Partial Fractions: Irreducible Quadratic Factors

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Textbook Reference

Opening Scenario

A quadratic factor $ax^2 + bx + c$ is irreducible over the reals when its discriminant $b^2 - 4ac < 0$ -- it has no real roots. Such factors cannot be split into linear factors with real coefficients. Over the denominator of a rational function, they require a numerator of the form $Ax + B$ (one degree below the quadratic) in the partial fraction decomposition.

The resulting integrals involve both a logarithm (from the $x$ part of the numerator) and an arctangent (from the constant part).

Quick Reference

Partial fractions for an irreducible quadratic factor $ax^2+bx+c$ (discriminant $< 0$):

$$\frac{P(x)}{(ax^2+bx+c)\cdots} = \frac{Ax+B}{ax^2+bx+c} + \cdots$$

Integration strategy for $\dfrac{Ax+B}{ax^2+bx+c}$:

1. Split: write $Ax + B = A_1(2ax + b) + B_1$ where $A_1 = A/(2a)$ and $B_1$ is the remainder.
2. The first part integrates as $A_1\ln(ax^2+bx+c)$ (since $d(ax^2+bx+c) = (2ax+b)\,dx$).
3. The second part requires completing the square in the denominator and using $\int\dfrac{1}{u^2+k^2}\,du = \dfrac{1}{k}\arctan\!\left(\dfrac{u}{k}\right) + C$.

Key Concepts

1. Identifying Irreducible Quadratics

A quadratic $ax^2+bx+c$ is irreducible over the reals when $b^2 - 4ac < 0$. Examples:

• $x^2 + 1$: discriminant $= -4 < 0$. Irreducible.
• $x^2 + x + 1$: discriminant $= 1 - 4 = -3 < 0$. Irreducible.
• $x^2 - 1 = (x-1)(x+1)$: discriminant $= 4 > 0$. Reducible -- split into linear factors instead.

2. Setting Up the Numerator Form

For a linear factor $(x-r)$, the numerator in the partial fraction is a constant $A$ (degree 0 = degree 1 minus 1). For a quadratic factor $ax^2+bx+c$, the numerator is degree 1, i.e., $Ax + B$.

3. Splitting the Numerator for Integration

To integrate $\dfrac{Ax+B}{ax^2+bx+c}$:

Write $Ax + B = \lambda(2ax + b) + \mu$ where $\lambda = \dfrac{A}{2a}$ and $\mu = B - \lambda b$.

Then: $$\int\frac{Ax+B}{ax^2+bx+c}\,dx = \lambda\int\frac{2ax+b}{ax^2+bx+c}\,dx + \mu\int\frac{1}{ax^2+bx+c}\,dx.$$

The first integral is $\lambda\ln(ax^2+bx+c) + C$ (using $u = ax^2+bx+c$, $du = (2ax+b)\,dx$; note: no absolute value needed since the discriminant is negative, so the quadratic is always positive). The second integral requires completing the square.

Worked Example

Evaluate $\displaystyle\int\frac{2x+3}{x^2+2x+5}\,dx$.

Step 1 -- Match numerator to derivative of denominator. The derivative of $x^2+2x+5$ is $2x+2$.

Write $2x+3 = (2x+2) + 1$.

Step 2 -- Split.

$\displaystyle\int\frac{2x+3}{x^2+2x+5}\,dx = \int\frac{2x+2}{x^2+2x+5}\,dx + \int\frac{1}{x^2+2x+5}\,dx$.

Step 3 -- First integral. $u = x^2+2x+5$, $du = (2x+2)\,dx$:

$\displaystyle\int\frac{du}{u} = \ln(x^2+2x+5) + C_1$.

Step 4 -- Second integral. Complete the square: $x^2+2x+5 = (x+1)^2 + 4$.

$\displaystyle\int\frac{1}{(x+1)^2+4}\,dx$. Let $u = x+1$, $du = dx$:

$\displaystyle\int\frac{du}{u^2+4} = \frac{1}{2}\arctan\!\left(\frac{u}{2}\right) + C_2 = \frac{1}{2}\arctan\!\left(\frac{x+1}{2}\right) + C_2$.

Step 5 -- Combine.

$$\int\frac{2x+3}{x^2+2x+5}\,dx = \ln(x^2+2x+5) + \frac{1}{2}\arctan\!\left(\frac{x+1}{2}\right) + C.$$

Full partial fraction example. Evaluate $\displaystyle\int\frac{x^2+1}{(x+1)(x^2+4)}\,dx$.

Decompose: $\dfrac{x^2+1}{(x+1)(x^2+4)} = \dfrac{A}{x+1} + \dfrac{Bx+C}{x^2+4}$.

Clear: $x^2+1 = A(x^2+4) + (Bx+C)(x+1)$.

$x = -1$: $2 = 5A \Rightarrow A = \frac{2}{5}$.

Equate $x^2$: $1 = A + B \Rightarrow B = \frac{3}{5}$.

Equate constants: $1 = 4A + C \Rightarrow C = 1 - \frac{8}{5} = -\frac{3}{5}$.

$$\int\left[\frac{2/5}{x+1} + \frac{(3/5)x - 3/5}{x^2+4}\right]dx = \frac{2}{5}\ln|x+1| + \frac{3}{10}\ln(x^2+4) - \frac{3}{10}\arctan\!\left(\frac{x}{2}\right) + C.$$

Common Errors Summary

Leveled Practice

Level 1 -- Identifying the Form

Problem 1. Is $x^2 + 2x + 3$ irreducible? Write the correct partial fraction form for $\dfrac{1}{(x-1)(x^2+2x+3)}$.

Level 2 -- Integration with Arctangent

Problem 2. Evaluate $\displaystyle\int\frac{3}{x^2+4x+8}\,dx$.

Mastery Checklist

• Determine whether a quadratic factor is irreducible using the discriminant
• Use $Ax+B$ (not just $A$) in the numerator for an irreducible quadratic factor
• Split the numerator to separate the logarithm piece from the arctangent piece
• Complete the square to evaluate the constant-numerator piece
• Apply the formula $\int\frac{1}{u^2+k^2}\,du = \frac{1}{k}\arctan(u/k) + C$ after completing the square

Mental Model

An irreducible quadratic factor represents a pair of complex conjugate roots. Over the reals, it cannot be split further. The numerator $Ax+B$ has two degrees of freedom to represent the contribution from this factor. The $A$ part (proportional to the derivative of the denominator) produces a logarithm via $u$-substitution. The $B$ part (a constant relative to the quadratic denominator) produces an arctangent via completing the square.

Connections

Looking back

• Completing the square: The core tool for handling the arctangent piece.
• Inverse trig derivatives: $\frac{d}{dx}\arctan(x) = \frac{1}{1+x^2}$ underlies the arctangent integral.

Looking ahead

• Improper rational functions (Section 7.4): Long division when degree of numerator is too high.
• Integration strategy (Section 7.5): Rational functions always admit partial fractions; knowing the full method is part of the complete toolkit.

Back to Techniques of Integration | Next: Improper Rational Functions