\). To see why the absolute value causes no difficulty, consider an interval where \(q(x)\) has a fixed sign. If \(q(x)>0\), the derivative is \(q'(x)/q(x)\) by the usual chain rule. If \(q(x)<0\), then \(|q(x)|=-q(x)\), so \[ \frac{d}{dx}\ln|q(x)| =\frac{1}{-q(x)}(-q'(x)) =\frac{q'(x)}{q(x)}. \] The formula therefore holds on any interval on which \(q\) does not vanish.

This is a substitution pattern in which the expression inside the logarithm is the denominator itself. The constant multiplier must remain in front of the logarithm: it is the factor \(\lambda\) from the numerator, not an optional adjustment.

A Coefficient Test for the Match

Suppose the numerator is \(Ax+B\), while the denominator is \(ax^2+bx+c\), with \(a\ne 0\). The numerator is proportional to the derivative precisely when there is a constant \(\lambda\) such that

Comparing the coefficients of \(x\) gives \(A=2a\lambda\), so \(\lambda=A/(2a)\). Comparing the constant terms then requires \(B=bA/(2a)\). Equivalently, the exact condition is

When this condition holds, the logarithmic antiderivative is

The coefficient test prevents a common oversight: matching only one term is not enough. Both the \(x\)-coefficient and the constant term must use the same multiplier. If the condition fails, the entire numerator is not a constant multiple of the denominator’s derivative. A general linear numerator can be separated into a derivative-matched part and a remaining constant part, but only the matched part gives the logarithmic term described here.

Worked Examples

When the Match Fails

Consider the numerator \(3x+2\) over the denominator \(3x^2+4x+7\). The derivative of the denominator is \(6x+4\). For \(3x+2\) to be a multiple of \(6x+4\), matching the \(x\)-terms would require a multiplier of \(1/2\). But \[ \frac12(6x+4)=3x+2, \] so in this particular case the match does hold. This illustrates why checking both terms matters: the constant term confirms the same multiplier.

By contrast, if the numerator were \(3x+1\), matching the \(x\)-terms would still suggest \(1/2\), but \[ \frac12(6x+4)=3x+2\ne3x+1. \] That numerator is not proportional to the denominator’s derivative. Applying the logarithmic formula to the whole fraction would therefore be incorrect. The derivative-matched logarithmic rule applies only when the equality of the two linear expressions is exact.

Common Mistake: Losing the Multiplier

A frequent error is to write \(\ln|q(x)|\) whenever a numerator resembles \(q'(x)\), without checking whether the two expressions are exactly equal. For example, consider \[ \int\frac{3x+2}{3x^2+4x+7}\,dx. \] The denominator derivative is \(6x+4\), and \[ 3x+2=\frac12(6x+4). \] The correct antiderivative is therefore \[ \frac12\ln|3x^2+4x+7|+C. \] If one incorrectly writes \(\ln|3x^2+4x+7|\), differentiation gives \[ \frac{6x+4}{3x^2+4x+7}, \] which is twice the required integrand. Differentiating the corrected answer gives \[ \frac12\cdot\frac{6x+4}{3x^2+4x+7} =\frac{3x+2}{3x^2+4x+7}, \] as required. Differentiation is a direct way to catch this coefficient error.

Why This Pattern Matters

In a partial-fraction decomposition, each term is integrated separately. A term with an irreducible quadratic denominator and a linear numerator may contribute a logarithm when its numerator is proportional to the denominator’s derivative. Recognizing that structure makes the integration efficient and also gives a reliable verification: differentiating the logarithmic term reproduces the original rational term.

The important distinction is between the form of the denominator and the relationship between numerator and derivative. An irreducible quadratic alone does not guarantee that the whole term integrates to a logarithm. The numerator must match \(q'(x)\) up to a constant factor. When it does, identify the factor, keep it in front of the logarithm, and check by differentiation.