Partial Fractions with Repeated Irreducible Quadratics

Why Repeated Quadratics Need Their Own Terms

A rational function with an irreducible quadratic in its denominator is not decomposed in the same way as one with only distinct linear factors. When the quadratic is repeated, the decomposition must include a term for each power of that quadratic. The numerators of those terms are linear, and integrating them may require completing the square.

The essential integration pattern is to separate a numerator into a part proportional to the derivative of the quadratic and a constant part. The derivative part often integrates directly by substitution; the remaining part leads to an arctangent for the first power or a reduction formula for higher powers.

The Decomposition Form

Let \(Q(x)\) be an irreducible quadratic over the real numbers. If a denominator contains \(Q(x)^m\), its partial-fraction contribution has the form below. Other factors in the denominator contribute their own terms.

Each numerator is at most linear because the degree of a numerator in a partial fraction must be less than the degree of its denominator factor. Here each \(Q(x)^k\) has degree \(2k\), but the partial-fraction numerator corresponding to the irreducible quadratic factor is linear.

For instance, if a denominator factors as \((x-r)Q(x)^2\), the decomposition has the form:

The coefficients are found by equating expressions after clearing denominators. The focus here is the form of the terms and how to integrate them once they are known.

Separate the Numerator Before Integrating

Suppose a term has the form \((px+q)/Q(x)^k\), where \(Q\) is quadratic. The derivative \(Q'(x)\) is linear, so the numerator can be written as a multiple of \(Q'(x)\) plus a constant. This exposes two different integration tasks.

For example, if \(Q(x)=ax^2+bx+c\), with \(a\ne 0\), then \(Q'(x)=2ax+b\), and:

The first part is directly related to the derivative of \(Q\). When \(k>1\), its integral is a power of \(Q\); when \(k=1\), it gives a logarithm. The constant part requires integrating \(1/Q(x)^k\).

To handle that remaining integral, complete the square. For a quadratic with positive leading coefficient and no real roots, we can write:

Thus \(1/Q(x)^k\) is \(A^{-k}\) times a reciprocal power of a shifted sum of squares. The first such integral is an arctangent; higher powers can be reduced to lower ones.

A Reduction Formula for Higher Powers

Write \(u=x-h\), and set \(D=u^2+d^2\). Define \(I_n=\int du/D^n\), with \(d>0\). The first case is:

For integers \(n\ge 2\), a reduction formula expresses \(I_n\) using \(I_{n-1}\):

The formula follows by differentiating \(u/D^{n-1}\). Since \(D'=2u\), the product and chain rules give:

Using \(u^2=D-d^2\), the numerator becomes \(2(n-1)d^2-(2n-3)D\). Therefore:

Solving this equation for \(1/D^n\) and integrating both sides gives the stated reduction formula. This derivation also explains why each step reduces the power by one. Repeated use eventually reaches \(I_1\), which is the arctangent case.

Worked Examples

A Common Mistake

A common error is to include only the highest power of a repeated quadratic in the decomposition. For example, one might incorrectly try to write \(1/Q(x)^2=A/Q(x)\), where \(A\) is a constant. Clearing denominators gives \(1=AQ(x)\), which cannot hold for a nonconstant quadratic \(Q\): the right side has degree two if \(A\ne0\), and is zero if \(A=0\).

The correction is to allow a term for each power, with its own linear numerator. A term with denominator \(Q^2\) may require both a \(1/Q\) term and a \(1/Q^2\) term elsewhere in the decomposition. After the correct form is set up, integrate each term separately and check the final antiderivative by differentiation.