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Integration By
       Partial Fractions
        () Ax
To find; ∫        dx
           P( x )
Integration By
       Partial Fractions
        () Ax
To find; ∫        dx
           P( x )
(1) If degA( x ) ≥ degP( x ) , perform a division
Integration By
        Partial Fractions
         ()Ax
To find; ∫        dx
           P( x )
(1) If degA( x ) ≥ degP( x ) , perform a division
( 2) If degA( x ) < degP( x ) , factorise P( x )
Integration By
        Partial Fractions
         ()Ax
To find; ∫        dx
           P( x )
(1) If degA( x ) ≥ degP( x ) , perform a division
( 2) If degA( x ) < degP( x ) , factorise P( x )
                                            A
    a) for linear factor ( x − a ) , write
                                           x−a
Integration By
        Partial Fractions
         ()Ax
To find; ∫        dx
           P( x )
(1) If degA( x ) ≥ degP( x ) , perform a division
( 2) If degA( x ) < degP( x ) , factorise P( x )
                                               A
    a) for linear factor ( x − a ) , write
                                            x−a
    b) for multiple linear factors ( x − a ) , write
                                                n


            A         B                  C
                +           ++
        ( x − a) ( x − a)            ( x − a) n
                          2
Integration By
        Partial Fractions
         ()Ax
To find; ∫        dx
           P( x )
(1) If degA( x ) ≥ degP( x ) , perform a division
( 2) If degA( x ) < degP( x ) , factorise P( x )
                                               A
    a) for linear factor ( x − a ) , write
                                            x−a
    b) for multiple linear factors ( x − a ) , write
                                                n


            A         B                  C
                +           ++
        ( x − a) ( x − a)            ( x − a) n
                          2

                                                       Ax + B
    c) for polynomial factors e.g. ax + bx + c, write 2
                                           2

                                                     ax + bx + c
x2
e.g. ( i ) ∫      dx
             x +1
x
              x2
e.g. ( i ) ∫      dx   x +1 x 2 + 0 x + 0
             x +1
                             x2 + x
x −1
              x2
e.g. ( i ) ∫      dx   x +1 x 2 + 0 x + 0
             x +1
                             x2 + x
                                −x+0
x −1
              x2
e.g. ( i ) ∫      dx   x +1 x 2 + 0 x + 0
             x +1
                             x2 + x
                                −x+0
                                − x −1
x −1
              x2
e.g. ( i ) ∫      dx   x +1 x 2 + 0 x + 0
             x +1
                             x2 + x
                                −x+0
                                − x −1
                                     1
x −1
              x2
e.g. ( i ) ∫      dx             x +1 x 2 + 0 x + 0
             x +1
                                       x2 + x
         = ∫ x −1 +
                       1
                           dx
                                         −x+0
                     x + 1
              
                                          − x −1
                                               1
x −1
               x2
e.g. ( i ) ∫      dx                 x +1 x 2 + 0 x + 0
             x +1
                                           x2 + x
         = ∫ x −1 +
                       1
                           dx
                                             −x+0
                     x + 1
               
                                              − x −1
             12
         = x − x + log( x + 1) + c
                                                   1
             2
x −1
               x2
e.g. ( i ) ∫      dx                 x +1 x 2 + 0 x + 0
             x +1
                                           x2 + x
         = ∫ x −1 +
                       1
                           dx
                                             −x+0
                     x + 1
               
                                              − x −1
             12
         = x − x + log( x + 1) + c
                                                   1
             2
            3dx
   ( ii ) ∫ 2
           x −x
x −1
               x2
e.g. ( i ) ∫      dx                 x +1 x 2 + 0 x + 0
             x +1
                                           x2 + x
         = ∫ x −1 +
                       1
                           dx
                                             −x+0
                     x + 1
               
                                              − x −1
             12
         = x − x + log( x + 1) + c
                                                   1
             2
            3dx
   ( ii ) ∫ 2
           x −x
              3dx
        =∫
            x( x − 1)
x −1
               x2
e.g. ( i ) ∫      dx                 x +1 x 2 + 0 x + 0
             x +1
                                           x2 + x
         = ∫ x −1 +
                       1
                           dx
                                             −x+0
                     x + 1
               
                                              − x −1
             12
         = x − x + log( x + 1) + c
                                                   1
             2
                                      A   B       3
            3dx
   ( ii ) ∫ 2                           +    =
                                      x x − 1 x( x − 1)
           x −x
              3dx
        =∫
            x( x − 1)
x −1
               x2
e.g. ( i ) ∫      dx                 x +1 x 2 + 0 x + 0
             x +1
                                           x2 + x
         = ∫ x −1 +
                       1
                           dx
                                             −x+0
                     x + 1
               
                                              − x −1
             12
         = x − x + log( x + 1) + c
                                                   1
             2
            3dx                      A     B        3
   ( ii ) ∫ 2                          +       =
                                     x x − 1 x( x − 1)
           x −x
              3dx                    A( x − 1) + Bx = 3
        =∫
            x( x − 1)
x −1
               x2
e.g. ( i ) ∫      dx                  x +1 x 2 + 0 x + 0
             x +1
                                            x2 + x
         = ∫ x −1 +
                       1
                           dx
                                              −x+0
                     x + 1
               
                                               − x −1
             12
         = x − x + log( x + 1) + c
                                                    1
             2
            3dx                        A     B        3
   ( ii ) ∫ 2                            +       =
                                       x x − 1 x( x − 1)
           x −x
              3dx                      A( x − 1) + Bx = 3
        =∫
            x( x − 1)
                                      x=0
                                     −A=3
                                      A = −3
x −1
               x2
e.g. ( i ) ∫      dx                  x +1 x 2 + 0 x + 0
             x +1
                                            x2 + x
         = ∫ x −1 +
                       1
                           dx
                                              −x+0
                     x + 1
               
