Неопределенные интегралы. Желтухин В.С. - 6 стр.

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6
Z
x
2
dx − 3
Z
x dx + 5
Z
dx = 2x
3
−
3
2
x
2
+ 5x + C.
Z
(6x
2
− 3x + 5) dx = 2x
3
−
3
2
x
2
+ 5x + C .
Z
(1 +
√
x )
4
dx
Z
(1 +
√
x )
4
dx =
Z
(1 + 4
√
x + 6x + 4x
√
x + x
2
) dx =
=
Z
dx + 4
Z
√
x dx + 6
Z
x dx + 4
Z
x
3/2
dx +
Z
x
2
dx =
= x +
8
3
x
3/2
+ 3x
2
+
8
5
x
5/2
+
1
3
x
3
+ C.
J =
Z
(x + 1)(x
2
− 3)
3x
2
dx
J =
Z
x
3
+ x
2
− 3x − 3
3x
2
dx =
Z
µ
1
3
x +
1
3
−
1
x
−
1
x
2
dx =
=
1
3
Z
x dx +
1
3
Z
dx −
Z
dx
x
−
Z
dx
x
2
=
=
1
6
x
2
+
1
3
x − ln x +
1
x
+ C.
Z
dx
x − a
= ln |x − a| + C.
Z
sin mx dx = −
1
m
cos mx + C.
Z
e
−3x
dx = −
1
3
e
−3x
+ C.
Z
dx
a
2
+ x
2
=
1
a
2
Z
dx
1 + (x/a)
2
=
1
a
arctg
x
a
+ C.
è íàïîñëåäîê âîñïîëüçóåìñÿ ï.ï. 2, 3 òàáë. 1:
              Z                  Z                Z
                                                                              3 2
          6        x2 dx − 3         x dx + 5             dx = 2x3 −            x + 5x + C.
                                                                              2
                       Z
                                                                            3 2
Òàêèì îáðàçîì,                 (6x2 − 3x + 5) dx = 2x3 −                      x + 5x + C. /
                                                                            2
                                                                Z
                                                                            √
Ï ð è ì å ð 2. Âû÷èñëèòü èíòåãðàë                                   (1 +          x )4 dx.
          Z                            Z
                      √                                   √              √
      .       (1 +         x )4 dx =       (1 + 4             x + 6x + 4x x + x2 ) dx =
          Z                Z                  Z                     Z                    Z
                               √                                            3/2
      =           dx + 4           x dx + 6       x dx + 4              x         dx +       x2 dx =

                        8 3/2           8        1
                     =x+  x + 3x2 + x5/2 + x3 + C. /
                        3               5        3
                                      Z
                                          (x + 1)(x2 − 3)
Ïðèì          å ð 3. Âû÷èñëèòü J =                        dx.
                                                3x2
               Z 3                       Z µ                    ¶
                  x + x2 − 3x − 3             1     1 1       1
  .J =                            dx =          x + − − 2 dx =
                        3x2                   3     3 x x
                 Z          Z       Z         Z
               1          1             dx       dx
          =        x dx +     dx −          −        =
               3          3              x       x2
               1 2 1              1
          =      x + x − ln x + + C. /
               6      3           x
      Ïðèìåðû íà ïðèìåíåíèå ïðàâèëà III:

Ï ð è ì å ð 4. .
      Z
           dx
 1)           = ln |x − a| + C.
          x−a
      Z
                                    1
 2)       sin mx dx = −               cos mx + C.
                                    m
      Z
                     1
 3)       e−3x dx = − e−3x + C. /
                     3
Ï ð è ì å ð 5. .
      Z                         Z
            dx      1                    dx       1      x
 1)               =                           2 =   arctg + C.
          a2 + x2   a2               1 + (x/a)    a      a

                                                      6

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