Integration, Partial Fractions, Formula, Irreducible Quadratic Factors, Worksheet - Calculus

Integration using partial fractions is a method to solve integrals of complex rational functions by breaking them down into simpler fractions whose integrals are easier to find, typically using logarithms. The process involves decomposing the integrand into partial fractions, solving for the unknown constants in the numerator of these fractions, and then integrating each simplified fraction.

💡Steps for Integration by Partial Fractions
• Ensure Proper Rational Function: The degree of the numerator must be less than the degree of the denominator. If it's not, perform polynomial long division first to obtain a proper fraction.
• Factor the Denominator: Factor the denominator of the rational function into its simplest components, such as distinct linear factors, repeated linear factors, irreducible quadratic factors, and repeated irreducible quadratic factors.
• Set up the Partial Fraction Decomposition: Write the original rational function as a sum of partial fractions, with appropriate numerators over each factor.
∘ For a distinct linear factor (ax+b), the term is A/(ax+b).
∘ For a repeated linear factor (ax+b)ⁿ, the terms are A₁/(ax+b) + A₂/(ax+b)² + ... + Aₙ/(ax+b)ⁿ.
∘ For an irreducible quadratic factor (ax²+bx+c), the term is (Ax+B)/(ax²+bx+c).
• Solve for the Constants: Equate the original numerator with the sum of the numerators of the partial fractions. Solve this equation for the unknown constants (A, B, etc.) by either substituting convenient values for x or by comparing coefficients of like powers of x.
• Integrate the Simpler Fractions: Substitute the found partial fractions back into the integral and integrate each term separately.
∘ Integrals of the form ∫(A/(ax+b)) dx typically result in A/a * ln|ax+b| + C.
∘ Add the constant of integration, C, for indefinite integrals.

💡Example
• To integrate (6x+13)/(x²+5x+6), you would first factor the denominator as (x+2)(x+3). Then, you would set up the decomposition:
(6x+13)/((x+2)(x+3)) = A/(x+2) + B/(x+3).
• Solving for A and B gives A = 7 and B = -1.
• So the integral becomes:
∫(7/(x+2) - 1/(x+3)) dx
• This can be integrated as: 7 ln|x+2| - ln|x+3| + C

💡Worksheets are provided in PDF format to further improve your understanding:
• Questions Worksheet: https://drive.google.com/file/d/1nyZAxMFIv3phTs8a9uQiZ7WKvrQ4fBKt/view?usp=drive_link
• Answers: https://drive.google.com/file/d/16wbbNzLH-O0LGu6SEvM5Nq9gNQlpoR-H/view?usp=drive_link

💡Chapters:
00:00 Integration using partial fractions, distinct factors
01:24 Repeated factors
03:42 Irreducible quadratic
05:21 Worked example

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