                                               − x −1
             12
         = x − x + log( x + 1) + c
                                                    1
             2
            3dx                        A     B        3
   ( ii ) ∫ 2                            +       =
                                       x x − 1 x( x − 1)
           x −x
              3dx                      A( x − 1) + Bx = 3
        =∫
            x( x − 1)
                                      x=0           x =1
                                     −A=3           B=3
                                      A = −3
x −1
               x2
e.g. ( i ) ∫      dx                  x +1 x 2 + 0 x + 0
             x +1
                                            x2 + x
         = ∫ x −1 +
                       1
                           dx
                                              −x+0
                     x + 1
               
                                               − x −1
             12
         = x − x + log( x + 1) + c
                                                    1
             2
            3dx                        A     B        3
   ( ii ) ∫ 2                            +       =
                                       x x − 1 x( x − 1)
           x −x
              3dx                      A( x − 1) + Bx = 3
        =∫
            x( x − 1)
                                      x=0           x =1
         − 3    3
      = ∫ +           dx           −A=3           B=3
          x ( x − 1) 
                                      A = −3
x −1
               x2
e.g. ( i ) ∫      dx                    x +1 x 2 + 0 x + 0
             x +1
                                              x2 + x
         = ∫ x −1 +
                       1
                           dx
                                                −x+0
                     x + 1
               
                                                 − x −1
             12
         = x − x + log( x + 1) + c
                                                      1
             2
            3dx                          A     B        3
   ( ii ) ∫ 2                              +       =
                                         x x − 1 x( x − 1)
           x −x
              3dx                        A( x − 1) + Bx = 3
        =∫
            x( x − 1)
                                        x=0           x =1
         − 3      3
      = ∫ +             dx           −A=3           B=3
          x ( x − 1) 
                                        A = −3
      = −3 log x + 3 log( x − 1) + c
x −1
               x2
e.g. ( i ) ∫      dx                    x +1 x 2 + 0 x + 0
             x +1
                                              x2 + x
         = ∫ x −1 +
                       1
                           dx
                                                −x+0
                     x + 1
               
                                                 − x −1
             12
         = x − x + log( x + 1) + c
                                                      1
             2
            3dx                          A     B        3
   ( ii ) ∫ 2                              +       =
                                         x x − 1 x( x − 1)
           x −x
              3dx                        A( x − 1) + Bx = 3
        =∫
            x( x − 1)
                                        x=0           x =1
         − 3      3
      = ∫ +             dx           −A=3           B=3
          x ( x − 1) 
                                        A = −3
      = −3 log x + 3 log( x − 1) + c
               x − 1
      = 3 log      +c
             
             x
x+5
( iii ) ∫ 2           dx
         x − 3 x − 10
x+5
( iii ) ∫ 2             dx
         x − 3 x − 10
                x+5
      =∫                   dx
          ( x − 5)( x + 2)
x+5                                              x+5
                                       A         B
( iii ) ∫ 2             dx                  +          =
                                   ( x − 5) ( x + 2) ( x − 5)( x + 2)
         x − 3 x − 10
                x+5
                                A( x + 2 ) + B( x − 5) = x + 5
      =∫                   dx
          ( x − 5)( x + 2)
x+5                                              x+5
                                       A         B
( iii ) ∫ 2             dx                  +          =
                                   ( x − 5) ( x + 2) ( x − 5)( x + 2)
         x − 3 x − 10
                x+5
                                A( x + 2 ) + B( x − 5) = x + 5
      =∫                   dx
          ( x − 5)( x + 2)                 x = −2
                                       − 7B = 3
                                                −3
                                           B=
                                                 7
x+5                                              x+5
                                       A         B
( iii ) ∫ 2             dx                  +          =
                                   ( x − 5) ( x + 2) ( x − 5)( x + 2)
         x − 3 x − 10
                x+5
                                A( x + 2 ) + B( x − 5) = x + 5
      =∫                   dx
          ( x − 5)( x + 2)                 x = −2              x=5
                                       − 7B = 3             7 A = 10
                                                −3                10
                                           B=                 A=
                                                 7                 7
x+5                                                    x+5
                                             A         B
( iii ) ∫ 2             dx                        +          =
                                         ( x − 5) ( x + 2) ( x − 5)( x + 2)
         x − 3 x − 10
                x+5
                                      A( x + 2 ) + B( x − 5) = x + 5
      =∫                   dx
          ( x − 5)( x + 2)                       x = −2              x=5
       10              3                   − 7B = 3             7 A = 10
   = ∫            −            dx
       7 ( x − 5 ) 7( x + 2 )                       −3                10
                                                 B=                 A=
                                                       7                 7
x+5                                                  x+5
                                           A         B
( iii ) ∫ 2             dx                      +          =
                                       ( x − 5) ( x + 2) ( x − 5)( x + 2)
         x − 3 x − 10
                x+5
                                    A( x + 2 ) + B( x − 5) = x + 5
      =∫                   dx
          ( x − 5)( x + 2)                     x = −2              x=5
       10              3                 − 7B = 3             7 A = 10
   = ∫            −             dx
                               
       7 ( x − 5 ) 7( x + 2 )                     −3                10
                                               B=                 A=
    10                3
   = log( x − 5) − log( x + 2 ) + c                  7                 7
     7                7
x+5                                                     x+5
                                              A         B
( iii ) ∫ 2             dx                         +          =
                                          ( x − 5) ( x + 2) ( x − 5)( x + 2)
         x − 3 x − 10
                x+5
                                       A( x + 2 ) + B( x − 5) = x + 5
      =∫                   dx
          ( x − 5)( x + 2)                        x = −2              x=5
          10              3                 − 7B = 3             7 A = 10
    = ∫              −             dx
                                  
