Partial Fractions With Irreducible Quadratic Terms

Why These Partial Fractions Need Two Kinds of Terms

A rational function can have a denominator that factors into irreducible quadratic polynomials. After partial-fraction decomposition, each such factor generally has a linear numerator. Integrating one term then requires two contributions: a logarithm from the part related to the denominator’s derivative, and an inverse tangent from the remaining constant part.

The key is to complete the square in each quadratic denominator. This puts the inverse-tangent contribution into a recognizable form. The sign of the quadratic’s leading coefficient matters: an irreducible quadratic with a negative leading coefficient is negative everywhere, not a positive square plus a positive constant. The formulas below handle either sign.

The Partial-Fraction Form

Suppose a proper rational function has a denominator that is a product of distinct irreducible real quadratics and has no linear factors. Its partial-fraction decomposition has the form

Here each \(q_i\) is irreducible, each \(a_i\ne 0\), and the numerator over each quadratic has degree less than two. The constants \(A_i\) and \(B_i\) are determined by multiplying through by the common denominator and matching coefficients. If the original rational function is improper, polynomial division is needed before applying this proper-fraction form.

Once the decomposition is found, integrate term by term. The same calculation applies to each term, so first consider a single irreducible quadratic \(q(x)=ax^2+bx+c\) and a linear numerator \(Ax+B\).

Completing the Square and Splitting the Numerator

Completing the square gives

Since \(b^2-4ac<0\), the quantity \(4ac-b^2\) is positive. Define

Then \(q(x)=a((x-h)^2+d)\). The bracket is positive for every real \(x\), so \(q\) has the same sign everywhere as \(a\). In particular, if \(a<0\), the completed-square expression is negative everywhere.

The derivative of \(q\) is \(q'(x)=2ax+b\). Write the numerator as a multiple of this derivative plus a constant:

This split is useful because the first part integrates to a logarithm. The constant part is integrated using the completed square.

The absolute value in the logarithm is necessary when \(q(x)<0\). There are no real zeros of \(q\), so this antiderivative is defined throughout the real line.

Why the Formula Works

The numerator split and the completed square give a direct derivation. Let \(\mu=B-\frac{Ab}{2a}\). Then

The first quotient has antiderivative \(\frac{A}{2a}\ln|q(x)|\), because the derivative of \(\ln|q(x)|\) is \(q'(x)/q(x)\) wherever \(q(x)\ne0\). For the second quotient, the substitution \(u=(x-h)/\sqrt d\) gives \(dx=\sqrt d\,du\), and therefore

Substituting back gives the stated formula. This derivation also explains why the sign of \(a\) must be retained in the inverse-tangent coefficient.

Worked Examples

A Common Sign Mistake

A common error is to assume that every irreducible quadratic can be written as a positive constant times a square plus a positive constant. For example, \(-x^2-1\) is irreducible, but it is \(-(x^2+1)\), not a positive square plus a positive constant. Dropping the negative factor changes the sign of the integrand and gives the wrong inverse-tangent contribution.

Using the Method Reliably

For a rational function with distinct irreducible quadratic factors, first obtain the partial-fraction decomposition, then integrate each term on its own. In each term, identify the derivative of the denominator, subtract the matching multiple from the numerator, and complete the square in the denominator. This separates the logarithmic and inverse-tangent parts without guessing.

Keep the logarithmic absolute value when the quadratic can be negative, and retain the sign of the leading coefficient in the inverse-tangent coefficient. Finally, differentiate the proposed antiderivative or recombine the partial fractions to check it. These checks catch both coefficient errors in the decomposition and sign errors in the integration.