Review: Integration Using Partial Fraction Decomposition

Partial fraction decomposition (PFD) is an algebraic technique used to integrate rational functions \(\frac{P(x)}{Q(x)}\) by breaking them down into a sum of simpler fractions that can be integrated using basic logarithm, power, or inverse tangent formulas.

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1. Prerequisite Checks & Setup

• Check the Degree: The degree of the numerator \(P(x)\) must be strictly less than the degree of the denominator \(Q(x)\) (\(\deg(P) < \deg(Q)\)). If \(\deg(P) \ge \deg(Q)\), perform polynomial long division first.
• Factor the Denominator: Factor \(Q(x)\) completely into real linear factors \((ax + b)\) and irreducible quadratic factors \((ax^2 + bx + c)\) where \(b^2 - 4ac < 0\).

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2. The Three Main Factor Types

Denominator Factor Type	Form in \(Q(x)\)	Term(s) in Decomposition	Typical Integral Result
Type 1: Distinct Linear	\((ax + b)\)	\(\frac{A}{ax + b}\)	Logarithm: \(\frac{A}{a}\ln\lvert ax + b\rvert\)
Type 2: Repeated Linear	\((ax + b)^k\)	\(\frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \dots + \frac{A_k}{(ax + b)^k}\)	Combination of \(\ln\lvert ax+b\rvert\) and Power Rule terms
Type 3: Irreducible Quadratic	\(ax^2 + bx + c\)	\(\frac{Ax + B}{ax^2 + bx + c}\)	Split into a \(\ln\) term (\(u\)-sub) and an \(\arctan\) term

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3. Illustrative Examples

Example 1: Distinct Linear Factors

Evaluate: \[\int \frac{5x - 3}{x^2 - 2x - 3}\,dx\]

• Factor denominator:
  \[x^2 - 2x - 3 = (x - 3)(x + 1)\]

• Set up partial fractions:
  \[\frac{5x - 3}{(x - 3)(x + 1)} = \frac{A}{x - 3} + \frac{B}{x + 1}\] \[5x - 3 = A(x + 1) + B(x - 3)\]

• Solve for coefficients:

  • Let \(x = 3\): \(5(3) - 3 = A(4) \implies 12 = 4A \implies A = 3\)
  • Let \(x = -1\): \(5(-1) - 3 = B(-4) \implies -8 = -4B \implies B = 2\)

• Integrate:
  \[\int \left(\frac{3}{x - 3} + \frac{2}{x + 1}\right)\,dx = 3\ln\lvert x - 3\rvert + 2\ln\lvert x + 1\rvert + C\]

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Example 2: Repeated Linear Factors

Evaluate: \[\int \frac{x^2 + 1}{(x - 1)^2 (x + 2)}\,dx\]

• Set up partial fractions:
  \[\frac{x^2 + 1}{(x - 1)^2 (x + 2)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{x + 2}\] \[x^2 + 1 = A(x - 1)(x + 2) + B(x + 2) + C(x - 1)^2\]

• Solve for coefficients:

  • Let \(x = 1\): \(1^2 + 1 = B(3) \implies B = \frac{2}{3}\)
  • Let \(x = -2\): \((-2)^2 + 1 = C(-3)^2 \implies 5 = 9C \implies C = \frac{5}{9}\)
  • Equate \(x^2\) coefficients: \(1 = A + C \implies 1 = A + \frac{5}{9} \implies A = \frac{4}{9}\)

• Integrate:
  \[\int \left(\frac{4/9}{x - 1} + \frac{2/3}{(x - 1)^2} + \frac{5/9}{x + 2}\right)\,dx\] \[= \frac{4}{9}\ln\lvert x - 1\rvert - \frac{2}{3(x - 1)} + \frac{5}{9}\ln\lvert x + 2\rvert + C\]

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Example 3: Distinct Irreducible Quadratic Factors

Evaluate: \[\int \frac{2x^2 - x + 4}{x(x^2 + 4)}\,dx\]

• Set up partial fractions:
  \[\frac{2x^2 - x + 4}{x(x^2 + 4)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 4}\] \[2x^2 - x + 4 = A(x^2 + 4) + (Bx + C)x\]

• Solve for coefficients:

  • Let \(x = 0\): \(4 = 4A \implies A = 1\)
  • Expand and collect like terms:
    \[2x^2 - x + 4 = (A + B)x^2 + Cx + 4A\]
  • Matching coefficients:
    • \(A + B = 2 \implies 1 + B = 2 \implies B = 1\)
    • \(C = -1\)

• Integrate:
  \[\int \left(\frac{1}{x} + \frac{x - 1}{x^2 + 4}\right)\,dx = \int \frac{1}{x}\,dx + \int \frac{x}{x^2 + 4}\,dx - \int \frac{1}{x^2 + 4}\,dx\] \[= \ln\lvert x\rvert + \frac{1}{2}\ln(x^2 + 4) - \frac{1}{2}\arctan\left(\frac{x}{2}\right) + C\]

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4. Practical Tips & Traps

• Linear numerators on quadratic factors: If the denominator factor is quadratic (\(ax^2 + bx + c\)), the numerator must be a linear polynomial (\(Ax + B\)), not just a constant.
• Heaviside cover-up method: For distinct linear factors, solve for each coefficient instantly by setting \(x\) to the root of that factor and covering up that term in the original denominator.
• Complete the square: If an irreducible quadratic has a middle term (e.g., \(x^2 + 2x + 5\)), complete the square to write it as \((x + 1)^2 + 4\) before integrating to cleanly yield the substitution and arctangent forms.