          7 ( x − 5 ) 7( x + 2 )                     −3                10
                                                  B=                 A=
       10                3
    = log( x − 5) − log( x + 2 ) + c                    7                 7
        7                7
           dx
 ( iv ) ∫ 3
         x +x
x+5                                                     x+5
                                              A         B
( iii ) ∫ 2             dx                         +          =
                                          ( x − 5) ( x + 2) ( x − 5)( x + 2)
         x − 3 x − 10
                x+5
                                       A( x + 2 ) + B( x − 5) = x + 5
      =∫                   dx
          ( x − 5)( x + 2)                        x = −2              x=5
          10              3                 − 7B = 3             7 A = 10
    = ∫              −             dx
                                  
          7 ( x − 5 ) 7( x + 2 )                     −3                10
                                                  B=                 A=
       10                3
    = log( x − 5) − log( x + 2 ) + c                    7                 7
        7                7
           dx
 ( iv ) ∫ 3
         x +x
               dx
    =∫
          x( x 2 + 1)
x+5                                                       x+5
                                              A          B
( iii ) ∫ 2             dx                         +            =
                                          ( x − 5) ( x + 2) ( x − 5)( x + 2)
         x − 3 x − 10
                x+5
                                       A( x + 2 ) + B( x − 5) = x + 5
      =∫                   dx
          ( x − 5)( x + 2)                        x = −2                x=5
          10              3                 − 7B = 3               7 A = 10
    = ∫              −             dx
                                  
          7 ( x − 5 ) 7( x + 2 )                      −3                  10
                                                  B=                   A=
       10                3
    = log( x − 5) − log( x + 2 ) + c                     7                   7
        7                7
                                                            A Bx + C           1
           dx                                                  +2         =
 ( iv ) ∫ 3                                                 x x + 1 x( x 2 + 1)
         x +x
                                               A( x 2 + 1) + ( Bx + C ) x = 1
               dx
    =∫
          x( x 2 + 1)
x+5                                                       x+5
                                              A          B
( iii ) ∫ 2             dx                         +            =
                                          ( x − 5) ( x + 2) ( x − 5)( x + 2)
         x − 3 x − 10
                x+5
                                       A( x + 2 ) + B( x − 5) = x + 5
      =∫                   dx
          ( x − 5)( x + 2)                        x = −2                x=5
          10              3                 − 7B = 3               7 A = 10
    = ∫              −             dx
                                  
          7 ( x − 5 ) 7( x + 2 )                      −3                  10
                                                  B=                   A=
       10                3
    = log( x − 5) − log( x + 2 ) + c                     7                   7
        7                7
                                                            A Bx + C           1
           dx                                                  +2         =
 ( iv ) ∫ 3                                                 x x + 1 x( x 2 + 1)
         x +x
                                               A( x 2 + 1) + ( Bx + C ) x = 1
               dx
    =∫
          x( x 2 + 1)
                                              x=0
                                           A =1
x+5                                                       x+5
                                              A          B
( iii ) ∫ 2             dx                         +            =
                                          ( x − 5) ( x + 2) ( x − 5)( x + 2)
         x − 3 x − 10
                x+5
                                       A( x + 2 ) + B( x − 5) = x + 5
      =∫                   dx
          ( x − 5)( x + 2)                        x = −2                x=5
          10              3                 − 7B = 3               7 A = 10
    = ∫              −             dx
                                  
          7 ( x − 5 ) 7( x + 2 )                      −3                  10
                                                  B=                   A=
       10                3
    = log( x − 5) − log( x + 2 ) + c                     7                   7
        7                7
                                                            A Bx + C           1
           dx                                                  +2         =
 ( iv ) ∫ 3                                                 x x + 1 x( x 2 + 1)
         x +x
                                               A( x 2 + 1) + ( Bx + C ) x = 1
               dx
    =∫
          x( x 2 + 1)                                               x=i
                                              x=0
                                                            − B + Ci = 1
                                              A =1
x+5                                                       x+5
                                              A          B
( iii ) ∫ 2             dx                         +            =
                                          ( x − 5) ( x + 2) ( x − 5)( x + 2)
         x − 3 x − 10
                x+5
                                       A( x + 2 ) + B( x − 5) = x + 5
      =∫                   dx
          ( x − 5)( x + 2)                        x = −2                x=5
          10              3                 − 7B = 3               7 A = 10
    = ∫              −             dx
                                  
          7 ( x − 5 ) 7( x + 2 )                      −3                  10
                                                  B=                   A=
       10                3
    = log( x − 5) − log( x + 2 ) + c                     7                   7
        7                7
                                                            A Bx + C           1
           dx                                                  +2         =
 ( iv ) ∫ 3                                                 x x + 1 x( x 2 + 1)
         x +x
                                               A( x 2 + 1) + ( Bx + C ) x = 1
               dx
    =∫
          x( x 2 + 1)                                               x=i
                                              x=0
                                                            − B + Ci = 1
                                              A =1
                                                               B = −1 C = 0
x+5                                                       x+5
                                              A          B
( iii ) ∫ 2             dx                         +            =
                                          ( x − 5) ( x + 2) ( x − 5)( x + 2)
         x − 3 x − 10
                x+5
                                       A( x + 2 ) + B( x − 5) = x + 5
      =∫                   dx
          ( x − 5)( x + 2)                        x = −2                x=5
          10               3                − 7B = 3               7 A = 10
    = ∫              −             dx
                                  
