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Integration Techniques, L’Hopital’s Rule, and Improper Integrals

8.5 Partial Fractions

Focus on distinct linear factors, integrate the decomposition, check decomposition coefficients.

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Ideas in this lesson

01Distinct linear factorsWorked steps · five questions02Integrate the decompositionWorked steps · five questions03Check decomposition coefficientsWorked steps · five questions
Idea 1 of 3

Distinct linear factors

Assign one numerator constant to each distinct linear factor. Clear denominators to solve for the constants.

Worked example 1

Write 1x(x+2)=Ax+Bx+2\dfrac{1}{x(x + 2)} = \dfrac{A}{x} + \dfrac{B}{x + 2}. Find AA.
  1. Multiply by x(x+2)x(x + 2): 1=A(x+2)+Bx1 = A(x + 2) + Bx.
  2. At x=0,A=12x = 0, A = \dfrac{1}{2}.
  3. At x=−2,B=−12x = -2, B = -\dfrac{1}{2}.
  4. A=12≈0.5A = \dfrac{1}{2} \approx 0.5

Answer: 12≈0.5\dfrac{1}{2} \approx 0.5

Worked example 2

Write 1x(x+3)=Ax+Bx+3\dfrac{1}{x(x + 3)} = \dfrac{A}{x} + \dfrac{B}{x + 3}. Find BB.
  1. Multiply by x(x+3)x(x + 3): 1=A(x+3)+Bx1 = A(x + 3) + Bx.
  2. At x=0,A=13x = 0, A = \dfrac{1}{3}.
  3. At x=−3,B=−13x = -3, B = -\dfrac{1}{3}.
  4. B=−13≈−0.33B = \dfrac{-1}{3} \approx -0.33

Answer: −13≈−0.33\dfrac{-1}{3} \approx -0.33

Idea 2 of 3

Integrate the decomposition

Each simple reciprocal linear term integrates to a logarithm, with its scale factor and domain restrictions.

Worked example 1

Evaluate ∫121x(x+2) dx\int_{1}^{2} \dfrac{1}{x(x + 2)}\,dx.
  1. 1x(x+2)=12[1x−1x+2]\dfrac{1}{x(x + 2)} = \dfrac{1}{2}[\dfrac{1}{x} - \dfrac{1}{x + 2}]
  2. F(x)=12ln⁡∣xx+2∣F(x) = \dfrac{1}{2}\ln |\dfrac{x}{x + 2}|
  3. F(2)−F(1)=ln⁡(64)2=ln⁡(64)2≈0.2F(2) - F(1) = \dfrac{\ln (\dfrac{6}{4})}{2} = \dfrac{\ln (\dfrac{6}{4})}{2} \approx 0.2

Answer: ln⁡(64)2≈0.2\dfrac{\ln (\dfrac{6}{4})}{2} \approx 0.2

Worked example 2

Evaluate ∫131x(x+3) dx\int_{1}^{3} \dfrac{1}{x(x + 3)}\,dx.
  1. 1x(x+3)=13[1x−1x+3]\dfrac{1}{x(x + 3)} = \dfrac{1}{3}[\dfrac{1}{x} - \dfrac{1}{x + 3}]
  2. F(x)=13ln⁡∣xx+3∣F(x) = \dfrac{1}{3}\ln |\dfrac{x}{x + 3}|
  3. F(3)−F(1)=ln⁡(126)3=ln⁡(126)3≈0.23F(3) - F(1) = \dfrac{\ln (\dfrac{12}{6})}{3} = \dfrac{\ln (\dfrac{12}{6})}{3} \approx 0.23

Answer: ln⁡(126)3≈0.23\dfrac{\ln (\dfrac{12}{6})}{3} \approx 0.23

Idea 3 of 3

Check decomposition coefficients

Expand or substitute into the cleared-denominator identity. Coefficients must reproduce the entire original numerator.

Worked example 1

Write 1x(x+2)=Ax+Bx+2\dfrac{1}{x(x + 2)} = \dfrac{A}{x} + \dfrac{B}{x + 2}. Find AA.
  1. Multiply by x(x+2)x(x + 2): 1=A(x+2)+Bx1 = A(x + 2) + Bx.
  2. At x=0,A=12x = 0, A = \dfrac{1}{2}.
  3. At x=−2,B=−12x = -2, B = -\dfrac{1}{2}.
  4. A=12≈0.5A = \dfrac{1}{2} \approx 0.5

Answer: 12≈0.5\dfrac{1}{2} \approx 0.5

Worked example 2

Write 1x(x+3)=Ax+Bx+3\dfrac{1}{x(x + 3)} = \dfrac{A}{x} + \dfrac{B}{x + 3}. Find BB.
  1. Multiply by x(x+3)x(x + 3): 1=A(x+3)+Bx1 = A(x + 3) + Bx.
  2. At x=0,A=13x = 0, A = \dfrac{1}{3}.
  3. At x=−3,B=−13x = -3, B = -\dfrac{1}{3}.
  4. B=−13≈−0.33B = \dfrac{-1}{3} \approx -0.33

Answer: −13≈−0.33\dfrac{-1}{3} \approx -0.33

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §8.5,p8.5, p. 542542. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.