Интегральное исчисление функции одной переменной. - 20 стр.

UptoLike

ВУЗ: 

МГТУ ГА | Москва

Рубрика: 

20 §1. îÅÏÐÒÅÄÅÌÅÎÎÙÊ ÉÎÔÅÇÒÁÌ. . .
IV. éÎÔÅÇÒÉÒÏ×ÁÎÉÅ ÄÉÆÆÅÒÅÎÃÉÁÌØÎÙÈ ÂÉÎÏÍÏ× ×ÉÄÁ:
Z
x
m
(a + bx
n
)
p
dx,
ÇÄÅ p =
α
β
; α, β ¡ ÃÅÌÙÅ ÞÉÓÌÁ; m, n ¡ ÐÒÏÉÚ×ÏÌØÎÙÅ ÞÉÓÌÁ.
äÁÎÎÙÊ ÉÎÔÅÇÒÁÌ ÉÎÔÅÇÒÉÒÕÅÔÓÑ ÌÉÛØ × ÓÌÅÄÕÀÝÉÈ ÔÒÅÈ ÓÌÕÞÁÑÈ:
1) ÅÓÌÉ p ÃÅÌÏÅ ÞÉÓÌÏ, ÔÏ ÉÓÐÏÌØÚÕÅÔÓÑ ÐÏÄÓÔÁÎÏ×ËÁ
x = t
N
, ÇÄÅ N ¡ ÏÂÝÉÊ ÚÎÁÍÅÎÁÔÅÌØ ÄÒÏÂÅÊ m É n;
2) ÅÓÌÉ
m+1
n
ÃÅÌÏÅ ÞÉÓÌÏ, ÔÏ ÉÓÐÏÌØÚÕÅÔÓÑ ÐÏÄÓÔÁÎÏ×ËÁ
a + bx
n
= t
β
;
3) ÅÓÌÉ
m+1
n
+ p ÃÅÌÏÅ ÞÉÓÌÏ, ÔÏ ÉÓÐÏÌØÚÕÅÔÓÑ ÐÏÄÓÔÁÎÏ×ËÁ
a + bx
n
= x
n
t
β
.
ðÒÉÍÅÒ 30.
Z
x
3
√
x
2
− 1
dx,
ÇÄÅ m = 3, n = 2. þÉÓÌÏ
m+1
n
=
3+1
2
= 2 ¡ ÃÅÌÏÅ, ÐÏÜÔÏÍÕ ÉÓÐÏÌØÚÕÅÍ
ÐÏÄÓÔÁÎÏ×ËÕ (1): x
2
− 1 = t
2
, x =
√
t
2
+ 1, dx =
t
√
1+t
2
dt, ÏÔÓÀÄÁ
Z
x
3
√
x
2
− 1
dx =
Z
(1 + t
2
)
3/2
t
·
t
(t
2
+ 1)
1/2
dt =
Z
(1 + t
2
) dt =
= t +
t
3
3
+ C = (x
2
− 1)
1/2
+
(x
2
− 1)
3/2
3
+ C.
ðÒÉÍÅÒ 31.
Z
dx
4
√
1 + x
4
=
Z
x
0
(1 + x
4
)
−1/4
dx,
ÚÄÅÓØ m = 0, n = 4, p = −
1
4
. ðÒÏ×ÅÒÉÍ ÕÓÌÏ×ÉÅ:
m+1
n
+ p ¡ ÃÅÌÏÅ ÞÉÓÌÏ ÉÌÉ
ÎÏÌØ.
m+1
n
+p =
0+1
4
−
1
4
= 0, ÐÏÜÔÏÍÕ ÉÓÐÏÌØÚÕÅÍ ÐÏÄÓÔÁÎÏ×ËÕ (2); ÚÄÅÓØ β = 4:
1 + x
4
= x
4
· t
4
, t =
4
q
1
x
4
+ 1 =
4
√
1+x
4
x
, x = (t
4
−1)
−1/4
, dx = −t
3
(t
4
−1)
−5/4
dt,
ÔÁË ÞÔÏ
4
√
1 + x
4
= tx = t · (t
4
− 1)
−1/4
,
Z
dx
4
√
1 + x
4
= −
Z
t
2
dt
t
4
− 1
=
1
4
Z
1
t + 1
−
1
t − 1
dt −
1
2
Z
dt
t
2
+ 1
=
=
1
4
ln
t + 1
t − 1
−
1
2
arctg t + C,
ÇÄÅ t =
4
√
1+x
4
x
.
20                                                §1. îÅÏÐÒÅÄÅÌÅÎÎÙÊ ÉÎÔÅÇÒÁÌ. . .