          7 ( x − 5 ) 7( x + 2 )                      −3                  10
                                                  B=                   A=
       10                 3
    = log( x − 5) − log( x + 2 ) + c                     7                   7
        7                 7
                                                            A Bx + C           1
            dx                                                 +2         =
 ( iv ) ∫ 3                                                 x x + 1 x( x 2 + 1)
         x +x
                                               A( x 2 + 1) + ( Bx + C ) x = 1
               dx
    =∫
          x( x 2 + 1)                                               x=i
                                              x=0
           1 − x  dx
    = ∫                                                    − B + Ci = 1
                                              A =1
                        
           x x + 1
                   2
                                                               B = −1 C = 0
x+5                                                       x+5
                                              A          B
( iii ) ∫ 2             dx                         +            =
                                          ( x − 5) ( x + 2) ( x − 5)( x + 2)
         x − 3 x − 10
                x+5
                                       A( x + 2 ) + B( x − 5) = x + 5
      =∫                   dx
          ( x − 5)( x + 2)                        x = −2                x=5
          10               3                − 7B = 3               7 A = 10
    = ∫              −             dx
                                  
          7 ( x − 5 ) 7( x + 2 )                      −3                  10
                                                  B=                   A=
       10                 3
    = log( x − 5) − log( x + 2 ) + c                     7                   7
        7                 7
                                                            A Bx + C           1
            dx                                                 +2         =
 ( iv ) ∫ 3                                                 x x + 1 x( x 2 + 1)
         x +x
                                               A( x 2 + 1) + ( Bx + C ) x = 1
               dx
    =∫
          x( x 2 + 1)                                               x=i
                                              x=0
           1 − x  dx
    = ∫                                                    − B + Ci = 1
                                              A =1
                        
           x x + 1
                   2
                                                               B = −1 C = 0
     = log x − log( x 2 + 1) + c
                  1
                  2
xdx
( v) ∫
         ( x + 1) 2 ( x 2 + 1)
xdx                                 Cx + D
                                     A        B                       x
( v) ∫                                   +          +2     =
         ( x + 1) 2 ( x 2 + 1)   ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1)
                                                  2


                                 A( x + 1)( x + 1) + B ( x       + 1) + ( Cx + D )( x + 1)
                                                                                             2
                                                                                                 =x
                                            2                2
xdx                                 Cx + D
                                     A        B                       x
( v) ∫                                   +          +2     =
         ( x + 1) 2 ( x 2 + 1)   ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1)
                                                  2


                                 A( x + 1)( x + 1) + B ( x       + 1) + ( Cx + D )( x + 1)
                                                                                             2
                                                                                                 =x
                                            2                2


                                                  x = −1
                                                2 B = −1
                                                      −1
                                                  B=
                                                       2
xdx                                 Cx + D
                                     A        B                       x
( v) ∫                                   +          +2     =
         ( x + 1) 2 ( x 2 + 1)   ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1)
                                                  2


                                 A( x + 1)( x + 1) + B ( x       + 1) + ( Cx + D )( x + 1)
                                                                                             2
                                                                                                 =x
                                            2                2


                                                                                 x=i
                                                  x = −1
                                                                       − 2C + 2 Di = i
                                                2 B = −1
                                                      −1
                                                  B=
                                                       2
xdx                                 Cx + D
                                     A        B                       x
( v) ∫                                   +          +2     =
         ( x + 1) 2 ( x 2 + 1)   ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1)
                                                  2


                                 A( x + 1)( x + 1) + B ( x       + 1) + ( Cx + D )( x + 1)
                                                                                             2
                                                                                                 =x
                                            2                2


                                                                                 x=i
                                                  x = −1
                                                                       − 2C + 2 Di = i
                                                2 B = −1
                                                                                    1
                                                                        C =0 D=
                                                      −1
                                                  B=                                 2
                                                       2
xdx                                 Cx + D
                                     A        B                       x
( v) ∫                                   +          +2     =
         ( x + 1) 2 ( x 2 + 1)   ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1)
                                                  2


                                 A( x + 1)( x + 1) + B ( x       + 1) + ( Cx + D )( x + 1)
                                                                                             2
                                                                                                 =x
                                            2                2


                                                                                 x=i
                                                  x = −1
                                                                       − 2C + 2 Di = i
                                                2 B = −1
                                                                                     1
                                                      −1                C =0 D=
                                                  B=                                 2
                                                       2
                                                                                x=0
                                                                       2A + B + D = 0
                                                                            11
                                                                        2A − + = 0
                                                                            22
                                                                                A=0
xdx                                 Cx + D
                                     A        B                       x
( v) ∫                                   +          +2     =
         ( x + 1) 2 ( x 2 + 1)   ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1)
                                                  2


                                 A( x + 1)( x + 1) + B ( x       + 1) + ( Cx + D )( x + 1)
                                                                                             2
                                                                                                 =x
                                            2                2


                                                                                 x=i
                                                  x = −1
    −1            1
= ∫            +         dx
    2( x + 1) 2( x + 1) 
                                                                       − 2C + 2 Di = i
                                                2 B = −1
              2    2
                                                                                     1
                                                      −1                C =0 D=
                                                  B=                                 2
                                                       2
                                                                                x=0
                                                                       2A + B + D = 0
                                                                            11
                                                                        2A − + = 0
                                                                            22
                                                                                A=0
xdx                                 Cx + D
                                     A        B                       x
( v) ∫                                   +          +2     =
         ( x + 1) 2 ( x 2 + 1)   ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1)
                                                  2


                                 A( x + 1)( x + 1) + B ( x       + 1) + ( Cx + D )( x + 1)
                                                                                             2
                                                                                                 =x
                                            2                2


                                                                                 x=i
                                                  x = −1
    −1            1
= ∫            +         dx
    2( x + 1) 2( x + 1) 
                                                                       − 2C + 2 Di = i
                                                2 B = −1
              2    2
                                                                                     1
                                                      −1                C =0 D=
                                                  B=
    1                   1                                                          2
        − ( x + 1) + 2
     ∫
                  −2
=                              dx                     2
                     ( x + 1) 
    2                                                                          x=0
                                                                       2A + B + D = 0
                                                                            11
                                                                        2A − + = 0
                                                                            22
                                                                                A=0
xdx                                 Cx + D
                                     A        B                       x
( v) ∫                                   +          +2     =
         ( x + 1) 2 ( x 2 + 1)   ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1)
                                                  2


