Linear Numerators Over Irreducible Quadratics

Why the Numerator Should Be Rewritten

A linear numerator over a quadratic denominator often contains two different kinds of contributions. One part may be proportional to the derivative of the denominator, producing a logarithm. What remains is a constant numerator, whose integral over an irreducible quadratic produces an inverse tangent.

The useful step is to separate these contributions before integrating. The method applies to a real quadratic with no real roots, so its denominator never vanishes on the real line. It works whether the leading coefficient is positive or negative.

Conditions and the Numerator Decomposition

Let \(Q(x)=ax^2+bx+c\), where \(a\ne 0\). Its discriminant is \(\Delta=b^2-4ac\). The quadratic is irreducible over the real numbers precisely when \(\Delta<0\). In that case \(Q(x)\ne 0\) for every real \(x\), and \(Q\) has the same sign as \(a\) everywhere.

The derivative of the denominator is \(Q'(x)=2ax+b\). For a linear numerator \(mx+n\), seek constants \(A\) and \(B\) such that \(mx+n=AQ'(x)+B\). Matching the coefficients of \(x\) and the constant terms gives \(2aA=m\) and \(bA+B=n\). Since \(a\ne0\), these equations have the solution below.

This identity splits the original integral into two simpler ones. The first has numerator \(Q'(x)\); the second has a constant numerator.

Because \(Q\) never vanishes, the logarithmic term is \(\ln|Q(x)|\). The absolute value is important when \(a<0\), since then \(Q(x)<0\) for every real \(x\).

The Reciprocal-Quadratic Term

For completeness, the inverse-tangent formula needed for the constant part follows directly from completing the square. Let \(D=-\Delta=4ac-b^2\), so \(D>0\). Set \(w=2ax+b\). Then \(dw=2a\,dx\), and the quadratic can be rewritten as \(Q(x)=(w^2+D)/(4a)\). Therefore:

The last equality follows by differentiating \(\arctan(w/\sqrt D)\), or by the standard derivative of the inverse tangent. This substitution does not require \(a>0\): the sign of \(a\) is already accounted for by \(dw=2a\,dx\).

Here \(C\) is an arbitrary constant. The formula combines the numerator decomposition with the reciprocal-quadratic integral. In particular, the inverse-tangent coefficient is the constant left after the decomposition, \(n-mb/(2a)\), multiplied by \(2/\sqrt{-\Delta}\).

Why the Formula Works

The key derivative identities explain both terms and verify the combined result. Since \(Q(x)\ne0\), differentiating \(\ln|Q(x)|\) gives \(Q'(x)/Q(x)\). For the inverse-tangent term, write \(D=-\Delta>0\) and \(w=2ax+b\). Its derivative is:

The final equality uses \(Q(x)=(w^2+D)/(4a)\), which holds for positive and negative \(a\). Thus the derivative of the proposed antiderivative is \(A Q'(x)/Q(x)+B/Q(x)\), where \(A=m/(2a)\) and \(B=n-mb/(2a)\). By the numerator identity, this is exactly \((mx+n)/Q(x)\).

Worked Examples

Common Mistake: Keeping Only the Logarithm

A frequent error is to notice the denominator’s derivative but assume the entire numerator is a multiple of it. For the first example, a mistaken calculation might write:

Differentiating this proposed answer gives \((3x+6)/(x^2+4x+13)\), not the original integrand with numerator \(3x+5\). The discrepancy is one unit in the numerator. The correction is to retain the leftover constant: \(3x+5=\frac32(2x+4)-1\). That \(-1\) produces the inverse-tangent term \(-\frac13\arctan((x+2)/3)\).

Another source of errors is changing the inverse-tangent coefficient when rewriting its argument. It is safest to keep the form \(\arctan((2ax+b)/\sqrt{-\Delta})\) until the formula has been applied. If the argument is then simplified, its derivative must still be accounted for in any direct check.

When This Method Is Useful

This method is especially useful when the numerator has degree one and the denominator is an irreducible quadratic. It identifies the two natural pieces directly, without requiring a more complicated substitution. The denominator’s derivative determines the logarithmic part; the remainder determines the size, sign, and presence of the inverse-tangent part. If the remainder is zero, the inverse-tangent contribution vanishes. If the numerator is already constant, the logarithmic contribution vanishes.

The discriminant condition is essential. If \(\Delta<0\), the denominator has no real zeros and the inverse-tangent formula applies on the whole real line. If \(\Delta\ge0\), the quadratic may have real zeros, and the integral can require a different form and careful attention to intervals that avoid those zeros. The formula here is specifically for irreducible real quadratics.