Integrating Repeated Irreducible Quadratic Factors

Why Repeated Quadratic Factors Need a Careful Split

A rational integrand with a repeated quadratic denominator can look more complicated than it is. Its numerator may contain one part that matches the derivative of the quadratic and another part that does not. The matching part integrates directly; the remainder leads to a reciprocal squared completed square.

For example, in an expression of the form \((mx+n)/q(x)^2\), the derivative \(q'(x)\) is linear. Rewriting the numerator using \(q'(x)\) exposes a direct antiderivative, while completing the square in \(q(x)\) puts the remaining integral into a standard form.

Completing the Square and Splitting the Numerator

Let \(q(x)=Ax^2+Bx+C\), where \(A\ne 0\). The quadratic is irreducible over the real numbers when its discriminant satisfies \(B^2-4AC<0\). Completing the square gives

The positivity of \(d^2\) is exactly what makes the quadratic irreducible. Also, \(q'(x)=2A(x-h)\). Thus a linear numerator can always be written as a multiple of \(q'(x)\) plus a constant:

This follows because \(\frac{m}{2A}q'(x)=m(x-h)=mx-mh\); adding \(mh+n\) gives \(mx+n\). For a squared denominator, the derivative-based term integrates by the reverse chain rule:

This identity separates the two tasks. The first term is already integrated. The remaining integral is handled by scaling the completed square and using a formula for the reciprocal of \((u^2+d^2)^2\).

The Reciprocal of a Squared Completed Square

For \(d>0\), define \(u=x-h\). The required basic integral is

Here \(C\) is an arbitrary constant of integration. This formula is valid for every real \(u\), since \(u^2+d^2\) is always positive when \(d>0\).

The scaling factor in the completed square matters. Since \(q(x)=A(u^2+d^2)\), it follows that

In particular, the factor is \(1/A^2\), which is positive whether \(A\) is positive or negative.

Why the Formula Works

A direct differentiation proves the formula and makes its two contributions clear. Differentiate its rational term:

For the inverse-tangent term, the chain rule gives

Writing the second derivative over the common denominator \(2d^2(u^2+d^2)^2\) gives a numerator of \(u^2+d^2\). Adding the two derivatives therefore yields

Thus differentiating the stated antiderivative gives exactly the integrand. The rational term and inverse-tangent term are both needed: together, rather than separately, they produce the squared denominator.

A Reliable Sequence of Steps

Worked Examples

Common Pitfalls

A frequent error is to treat every numerator as if it were a multiple of the denominator’s derivative. For instance, with numerator \(3u-1\) over \((u^2+4)^2\), a mistaken calculation might write

This drops the constant \(-1\). Differentiating the proposed answer gives \(3u/(u^2+4)^2\), not \((3u-1)/(u^2+4)^2\). The correction is to keep the remainder and integrate it separately:

Another error is to lose the leading coefficient when completing the square. If \(q(x)=A(u^2+d^2)\), then \(q(x)^2=A^2(u^2+d^2)^2\); the residual integral therefore carries \(1/A^2\). Finally, the squared-quadratic formula requires \(d>0\). If the quadratic is not irreducible, this completed-square form with positive \(d^2\) does not apply.

When This Split Is Useful

Repeated irreducible quadratic factors arise when rational integrands contain powers of a quadratic that has no real roots. The numerator split is useful because it isolates the portion directly related to the denominator’s rate of change. Completing the square then reveals the constant-remainder integral as a reciprocal squared distance from a shifted center, whose antiderivative combines a rational term with an inverse tangent.

For a squared denominator, the process is complete once both pieces have been integrated. With higher powers, the same numerator split can still isolate a derivative-based term, but the remaining reciprocal power is a distinct integral that may require additional methods. It is important not to apply the squared-denominator formula to a different power.