                                 A( x + 1)( x + 1) + B ( x       + 1) + ( Cx + D )( x + 1)
                                                                                             2
                                                                                                 =x
                                            2                2


                                                                                 x=i
                                                  x = −1
    −1            1
= ∫            +         dx
    2( x + 1) 2( x + 1) 
                                                                       − 2C + 2 Di = i
                                                2 B = −1
              2    2
                                                                                     1
                                                      −1                C =0 D=
                                                  B=
    1                   1                                                          2
        − ( x + 1) + 2
     ∫
                  −2
=                              dx                     2
                     ( x + 1) 
    2                                                                          x=0
    − ( x + 1) −1      −1                                            2A + B + D = 0
 1
=                 + tan x  + c
 2  −1                                                                    11
                                                                        2A − + = 0
                                                                            22
                                                                                A=0
xdx                                 Cx + D
                                     A        B                       x
( v) ∫                                   +          +2     =
         ( x + 1) 2 ( x 2 + 1)   ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1)
                                                  2


                                 A( x + 1)( x + 1) + B ( x       + 1) + ( Cx + D )( x + 1)
                                                                                             2
                                                                                                 =x
                                            2                2


                                                                                 x=i
                                                  x = −1
    −1            1
= ∫            +         dx
    2( x + 1) 2( x + 1) 
                                                                       − 2C + 2 Di = i
                                                2 B = −1
              2    2
                                                                                     1
                                                      −1                C =0 D=
                                                  B=
    1                   1                                                          2
        − ( x + 1) + 2
     ∫
                  −2
=                              dx                     2
                     ( x + 1) 
    2                                                                          x=0
    − ( x + 1) −1      −1                                            2A + B + D = 0
 1
=                 + tan x  + c
 2  −1                                                                    11
                                                                        2A − + = 0
                                                                            22
 1 1                
          + tan −1 x  + c
=                                                                              A=0
 2  x +1            
xdx                                 Cx + D
                                     A        B                       x
( v) ∫                                   +          +2     =
         ( x + 1) 2 ( x 2 + 1)   ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1)
                                                  2


                                 A( x + 1)( x + 1) + B ( x       + 1) + ( Cx + D )( x + 1)
                                                                                             2
                                                                                                 =x
                                            2                2


                                                                                 x=i
                                                  x = −1
    −1            1
= ∫            +         dx
    2( x + 1) 2( x + 1) 
                                                                       − 2C + 2 Di = i
                                                2 B = −1
              2    2
                                                                                     1
                                                      −1                C =0 D=
                                                  B=
    1                   1                                                          2
        − ( x + 1) + 2
     ∫
                  −2
=                              dx                     2
                     ( x + 1) 
    2                                                                          x=0
    − ( x + 1) −1      −1                                            2A + B + D = 0
 1
=                 + tan x  + c
 2  −1                                                                    11
                                                                        2A − + = 0
                                                                            22
                                         Exercise 2G;
 1 1                
          + tan −1 x  + c
=                                                                              A=0
                                         1, 3, 5, 7 to 21
 2  x +1            

