Techniques of Integration · Partial Fractions
Partial Fractions: Distinct Linear
Decompose a rational function with distinct linear factors in the denominator into simpler fractions that can be integrated individually.
- 2 worked examples
- 16 in Techniques of Integration
Worked examples
- 1/((x-1)(x+2)) = A/(x-1) + B/(x+2)
- Multiply through: 1 = A(x+2) + B(x-1)
- x = 1: 1 = 3A → A = 1/3. x = -2: 1 = -3B → B = -1/3
- ∫ [(1/3)/(x-1) - (1/3)/(x+2)] dx = (1/3)ln|x-1| - (1/3)ln|x+2| + C
Answer: (1/3)ln|(x-1)/(x+2)| + C
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Common questions
What is the Partial Fractions: Distinct Linear?
Decompose a rational function with distinct linear factors in the denominator into simpler fractions that can be integrated individually. It is one of the techniques of integration formulas in the CalcRef reference.
When does the Partial Fractions: Distinct Linear apply?
The Partial Fractions: Distinct Linear holds under this condition: Degree of P(x) must be less than the degree of the denominator.
How do you use the Partial Fractions: Distinct Linear?
Worked example. Find ∫ 1/((x-1)(x+2)) dx. 1/((x-1)(x+2)) = A/(x-1) + B/(x+2). Multiply through: 1 = A(x+2) + B(x-1). x = 1: 1 = 3A → A = 1/3. x = -2: 1 = -3B → B = -1/3. ∫ [(1/3)/(x-1) - (1/3)/(x+2)] dx = (1/3)ln|x-1| - (1/3)ln|x+2| + C. Answer: (1/3)ln|(x-1)/(x+2)| + C.
Practice
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