     IV. éÎÔÅÇÒÉÒÏ×ÁÎÉÅ ÄÉÆÆÅÒÅÎÃÉÁÌØÎÙÈ ÂÉÎÏÍÏ× ×ÉÄÁ:
                           Z
                              xm(a + bxn )p dx,

ÇÄÅ p = αβ ; α, β ¡ ÃÅÌÙÅ ÞÉÓÌÁ; m, n ¡ ÐÒÏÉÚ×ÏÌØÎÙÅ ÞÉÓÌÁ.
   äÁÎÎÙÊ ÉÎÔÅÇÒÁÌ ÉÎÔÅÇÒÉÒÕÅÔÓÑ ÌÉÛØ × ÓÌÅÄÕÀÝÉÈ ÔÒÅÈ ÓÌÕÞÁÑÈ:
   1) ÅÓÌÉ p ÃÅÌÏÅ ÞÉÓÌÏ, ÔÏ ÉÓÐÏÌØÚÕÅÔÓÑ ÐÏÄÓÔÁÎÏ×ËÁ
                x = tN , ÇÄÅ N ¡ ÏÂÝÉÊ ÚÎÁÍÅÎÁÔÅÌØ ÄÒÏÂÅÊ m É n;
               m+1
     2) ÅÓÌÉ    n    ÃÅÌÏÅ ÞÉÓÌÏ, ÔÏ ÉÓÐÏÌØÚÕÅÔÓÑ ÐÏÄÓÔÁÎÏ×ËÁ
                                       a + bxn = tβ ;
               m+1
     3) ÅÓÌÉ    n    + p ÃÅÌÏÅ ÞÉÓÌÏ, ÔÏ ÉÓÐÏÌØÚÕÅÔÓÑ ÐÏÄÓÔÁÎÏ×ËÁ
                                      a + bxn = xn tβ .
     ðÒÉÍÅÒ 30.
                                   x3
                                      Z
                                √        dx,
                                  x2 − 1
ÇÄÅ m = 3, n = 2. þÉÓÌÏ m+1     = 3+1 = 2 ¡ ÃÅÌÏÅ, ÐÏÜÔÏÍÕ ÉÓÐÏÌØÚÕÅÍ
                  2       2
                             n  √ 2            t
                                  2
ÐÏÄÓÔÁÎÏ×ËÕ (1): x − 1 = t , x = t + 1, dx = √1+t   dt, ÏÔÓÀÄÁ
                                                  2



        x3                  (1 + t2 )3/2       t
  Z                     Z                                 Z
      √       dx =                       · 2      1/2
                                                      dt = (1 + t2 ) dt =
       x2 − 1                   t         (t + 1)
                                              t3          2    1/2    (x2 − 1)3/2
                                        = t + + C = (x − 1) +                     + C.
                                              3                            3
     ðÒÉÍÅÒ 31.
                            dx
                            Z       Z
                        √
                        4
                                   = x0(1 + x4)−1/4 dx,
                          1+x    4

ÚÄÅÓØ m = 0, n = 4, p = − 14 . ðÒÏ×ÅÒÉÍ ÕÓÌÏ×ÉÅ: m+1
                                                  n + p ¡ ÃÅÌÏÅ ÞÉÓÌÏ ÉÌÉ
      m+1      0+1 1
ÎÏÌØ. n +p = 4 −   q4 = 0, ÐÏÜÔÏÍÕ ÉÓÐÏÌØÚÕÅÍ ÐÏÄÓÔÁÎÏ×ËÕ (2); ÚÄÅÓØ β = 4:
                                      √
                                      4   4
1 + x4 = x4 · t4, t = 4 x14 + 1 = 1+x         4
                                    x , x = (t − 1)
                                                    −1/4
                                                         , dx = −t3 (t4 − 1)−5/4 dt,
         √
ÔÁË ÞÔÏ 4 1 + x4 = tx = t · (t4 − 1)−1/4,
                     Z 2           Z              
        dx              t dt     1      1       1            1    dt
  Z                                                            Z
     √         =−              =           −          dt  −              =
     4
       1 + x4          t4 − 1 4        t+1 t−1               2 t2 + 1
                                                  1       t+1     1
                                                = ln             − arctg t + C,
                                                  4       t−1     2
          √
          4
              1+x4
ÇÄÅ t =        x .

Страницы