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X2 T05 06 Partial Fractions

  • 1. Integration By Partial Fractions () Ax To find; ∫ dx P( x )
  • 2. Integration By Partial Fractions () Ax To find; ∫ dx P( x ) (1) If degA( x ) ≥ degP( x ) , perform a division
  • 3. Integration By Partial Fractions ()Ax To find; ∫ dx P( x ) (1) If degA( x ) ≥ degP( x ) , perform a division ( 2) If degA( x ) < degP( x ) , factorise P( x )
  • 4. Integration By Partial Fractions ()Ax To find; ∫ dx P( x ) (1) If degA( x ) ≥ degP( x ) , perform a division ( 2) If degA( x ) < degP( x ) , factorise P( x ) A a) for linear factor ( x − a ) , write x−a
  • 5. Integration By Partial Fractions ()Ax To find; ∫ dx P( x ) (1) If degA( x ) ≥ degP( x ) , perform a division ( 2) If degA( x ) < degP( x ) , factorise P( x ) A a) for linear factor ( x − a ) , write x−a b) for multiple linear factors ( x − a ) , write n A B C + ++ ( x − a) ( x − a) ( x − a) n 2
  • 6. Integration By Partial Fractions ()Ax To find; ∫ dx P( x ) (1) If degA( x ) ≥ degP( x ) , perform a division ( 2) If degA( x ) < degP( x ) , factorise P( x ) A a) for linear factor ( x − a ) , write x−a b) for multiple linear factors ( x − a ) , write n A B C + ++ ( x − a) ( x − a) ( x − a) n 2 Ax + B c) for polynomial factors e.g. ax + bx + c, write 2 2 ax + bx + c
  • 7. x2 e.g. ( i ) ∫ dx x +1
  • 8. x x2 e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0 x +1 x2 + x
  • 9. x −1 x2 e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0 x +1 x2 + x −x+0
  • 10. x −1 x2 e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0 x +1 x2 + x −x+0 − x −1
  • 11. x −1 x2 e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0 x +1 x2 + x −x+0 − x −1 1
  • 12. x −1 x2 e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0 x +1 x2 + x = ∫ x −1 + 1  dx  −x+0 x + 1  − x −1 1
  • 13. x −1 x2 e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0 x +1 x2 + x = ∫ x −1 + 1  dx  −x+0 x + 1  − x −1 12 = x − x + log( x + 1) + c 1 2
  • 14. x −1 x2 e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0 x +1 x2 + x = ∫ x −1 + 1  dx  −x+0 x + 1  − x −1 12 = x − x + log( x + 1) + c 1 2 3dx ( ii ) ∫ 2 x −x
  • 15. x −1 x2 e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0 x +1 x2 + x = ∫ x −1 + 1  dx  −x+0 x + 1  − x −1 12 = x − x + log( x + 1) + c 1 2 3dx ( ii ) ∫ 2 x −x 3dx =∫ x( x − 1)
  • 16. x −1 x2 e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0 x +1 x2 + x = ∫ x −1 + 1  dx  −x+0 x + 1  − x −1 12 = x − x + log( x + 1) + c 1 2 A B 3 3dx ( ii ) ∫ 2 + = x x − 1 x( x − 1) x −x 3dx =∫ x( x − 1)
  • 17. x −1 x2 e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0 x +1 x2 + x = ∫ x −1 + 1  dx  −x+0 x + 1  − x −1 12 = x − x + log( x + 1) + c 1 2 3dx A B 3 ( ii ) ∫ 2 + = x x − 1 x( x − 1) x −x 3dx A( x − 1) + Bx = 3 =∫ x( x − 1)
  • 18. x −1 x2 e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0 x +1 x2 + x = ∫ x −1 + 1  dx  −x+0 x + 1  − x −1 12 = x − x + log( x + 1) + c 1 2 3dx A B 3 ( ii ) ∫ 2 + = x x − 1 x( x − 1) x −x 3dx A( x − 1) + Bx = 3 =∫ x( x − 1) x=0 −A=3 A = −3
  • 19. x −1 x2 e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0 x +1 x2 + x = ∫ x −1 + 1  dx  −x+0 x + 1  − x −1 12 = x − x + log( x + 1) + c 1 2 3dx A B 3 ( ii ) ∫ 2 + = x x − 1 x( x − 1) x −x 3dx A( x − 1) + Bx = 3 =∫ x( x − 1) x=0 x =1 −A=3 B=3 A = −3
  • 20. x −1 x2 e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0 x +1 x2 + x = ∫ x −1 + 1  dx  −x+0 x + 1  − x −1 12 = x − x + log( x + 1) + c 1 2 3dx A B 3 ( ii ) ∫ 2 + = x x − 1 x( x − 1) x −x 3dx A( x − 1) + Bx = 3 =∫ x( x − 1) x=0 x =1 − 3 3 = ∫ +  dx −A=3 B=3  x ( x − 1)  A = −3
  • 21. x −1 x2 e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0 x +1 x2 + x = ∫ x −1 + 1  dx  −x+0 x + 1  − x −1 12 = x − x + log( x + 1) + c 1 2 3dx A B 3 ( ii ) ∫ 2 + = x x − 1 x( x − 1) x −x 3dx A( x − 1) + Bx = 3 =∫ x( x − 1) x=0 x =1 − 3 3 = ∫ +  dx −A=3 B=3  x ( x − 1)  A = −3 = −3 log x + 3 log( x − 1) + c
