Практикум по высшей математике с основами математической статистики. Фетисов Ю.М - 24 стр.

                  x +4 =t
    x +1
         ⋅ dx = x =(t −4 )
                                      (t −4 )2 +1 ⋅ 2 ⋅ (t −4 )⋅ dt =
∫                                   =∫
                           2

    x +4                                   t
                dx =2 ⋅ (t −4 )⋅ dt

=2 ⋅ ∫
          (t−4t +17 )⋅ (t −4 )⋅ dt
               2
                                    =2 ⋅ ∫
                                                t 3 −4t 2 +17t −4t 2 +16t −20
                                                                              ⋅ dt =
                    t                                          t
       t 3 −8t 2 +33t −20                 �                 20 �
=2 ⋅ ∫                      ⋅ dt =2 ∫� t 2 −8t +33 − � ⋅ dt =2 ⋅ ∫t 2 ⋅ dt −
                 t                          �                t �
                                               3        2
                                 dt 2t 16t
−16 ⋅ ∫t ⋅ dt +66 ⋅ ∫dt −40 ⋅ ∫ =                  −      +66t −40 ⋅ ln t +C =
                                   t          3       2

=
      (              )
 2 ⋅ x +4 3 16 ⋅ x +4 2
                 −
                                 (         )        (
                                     +66 ⋅ x +4 −40 ln x +4 +C =  )           (               )
          3               2
 2 �                                  �
                                                        (                     )
          3
= ⋅ �� x 2 +3 x ⋅ 4 +3 x ⋅16 +64 �� +8 ⋅ x +8 x +16 +66 x +264 −
 3 �                                    �

          (              )                                                (              )
                              3
                         2
−40 ln x +4 +C = ⋅ x 2 +16 x +138 x −40 ln x +4 +C1 ,
                         3


                                                            (         )
                             3
          2
    Ответ: ⋅ x 2 +16 x +138 x −40 ln                            x +4 +C .
          3

    Пример 19. Найти неопределенный интеграл                                       ∫x ⋅ e
                                                                                             3x
                                                                                                  dx .

  Решение. Воспользуемся формулой интегрирования по частям. Для этого
обозначим x через u, а e2xdx через dv:

                     u =x            dv =e 3 x dx
                                                                                      x ⋅ e3x 1
∫x ⋅ e        dx =                                                                  =        − ⋅ ∫e 3 x dx =
         3x
                                                                              3x
                                          1                   e
                     du =dx v =∫e 3 x dx = ⋅ ∫e 3 x d (3 x ) =                           3    3
                                          3                    3

              x ⋅ e3 x 1 e3x
      =               − ⋅    +C.
                 3     3 3


                                           Определенный интеграл
                                                                                     0
                                                                                             x ⋅ dx
    Пример 20. Вычислить определенный интеграл:                                     ∫
                                                                                    −4 1 −2 x
                                                                                              .




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