Reducing Polynomial Numerators Before Integration

Improper Rational Functions

A rational function is improper when the degree of the numerator is at least the degree of the denominator. The reciprocal and linear-numerator formulas for irreducible quadratics apply only to proper fractions, so an improper fraction must be divided first. The division produces a polynomial quotient, which is integrated by the power rule, plus a remainder of lower degree over the quadratic, which is integrated by the logarithm and arctangent methods.

With a quadratic denominator, the remainder has degree at most one, so it can always be written as \(px+q\). The decomposition of this remainder is the second key step, and it is what links the present tutorial to the previous one.

Two Ways to Divide

Long division works for any numerator, but for simple numerators a shortcut is faster: add and subtract the denominator. For example, \(x^2=(x^2+4)-4\), so \(\frac{x^2}{x^2+4}=1-\frac{4}{x^2+4}\). In general, rewrite the numerator so that a multiple of the denominator appears, then cancel it. This numerator decomposition is the technique used in the earlier tutorials for \(\frac{x^2}{1+x^2}\), and it works whenever the leading coefficients are easy to match.

Worked Examples

A Reliable Procedure

Common Mistakes

The leading mistake is to skip the division and apply the arctangent formula to an improper fraction. Another is to drop the remainder or to lose its sign: in the second example the remainder is \(-x-5\), and the minus sign carries into the logarithm and arctangent. A third is to forget the factor \(\frac12\) in the logarithm term when the numerator is split with \(2x+b\).

Interpreting the Result

For large \(|x|\), the proper fraction part tends to zero, so the integrand is close to the polynomial quotient and the antiderivative is close to the polynomial's antiderivative, with a logarithmic correction. This explains why every answer above begins with a polynomial term and then adds slowly varying terms. It also gives a quick check: the leading term of your answer must match the integral of the quotient.

A Numerical Spot Check

Check \(\int_0^1\frac{x^3}{x^2+1}\,dx\approx0.153\) with Simpson's rule: the heights are \(0\), \(\frac{0.125}{1.25}=0.1\), and \(\frac12\). Then \(\frac{0.5}{3}(0+0.4+0.5)=0.15\), which agrees to two digits with \(0.1534\).

Practice Problems with Answers

Try each problem before reading the answer. Every answer below was checked by differentiation or by numerical integration.

1. Evaluate \(\int_0^2\frac{x^2}{x^2+4}\,dx\). Hint: write \(x^2=(x^2+4)-4\). Answer: \(2-\frac\pi2\)
2. Find an antiderivative of \(\frac{x^2+x}{x^2+2x+5}\). Hint: the remainder is \(-x-5\). Answer: \(x-\frac12\ln(x^2+2x+5)-2\arctan\frac{x+1}{2}\)
3. Evaluate \(\int_0^1\frac{x^3}{x^2+1}\,dx\). Hint: write \(x^3=x(x^2+1)-x\). Answer: \(\frac12-\frac12\ln2\)
4. Find an antiderivative of \(\frac{2x^2-3x+1}{x^2-2x+2}\). Hint: the remainder is \(x-3\). Answer: \(2x+\frac12\ln(x^2-2x+2)-2\arctan(x-1)\)

The General Formula for the Remainder

For a proper fraction with a monic quadratic denominator, \(\int\frac{px+q}{x^2+bx+c}\,dx=\frac p2\ln(x^2+bx+c)+\Bigl(q-\frac{pb}{2}\Bigr)\int\frac{dx}{x^2+bx+c}\). The first term comes from the part of the numerator equal to \(\frac p2\) times the derivative \(2x+b\) of the denominator, and the second term takes whatever is left. For an irreducible denominator, the last integral is the arctangent of the previous tutorial. This formula is the backbone of the present method: after division, every remainder is processed by it.

Checking by Recombination

An efficient check recombines the pieces over the common denominator. For the second example, let \(Q=x^2+2x+5=(x+1)^2+4\). The derivative of the answer is \(1-\frac{x+1}{Q}-\frac{1}{1+\frac{(x+1)^2}{4}}\), and the last fraction equals \(\frac{4}{Q}\). So the derivative is \(\frac{Q-(x+1)-4}{Q}=\frac{x^2+2x+5-x-5}{Q}=\frac{x^2+x}{Q}\), the original integrand. When a recombination check is awkward, compare numerical values of the integrand and of a finite-difference estimate of the antiderivative's derivative.

Domain and Continuity

The denominators in this tutorial are irreducible quadratics, which never vanish for real \(x\). The integrand is therefore continuous on the entire real line, and the logarithm \(\ln(x^2+bx+c)\) has a positive argument, so no absolute values are needed. This makes the whole family friendlier than rational functions with real roots, where the logarithms need absolute values and intervals must avoid the poles.

Where This Fits in the Course

This tutorial belongs to the module “Completing the Square Techniques”. The tutorial just before it is “Linear Numerators Over Irreducible Quadratics,” and the one that follows is “Definite Integrals of Inverse-Tangent Forms.” It combines the linear-numerator method from the previous tutorial with polynomial division. The following tutorial applies the same ideas to definite integrals, where the exact values of the arctangent at standard angles make the answers compact. Practice the division step until it is automatic, since an error there propagates into every later term of the answer.