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Partial Fractions

Partial Fractions. Module 7 Lecture 2. Linear factors. Important: The repeated factor occurs twice on the RHS, once as power 1 and once as power 2. 3. =. +. +. -. +. -. (. x. 1. )(. x. 2. ). x. 1. x. 2. A repeated factor. 3. =. +. +. +. -. 2. 2. x. 1. x. 2. +.

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Partial Fractions

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  1. Partial Fractions Module 7 Lecture 2

  2. Linear factors Important:The repeated factor occurs twice on the RHS, once as power 1 and once as power 2. 3 = + + - + - ( x 1 )( x 2 ) x 1 x 2 A repeated factor 3 = + + + - 2 2 x 1 x 2 + - + ( x 1 ) ( x 2 ) ( x 1 ) A quadratic factor 3 = + - 2 2 x 2 + - + ( x 1 ) ( x 2 ) ( x 1 ) Recap: Module 7 – Lecture 1 Numerators of linear factors and the highest power factor of a repeated factor are found by the cover up rule.

  3. Linear factors 3 = + + - + - ( x 1 )( x 2 ) x 1 x 2 Repeated factor The numerator of the lower power is a number A A 3 = + + + - 2 2 x 1 x 2 + - + ( x 1 ) ( x 2 ) ( x 1 ) Quadratic factor 3 + Ax B = + - 2 2 x 2 + - ( x 1 ) ( x 2 ) + ( x 1 ) Recap: Module 7 – Lecture 1 Numerators of linear factors and the highest power factor of a repeated factor are found by the cover up rule. The numerator of the quadratic term is Ax+B as this is the most general non-cancelling numerator

  4. 3 A -1 1 = + + A -1 1 2 + - 3 ( x 1 ) ( x 2 ) + - 2 = + + x 1 (x 2) + 3 ( x 1 ) 2 2 2 + + + - - - ( ( ( x x x 1 1 1 ) ) ) ( ( ( x x x 2 2 2 ) ) ) + 2 - x 1 (x 2) + 3 ( x 1 ) None A None 1/3 = + 0 1/3 Example (i) Multiply through by the LHS denominator & cancel (ii) Tidy up (iii) Equate coefficients (x2 and/or the constant term) Equate x2 A

  5. INTEGRATION of PARTIAL FRACTIONS

  6. USING GIVEN INTEGRALS FROM TABLES FOR INTEGRATION

  7. Using Tables Of Integrals There are 2 types of rational function integrals that you will meet in the work on partial fractions which you can work out from your green tables.

  8. Type 1 (Log) Integrals with a denominator of power 1 that can be put in the form

  9. Type 1 (Log) Example 1 2 dx dx 2 2 3 dx dx 2 3 dx ò ò ò ò ò = = = = 2 - - -1 - - 3 x 1 3 x 1 3 3 x x 1 3 3 x 1 Constant out in front Denominator differentiates to 3 Get 3 in numerator Correct Integrate

  10. Type 1 (Log) Example 2 dx xdx 5 xdx 5 5 5x 6x dx 6x dx ò ò ò ò ò = = = = 5 3 x2+4 3x2+4 3x2+4 3x2+4 6 3x2+4 Constant out in front Denominator differentiates to 6x Get 6x in numerator Correct Integrate

  11. 1 2 Type 1 (Log) Example 3 (x+1) dx ò x2+2x+5 (x+1) dx 2(x+1) dx 2(x+1) dx ò ò ò = = = x2+2x+5 x2+2x+5 x2+2x+5 = Correct Integrate Get 2(x+1) in numerator Denominator differentiates to 2(x+1)

  12. 1 1 x arctan 2 2 a a + a x Type 2 (arc tan) Here is the table entry function integral Method: Write denominator in form a2 + x2

  13. dx 1 dx 1 3 3 x 1 3 x ò ò = = ´ = + arctan arctan c 4 2 9 9 2 2 6 2 + 9 x 4 1 1 x x 2 + x arctan arctan 9 a a a a Factor out the 9 Tidy up 2 a is now 3 Type 2 (arc tan) Example 1 Method: Write denominator in form a2 + x2 to get

  14. 1 dx 1 2 1 dx ò ò = ´ = + = arctan arctan c 2 x 2 x 1 2 4 4 1 2 + 4x 1 1 1 x x 2 + x arctan arctan 4 a a a a Factor out the 4 Tidy up 1 a is now 2 Type 2 (arc tan) Example 2 Method: Write denominator in form a2 + x2 to get

  15. 3 3dx x ò + = arctan c 2 + x 2 1 1 x x arctan arctan a a a a a is now 2 2 Type 2 (arc tan) Example 3 Method: Write denominator in form a2 + x2 to get

  16. Split up - 3 x 1 3 x 1 1 3 1 x ò ò ò ò 2 = - - = + + + dx dx dx dx ln( x 4 ) arctan c 2 2 2 2 2 2 2 + + + + x 4 x 4 x x 4 4 Type 2 Type 1 x 2 3 ò dx 2 + 2 x 4 Mixed Type - Example 1

  17. Split up - 3 x 1 3 x 1 1 1 3x dx ò ò ò = = + - + - dx dx ln( 9x2 4 ) arctan c 2 6 6 2 + + + 9x2 4 9x2 4 9x 4 Type 2 Type 1 2 a = 3 18x 3 1 1 ò ò dx - dx 2 9 + 18 9x 4 2 + x 4 9 Mixed Type - Example 2

  18. Integrals arising from partial fractions A partial fraction problem can have

  19. - 4 x 2 ò dx 2 + x 4 4 2 x dx ò ò = - dx 2 2 2 2 + + x 4 x 4 1 x 2 = + - ´ 2 ln( x 4 ) 2 arctan 2 2 x 2 = + - 2 ln( x 4 ) arctan 2 - dx 3 3 - - ò ò 2 1 = - = - = 3 3 ( x 2 ) dx ( x 2 ) - - 2 1 x 2 - ( x 2 ) Example Integrate the following partial fraction expression

  20. +c x 2 + + - 2 ln( x 4 ) arctan 2 3 - - x 2 So

  21. Example Integrate the following partial fraction expression

  22. Partial fractions 1 Partial fractions

  23. Partial fractions 2 Partial fractions

  24. Partial fractions 3 Partial fractions

  25. Partial fractions 4 Partial fractions - preliminary division

  26. 4 2 - - - x 1 3 x 2 x 1 4 5 = + = + + x x - + 2 2 3 ( x 1 ) 3 ( x 2 ) - + - + ( x 1 ) ( x 2 ) ( x 1 ) ( x 2 ) Partial fractions Note the missing fraction has a numerator of zero

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