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Unit 2: Integration by Partial Fraction




          2.3.4 Repeated Quadratic Factors                                                      Notes

          To each irreducible quadratic factor, occurring n times in the denominator of a proper rational
          fraction, there corresponds a sum of n partial fractions of the form.

                         2
                                                   2
            A x B 1    A x b               A x b
                                                     
               
                           
             1
             2
                                                  2
                        2
                 
           9x   bx c  (9x   bx   ) c  2     (9x   bx   ) c  2
              Task  Make distinction between Distinct Linear Factors and Repeated Linear Factors.
                                    x   5
                 Example: Evaluate    2  dx
                                 x   5x   6
          Solution:
                 x   5    A      B
          Let                 
              x    2 (x   3)  (x  2)  (x  3)
                        
           x   5   A (x   2) B (x  3)                                        ....(1)
          Putting  x   in (1)
                    3
                              B
                          3     B   3

                   x   5       2    3  
          Now,    2    dx              dx
                 x   5x   6    (x  2)  (x  3) 

                             
                    2log(x   2) 3log(x   3)
                            2
                    log(x  2)   log(x   3) 3
                      (x  3) 3
                    log
                      (x   2) 2

                                    1
                 Example: Evaluate   2  dx
                                  ( x x   1)
          Solution:

                 1            1       A    B     C
          Let   2   dx                    
                ( x x   1)  ( x x   1)(x   1)  x  (x   1)  x   1
                                   
                          
           1   A (x   1)(x  1) Bx (x   1) cx (x   1)
          Putting x = 0, then
                      
                  1   A ( 1)(1)   A   1
          Putting x=a, then
                               1
                  1   B .1.(2)   B 
                               2





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