  • 22. x −1 x2 e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0 x +1 x2 + x = ∫ x −1 + 1  dx  −x+0 x + 1  − x −1 12 = x − x + log( x + 1) + c 1 2 3dx A B 3 ( ii ) ∫ 2 + = x x − 1 x( x − 1) x −x 3dx A( x − 1) + Bx = 3 =∫ x( x − 1) x=0 x =1 − 3 3 = ∫ +  dx −A=3 B=3  x ( x − 1)  A = −3 = −3 log x + 3 log( x − 1) + c x − 1 = 3 log +c  x
  • 23. x+5 ( iii ) ∫ 2 dx x − 3 x − 10
  • 24. x+5 ( iii ) ∫ 2 dx x − 3 x − 10 x+5 =∫ dx ( x − 5)( x + 2)
  • 25. x+5 x+5 A B ( iii ) ∫ 2 dx + = ( x − 5) ( x + 2) ( x − 5)( x + 2) x − 3 x − 10 x+5 A( x + 2 ) + B( x − 5) = x + 5 =∫ dx ( x − 5)( x + 2)
  • 26. x+5 x+5 A B ( iii ) ∫ 2 dx + = ( x − 5) ( x + 2) ( x − 5)( x + 2) x − 3 x − 10 x+5 A( x + 2 ) + B( x − 5) = x + 5 =∫ dx ( x − 5)( x + 2) x = −2 − 7B = 3 −3 B= 7
  • 27. x+5 x+5 A B ( iii ) ∫ 2 dx + = ( x − 5) ( x + 2) ( x − 5)( x + 2) x − 3 x − 10 x+5 A( x + 2 ) + B( x − 5) = x + 5 =∫ dx ( x − 5)( x + 2) x = −2 x=5 − 7B = 3 7 A = 10 −3 10 B= A= 7 7
  • 28. x+5 x+5 A B ( iii ) ∫ 2 dx + = ( x − 5) ( x + 2) ( x − 5)( x + 2) x − 3 x − 10 x+5 A( x + 2 ) + B( x − 5) = x + 5 =∫ dx ( x − 5)( x + 2) x = −2 x=5  10 3 − 7B = 3 7 A = 10 = ∫ −  dx  7 ( x − 5 ) 7( x + 2 )  −3 10 B= A= 7 7
  • 29. x+5 x+5 A B ( iii ) ∫ 2 dx + = ( x − 5) ( x + 2) ( x − 5)( x + 2) x − 3 x − 10 x+5 A( x + 2 ) + B( x − 5) = x + 5 =∫ dx ( x − 5)( x + 2) x = −2 x=5  10 3 − 7B = 3 7 A = 10 = ∫ − dx   7 ( x − 5 ) 7( x + 2 )  −3 10 B= A= 10 3 = log( x − 5) − log( x + 2 ) + c 7 7 7 7
  • 30. x+5 x+5 A B ( iii ) ∫ 2 dx + = ( x − 5) ( x + 2) ( x − 5)( x + 2) x − 3 x − 10 x+5 A( x + 2 ) + B( x − 5) = x + 5 =∫ dx ( x − 5)( x + 2) x = −2 x=5  10 3 − 7B = 3 7 A = 10 = ∫ − dx   7 ( x − 5 ) 7( x + 2 )  −3 10 B= A= 10 3 = log( x − 5) − log( x + 2 ) + c 7 7 7 7 dx ( iv ) ∫ 3 x +x
  • 31. x+5 x+5 A B ( iii ) ∫ 2 dx + = ( x − 5) ( x + 2) ( x − 5)( x + 2) x − 3 x − 10 x+5 A( x + 2 ) + B( x − 5) = x + 5 =∫ dx ( x − 5)( x + 2) x = −2 x=5  10 3 − 7B = 3 7 A = 10 = ∫ − dx   7 ( x − 5 ) 7( x + 2 )  −3 10 B= A= 10 3 = log( x − 5) − log( x + 2 ) + c 7 7 7 7 dx ( iv ) ∫ 3 x +x dx =∫ x( x 2 + 1)
  • 32. x+5 x+5 A B ( iii ) ∫ 2 dx + = ( x − 5) ( x + 2) ( x − 5)( x + 2) x − 3 x − 10 x+5 A( x + 2 ) + B( x − 5) = x + 5 =∫ dx ( x − 5)( x + 2) x = −2 x=5  10 3 − 7B = 3 7 A = 10 = ∫ − dx   7 ( x − 5 ) 7( x + 2 )  −3 10 B= A= 10 3 = log( x − 5) − log( x + 2 ) + c 7 7 7 7 A Bx + C 1 dx +2 = ( iv ) ∫ 3 x x + 1 x( x 2 + 1) x +x A( x 2 + 1) + ( Bx + C ) x = 1 dx =∫ x( x 2 + 1)
  • 33. x+5 x+5 A B ( iii ) ∫ 2 dx + = ( x − 5) ( x + 2) ( x − 5)( x + 2) x − 3 x − 10 x+5 A( x + 2 ) + B( x − 5) = x + 5 =∫ dx ( x − 5)( x + 2) x = −2 x=5  10 3 − 7B = 3 7 A = 10 = ∫ − dx   7 ( x − 5 ) 7( x + 2 )  −3 10 B= A= 10 3 = log( x − 5) − log( x + 2 ) + c 7 7 7 7 A Bx + C 1 dx +2 = ( iv ) ∫ 3 x x + 1 x( x 2 + 1) x +x A( x 2 + 1) + ( Bx + C ) x = 1 dx =∫ x( x 2 + 1) x=0 A =1
  • 34. x+5 x+5 A B ( iii ) ∫ 2 dx + = ( x − 5) ( x + 2) ( x − 5)( x + 2) x − 3 x − 10 x+5 A( x + 2 ) + B( x − 5) = x + 5 =∫ dx ( x − 5)( x + 2) x = −2 x=5  10 3 − 7B = 3 7 A = 10 = ∫ − dx   7 ( x − 5 ) 7( x + 2 )  −3 10 B= A= 10 3 = log( x − 5) − log( x + 2 ) + c 7 7 7 7 A Bx + C 1 dx +2 = ( iv ) ∫ 3 x x + 1 x( x 2 + 1) x +x A( x 2 + 1) + ( Bx + C ) x = 1 dx =∫ x( x 2 + 1) x=i x=0 − B + Ci = 1 A =1
  • 35. x+5 x+5 A B ( iii ) ∫ 2 dx + = ( x − 5) ( x + 2) ( x − 5)( x + 2) x − 3 x − 10 x+5 A( x + 2 ) + B( x − 5) = x + 5 =∫ dx ( x − 5)( x + 2) x = −2 x=5  10 3 − 7B = 3 7 A = 10 = ∫ − dx   7 ( x − 5 ) 7( x + 2 )  −3 10 B= A= 10 3 = log( x − 5) − log( x + 2 ) + c 7 7 7 7 A Bx + C 1 dx +2 = ( iv ) ∫ 3 x x + 1 x( x 2 + 1) x +x A( x 2 + 1) + ( Bx + C ) x = 1 dx =∫ x( x 2 + 1) x=i x=0 − B + Ci = 1 A =1 B = −1 C = 0
