Integrating Linear Numerators Over Quadratics

Why the Numerator and Denominator Fit Together

A rational term with a linear numerator over an irreducible quadratic has a useful structure: the derivative of the denominator is itself linear. Part of the numerator can therefore be integrated by the logarithm rule. What remains is a constant divided by the quadratic, which leads to an arctangent after completing the square.

This method is useful when integrating terms such as \((mx+n)/(ax^2+bx+c)\), where the quadratic has no real roots. It produces a compact formula, but the coefficient of the arctangent depends on the quadratic’s scale. Keeping track of that scale is essential.

For such a quadratic, \(4ac-b^2>0\). Its denominator never vanishes on the real line, so the rational function is defined for every real \(x\). We will use the positive number \(S=\sqrt{4ac-b^2}\) to keep the formula readable.

Separate the Derivative Part

Let \(m,n,a,b,c\) be real constants, with \(a\ne 0\) and \(b^2-4ac<0\). The denominator and its derivative are

Because \(q'\) is linear, we can express any linear numerator \(mx+n\) as a multiple of \(q'\) plus a constant. Matching the coefficient of \(x\) determines that multiple; subtracting it leaves the constant remainder.

Write \(R=n-\frac{mb}{2a}\) for the remainder. Dividing by \(q(x)\) separates the integrand into two terms: \(\frac{m}{2a}\frac{q'(x)}{q(x)}\) and \(\frac{R}{q(x)}\). The first integrates to a logarithm. To integrate the second, complete the square in the quadratic.

Deriving the General Formula

Set \(S=\sqrt{4ac-b^2}>0\). Completing the square, or expanding the right-hand side, gives

In the remaining integral, substitute \(u=2ax+b\). Then \(du=2a\,dx\). Using the standard antiderivative of \(1/(u^2+S^2)\) yields the following result.

Here \(C\) is an arbitrary constant. Since the quadratic has no real zeros, it has a constant sign. The absolute value in the logarithm makes the formula valid whether that sign is positive or negative.

For clarity, the same result can be written using the remainder \(R=n-\frac{mb}{2a}\): the arctangent coefficient is \(2R/S\). The two forms agree because \(2R=2n-\frac{mb}{a}\).

Why the Formula Works

The argument establishes the formula directly. First, the numerator identity gives

Because \(q\) is nonzero on the real line, the derivative of \(\ln|q(x)|\) is \(q'(x)/q(x)\). For the other term, the completed-square identity and substitution \(u=2ax+b\) give

Multiplying this last result by \(R\) produces the arctangent term \(2R/S\). Together, the two integrated parts give the stated expression. This also explains why both the constant remainder and the positive square root \(S\) matter: the former sets the arctangent’s size, and the latter sets its scale.

Worked Examples

A Common Scaling Mistake

In the first example, completing the square gives \(x^2+4x+13=(x+2)^2+9\). The correct remainder integral is

A common error is to omit the factor \(1/3\) and write just \(\arctan((x+2)/3)\). That is incorrect: differentiating the proposed expression gives \(3/((x+2)^2+9)\), three times the desired integrand. With the factor \(1/3\), the derivative is \(1/((x+2)^2+9)\). In the full example the remainder is \(2\), so its contribution is \(\frac{2}{3}\arctan((x+2)/3)\), not \(2\arctan((x+2)/3)\).

When to Use This Method

Use this formula when the denominator is a single irreducible real quadratic and the numerator has degree at most one. If the numerator is not visibly a multiple of the denominator’s derivative plus a constant, solve for the two coefficients by matching powers of \(x\). The identity used here makes that split explicit.

The method requires \(a\ne0\) and \(b^2-4ac<0\). If the quadratic has real roots, it is not an irreducible quadratic of this type, and the arctangent formula derived here does not apply in this form. For the stated conditions, \(S\) is real and positive, the denominator has no real zeros, and both terms of the antiderivative are defined throughout the real line.

As a final check on any calculation, differentiate the proposed result and combine its terms over the original denominator. The numerator must return exactly to \(mx+n\); this catches errors in either the derivative-matching coefficient or the arctangent scale.