I am trying to decompose the following rational function: 1 ( x 2 − 1 ) 2 in partial fractions (in order to untegrate it later).I have notices that ( x 2 − 1 ) 2 = ( x + 1 ) 2 ( x − 1 ) 2 Therefore ∃ A , B , C , D s.t: 1 ( x 2 − 1 ) 2 = 1 ( x − 1 ) 2 ( x + 1 ) 2 = A x + B ( x + 1 ) 2 + C x + D ( x − 1 ) 2 We then have A x + B = 1 ( x − 1 ) 2 | x = − 1 ⟹ B − A = 1 / 4Same thing for C x + D: C x + D = 1 ( x + 1 ) 2 | x = 1 ⟹ C + D = 1 / 4How do I find A, B, C, D from here?

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I am trying to decompose the following rational function:             1              (                  x          2                −        1                  )          2                     in partial fractions (in order to untegrate it later).I have notices that   (      x    2    −  1      )    2    =  (  x  +  1      )    2    (  x  −  1      )    2  Therefore   ∃  A  ,  B  ,  C  ,  D s.t:            1              (                  x          2                −        1                  )          2                      =            1              (        x        −        1                  )          2                (        x        +        1                  )          2                      =                    A        x        +        B                    (        x        +        1                  )          2                      +                    C        x        +        D                    (        x        −        1                  )          2                    We then have  A  x  +  B  =                                    1                      (            x            −            1                          )              2                                          |              x      =      −      1          ⟹    B  −  A  =  1      /    4Same thing for   C  x  +  D:  C  x  +  D  =                                    1                      (            x            +            1                          )              2                                          |              x      =      1          ⟹    C  +  D  =  1      /    4How do I find A, B, C, D from here?

kpgt1z · Accepted Answer

We have  1  =  (  x  −  1      )    2    (  A  x  +  B  )  +  (  x  +  1      )    2    (  C  x  +  D  )You already have  B  −  A  =  C  +  D  =      1    4  Substituting   x  =  0 gives  1  =  B  +  DSubstituting   x  =  2 gives  1  =  2  A  +  B  +  9  (  2  C  +  D  )Solving these gives  A  =      1    4    ,    B  =      1    2    ,    C  =  −      1    4    ,    D  =      1    2

Leah Pope · Accepted Answer

Consider:      1          (              x        2            −      1              )        2              =      1          (      x      +      1              )        2            (      x      −      1              )        2              =      A          x      +      1        +      B          (      x      +      1              )        2              +      C          x      −      1        +      D          (      x      −      1              )        2            Try to decompose the above

I am trying to decompose the following rational function: <mstyle displaystyle="true" scriptleve

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