Harolds Partial Fractions Cheat Sheet

University:
University of California, Berkeley
Course:
MATH 54 | Linear Algebra and Differential Equations
Academic year:

2022
Views:

24

Pages:

2
Author:

jaxbrokenhart4z4x

Harold’s Partial Fractions Cheat Sheet Partial Fractions Condition Preparation Case I: Simple linear (𝟏𝒔𝒕 degree) Case II: Multiple degree linear (𝟏𝒔𝒕 degree) Case III: Simple quadratic (𝟐𝒏𝒅 degree) Case IV: Multiple degree quadratic (𝟐𝒏𝒅 degree) (http://en.wikipedia.org/wiki/Partial_fraction_deco mposition) 𝑃(𝑥) 𝑎𝑥 𝑛 + ⋯ + 𝑏 𝑓(𝑥) = = 𝑄(𝑥) 𝑐𝑥 𝑚 + ⋯ + 𝑑 where 𝑃(𝑥 ) 𝑎𝑛𝑑 𝑄(𝑥 ) are polynomials Case 1: 𝑛 ≥ 𝑚, Peform long division first Case 2: 𝑛 < 𝑚, Proceed to the cases below 𝐴 (𝑎𝑥 + 𝑏) 𝐴 𝐵 𝐶 + + (𝑎𝑥 + 𝑏)3 (𝑎𝑥 + 𝑏) (𝑎𝑥 + 𝑏)2 𝐴𝑥 + 𝐵 (𝑎𝑥 2 + 𝑏𝑥 + 𝑐) 𝐴𝑥 + 𝐵 𝐶𝑥 + 𝐷 𝐸𝑥 + 𝐹 + + (𝑎𝑥 2 + 𝑏𝑥 + 𝑐) (𝑎𝑥 2 + 𝑏𝑥 + 𝑐)2 (𝑎𝑥 2 + 𝑏𝑥 + 𝑐)3 𝑃(𝑥) (𝑎𝑥 + 𝑏)(𝑐𝑥 + 𝑑)2 (𝑒𝑥 2 + 𝑓𝑥 + 𝑔) Example Expansion = Typical Solution for Cases I & II Typical Solution for Cases III & IV Steps to Solve 1. Write down problem 2. Check if long division is needed 3. Factor denominator of function 4. Expand function with A, B, Cs 5. Get a common denominator 6. Focus on numerator 7. Expand/distribute/FOIL 𝐴 𝐵 𝐶 𝐷𝑥 + 𝐸 + + + (𝑎𝑥 + 𝑏) (𝑐𝑥 + 𝑑) (𝑐𝑥 + 𝑑)2 (𝑒𝑥 2 + 𝑓𝑥 + 𝑔) 𝑎 𝑑𝑥 = 𝑎 ln|𝑥 + 𝑏| + 𝐶 𝑥+𝑏 𝑎 𝑥 𝑎 𝑑𝑥 = 𝑡𝑎𝑛−1 ( ) + 𝐶 ∫ 2 𝑏 𝑥 + 𝑏2 𝑏 ∫ Example 5𝑥 + 1 𝑑𝑥 2𝑥 2 − 𝑥 − 1 Not needed since degree of numerator (top) is less than degree of denominator (bottom) 5𝑥 + 1 (2𝑥 + 1)(𝑥 − 1) 5𝑥 + 1 𝐴 𝐵 = + (2𝑥 + 1)(𝑥 − 1) (2𝑥 + 1) (𝑥 − 1) 𝐴(𝑥 − 1) 𝐵(2𝑥 + 1) = + (2𝑥 + 1)(𝑥 − 1) (2𝑥 + 1)(𝑥 − 1) 5𝑥 + 1 = 𝐴(𝑥 − 1) + 𝐵(2𝑥 + 1) 5𝑥 + 1 = 𝐴𝑥 − 𝐴 + 2𝐵𝑥 + 𝐵 ∫ 8. Separate 9. Factor 10. Setup system of equations 11. Solve system of equations 5𝑥 + 1 = 𝐴𝑥 + 2𝐵𝑥 − 𝐴 + 𝐵 (5)𝑥 + (1) = (𝐴 + 2𝐵)𝑥 + (−𝐴 + 𝐵) 𝐴 + 2𝐵 = 5 −𝐴 + 𝐵 = 1 𝐵 =𝐴+1 𝐴 + 2(𝐴 + 1) = 5 𝐴 + 2𝐴 + 2 = 5 3𝐴 = 3 𝐴=1 𝐵 =𝐴+1=1+1=2 𝑨=𝟏 𝑩=𝟐 𝐴 + 2𝐵 = 5 + [−𝐴 + 𝐵 = 1] ------------------------3𝐵 = 6 𝑩=𝟐 a. Substitution method b. Elimination method 𝐴 + 2𝐵 = 5 −2 [−𝐴 + 𝐵 = 1] ------------------------3𝐴 = 3 𝑨=𝟏 𝐴 𝐵 𝑘] = [ 1 2 5] | [ | −1 1 1 𝐴 𝐵 𝑘 Use TI-84 𝑟𝑟𝑒𝑓() function c. Matrix method =[ 1 0 | 1] 0 1 2 𝑨=𝟏 𝑩=𝟐 12. Reassemble newly expanded function 13. Restate problem with expanded function 14. Integrate restated problem 5𝑥 + 1 1 2 = + (2𝑥 + 1)(𝑥 − 1) (2𝑥 + 1) (𝑥 − 1) 1 2 𝑑𝑥 ∫ 𝑑𝑥 + ∫ 𝑥−1 2𝑥 + 1 1 𝑑𝑥 ∫ 2𝑥 + 1 𝑢 = 2𝑥 + 1 𝑑𝑢 = 2 𝑑𝑥 1 1 1 1 1 1 ∫ 2 𝑑𝑥 = ∫ 𝑑𝑢 = ln|𝑢| = ln|2𝑥 + 1| 2 𝑢 2 𝑢 2 2 1 ln|2𝑥 + 1| + 2 ln|𝑥 − 1| + 𝐶 2 = ln √|2𝑥 + 1| + ln(𝑥 − 1)2 + 𝐶 = 15. Simplify 16. DONE = ln [√|2𝑥 + 1| (𝑥 − 1)2 ] + 𝐶