  • 36. x+5 x+5 A B ( iii ) ∫ 2 dx + = ( x − 5) ( x + 2) ( x − 5)( x + 2) x − 3 x − 10 x+5 A( x + 2 ) + B( x − 5) = x + 5 =∫ dx ( x − 5)( x + 2) x = −2 x=5  10 3 − 7B = 3 7 A = 10 = ∫ − dx   7 ( x − 5 ) 7( x + 2 )  −3 10 B= A= 10 3 = log( x − 5) − log( x + 2 ) + c 7 7 7 7 A Bx + C 1 dx +2 = ( iv ) ∫ 3 x x + 1 x( x 2 + 1) x +x A( x 2 + 1) + ( Bx + C ) x = 1 dx =∫ x( x 2 + 1) x=i x=0  1 − x  dx = ∫ − B + Ci = 1 A =1   x x + 1 2 B = −1 C = 0
  • 37. x+5 x+5 A B ( iii ) ∫ 2 dx + = ( x − 5) ( x + 2) ( x − 5)( x + 2) x − 3 x − 10 x+5 A( x + 2 ) + B( x − 5) = x + 5 =∫ dx ( x − 5)( x + 2) x = −2 x=5  10 3 − 7B = 3 7 A = 10 = ∫ − dx   7 ( x − 5 ) 7( x + 2 )  −3 10 B= A= 10 3 = log( x − 5) − log( x + 2 ) + c 7 7 7 7 A Bx + C 1 dx +2 = ( iv ) ∫ 3 x x + 1 x( x 2 + 1) x +x A( x 2 + 1) + ( Bx + C ) x = 1 dx =∫ x( x 2 + 1) x=i x=0  1 − x  dx = ∫ − B + Ci = 1 A =1   x x + 1 2 B = −1 C = 0 = log x − log( x 2 + 1) + c 1 2
  • 38. xdx ( v) ∫ ( x + 1) 2 ( x 2 + 1)
  • 39. xdx Cx + D A B x ( v) ∫ + +2 = ( x + 1) 2 ( x 2 + 1) ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1) 2 A( x + 1)( x + 1) + B ( x + 1) + ( Cx + D )( x + 1) 2 =x 2 2
  • 40. xdx Cx + D A B x ( v) ∫ + +2 = ( x + 1) 2 ( x 2 + 1) ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1) 2 A( x + 1)( x + 1) + B ( x + 1) + ( Cx + D )( x + 1) 2 =x 2 2 x = −1 2 B = −1 −1 B= 2
  • 41. xdx Cx + D A B x ( v) ∫ + +2 = ( x + 1) 2 ( x 2 + 1) ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1) 2 A( x + 1)( x + 1) + B ( x + 1) + ( Cx + D )( x + 1) 2 =x 2 2 x=i x = −1 − 2C + 2 Di = i 2 B = −1 −1 B= 2
  • 42. xdx Cx + D A B x ( v) ∫ + +2 = ( x + 1) 2 ( x 2 + 1) ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1) 2 A( x + 1)( x + 1) + B ( x + 1) + ( Cx + D )( x + 1) 2 =x 2 2 x=i x = −1 − 2C + 2 Di = i 2 B = −1 1 C =0 D= −1 B= 2 2
  • 43. xdx Cx + D A B x ( v) ∫ + +2 = ( x + 1) 2 ( x 2 + 1) ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1) 2 A( x + 1)( x + 1) + B ( x + 1) + ( Cx + D )( x + 1) 2 =x 2 2 x=i x = −1 − 2C + 2 Di = i 2 B = −1 1 −1 C =0 D= B= 2 2 x=0 2A + B + D = 0 11 2A − + = 0 22 A=0
  • 44. xdx Cx + D A B x ( v) ∫ + +2 = ( x + 1) 2 ( x 2 + 1) ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1) 2 A( x + 1)( x + 1) + B ( x + 1) + ( Cx + D )( x + 1) 2 =x 2 2 x=i x = −1  −1 1 = ∫ +  dx  2( x + 1) 2( x + 1)  − 2C + 2 Di = i 2 B = −1 2 2 1 −1 C =0 D= B= 2 2 x=0 2A + B + D = 0 11 2A − + = 0 22 A=0
  • 45. xdx Cx + D A B x ( v) ∫ + +2 = ( x + 1) 2 ( x 2 + 1) ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1) 2 A( x + 1)( x + 1) + B ( x + 1) + ( Cx + D )( x + 1) 2 =x 2 2 x=i x = −1  −1 1 = ∫ +  dx  2( x + 1) 2( x + 1)  − 2C + 2 Di = i 2 B = −1 2 2 1 −1 C =0 D= B= 1 1 2 − ( x + 1) + 2 ∫ −2 =  dx 2 ( x + 1)  2 x=0 2A + B + D = 0 11 2A − + = 0 22 A=0
  • 46. xdx Cx + D A B x ( v) ∫ + +2 = ( x + 1) 2 ( x 2 + 1) ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1) 2 A( x + 1)( x + 1) + B ( x + 1) + ( Cx + D )( x + 1) 2 =x 2 2 x=i x = −1  −1 1 = ∫ +  dx  2( x + 1) 2( x + 1)  − 2C + 2 Di = i 2 B = −1 2 2 1 −1 C =0 D= B= 1 1 2 − ( x + 1) + 2 ∫ −2 =  dx 2 ( x + 1)  2 x=0  − ( x + 1) −1 −1  2A + B + D = 0 1 = + tan x  + c 2  −1  11 2A − + = 0 22 A=0
  • 47. xdx Cx + D A B x ( v) ∫ + +2 = ( x + 1) 2 ( x 2 + 1) ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1) 2 A( x + 1)( x + 1) + B ( x + 1) + ( Cx + D )( x + 1) 2 =x 2 2 x=i x = −1  −1 1 = ∫ +  dx  2( x + 1) 2( x + 1)  − 2C + 2 Di = i 2 B = −1 2 2 1 −1 C =0 D= B= 1 1 2 − ( x + 1) + 2 ∫ −2 =  dx 2 ( x + 1)  2 x=0  − ( x + 1) −1 −1  2A + B + D = 0 1 = + tan x  + c 2  −1  11 2A − + = 0 22 1 1  + tan −1 x  + c = A=0 2  x +1 
  • 48. xdx Cx + D A B x ( v) ∫ + +2 = ( x + 1) 2 ( x 2 + 1) ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1) 2 A( x + 1)( x + 1) + B ( x + 1) + ( Cx + D )( x + 1) 2 =x 2 2 x=i x = −1  −1 1 = ∫ +  dx  2( x + 1) 2( x + 1)  − 2C + 2 Di = i 2 B = −1 2 2 1 −1 C =0 D= B= 1 1 2 − ( x + 1) + 2 ∫ −2 =  dx 2 ( x + 1)  2 x=0  − ( x + 1) −1 −1  2A + B + D = 0 1 = + tan x  + c 2  −1  11 2A − + = 0 22 Exercise 2G; 1 1  + tan −1 x  + c = A=0 1, 3, 5, 7 to 21 2  x +1 