Ky-thuat-bien-doi-vi-phan-de-tim-nguyen-ham

1. Kß THUŠT BI˜N ÊI VI PH…N š TœM NGUY–N H€M
So¤n th£o bði: L¶ Trung T½n, Tê: To¡n, Tr÷íng THPT Hçng Ngü 2
I. Tâm t­t l½ thuy¸t
Trong s¡ch gi¡o khoa ¤i sè v Gi£i t½ch lîp 11 cì b£n câ ành ngh¾a vi ph¥n nh÷ sau
ành ngh¾a 1. Cho h m sè y = f(x) x¡c ành tr¶n kho£ng (a; b) v câ ¤o h m t¤i
x 2 (a; b). Gi£ sû x l sè gia cõa x. Ta gåi f0(x):x l vi ph¥n cõa h m sè y = f(x) t¤i x
ùng vîi sè gia x. Kþ hi»u dy = df(x) = f0(x):x
Chó þ: Vîi h m sè y = x ta câ dx = x. N¶n dy = df(x) = f0(x)dx = y0dx
Düa v o c¡c t½nh ch§t cõa ¤o h m, ta ÷ñc c¡c t½nh ch§t cho vi ph¥n nh÷ sau
T½nh ch§t. Cho u = u(x); v = v(x) l hai h m sè câ ¤o h m t¤i iºm x thuëc kho£ng
x¡c ành, ta câ
d(u + v) = du + dv (1)
d(u v) = du dv (2)
d(u:v) = vdu + udv , vdu = d(u:v) udv (3)
d
u
v
=
vdu udv
v2
,
vdu
v2 = d
u
v
+
udv
v2 v = v(x)6= 0 (4)
C¡c vi ph¥n °c bi»t c¦n nhî:
Cho a6= 0; b 2 R, ta câ:
xdx = d
x+1
+ 1
Mð rëng
! (ax + b)dx = d
(ax + b)+1
a( + 1)
vîi 6= 1.
1
x
dx = d (ln jxj)
Mð rëng
!
1
ax + b
dx = d
ln jax + bj
a
.
exdx = d (ex)
Mð rëng
! eax+bdx = d
eax+b
a
.
sin xdx = d (cos x)
Mð rëng
! sin (ax + b)dx = d
cos (ax + b)
a
.
cos xdx = d (sin x)
Mð rëng
! cos (ax + b)dx = d
sin (ax + b)
a
.
1
cos2 x
dx = d (tan x)
Mð rëng
!
1
cos2 (ax + b)
dx = d
tan (ax + b)
a
.
1
sin2 x
dx = d (cot x)
Mð rëng
!
1
sin2 (ax + b)
dx = d
cot (ax + b)
a
.
dx
p
x2 + a
= d
ln

3. x +

5. p
x2 + a
p
x2 + adx = d
x
p
x2 + a + a ln

7. x +

9. p
x2 + a
2
!
1

10. C¡c cæng thùc nguy¶n h m:
F
Z
du = u + C F
Z
du
u
= ln juj + C
II. C¡c th½ dö minh håa:
Th½ dö 1. T¼m nguy¶n h m :
Z
(2x + 1) cos 2xdx
Líi gi£i.
Ta câ:
(2x + 1) cos 2xdx = (2x + 1)d
sin 2x
2
= d
(2x + 1)
sin 2x
2
sin 2x
2
d(2x + 1)
= d
(2x + 1)
sin 2x
2
sin 2xdx
= d
(2x + 1)
sin 2x
2
+ d
cos 2x
2
= d
(2x + 1)
sin 2x
2
+
cos 2x
2
Vªy,
Z
(2x + 1) cos 2xdx =
Z
d
(2x + 1)
sin 2x
2
+
cos 2x
2
= (2x + 1)
sin 2x
2
+
cos 2x
2
+ C.
Th½ dö 2. T¼m nguy¶n h m :
Z
x2exdx
Líi gi£i.
Ta câ:
x2exdx = x2d (ex)
= d (x2ex) exd(x2)
= d (x2ex) 2xexdx
= d (x2ex) xd(2ex)
= d (x2ex) (d(2xex) 2exdx)
= d (x2ex 2xex) + d(2ex)
Vªy,
Z
x2exdx =
Z
d
x2ex 2xex
+ d(2ex)
=
Z
d
ex(x2 2x + 2)
= ex(x2 2x + 2) + C.
Th½ dö 3. T¼m nguy¶n h m :
Z
x ln xdx
Líi gi£i.
Ta câ:
x ln xdx = ln xd
x2
2
= d
x2
2
ln x
x2
2
d (ln x) = d
x2
2
ln x
x
2
dx = d
x2
2
ln x
d
x2
4
Vªy,
Z
x ln xdx =
Z
d
x2
2
ln x
d
x2
4
=
Z
d
x2
2
ln x
x2
4
=
x2
2
ln x
x2
4
+ C.
2

11. Th½ dö 4. T¼m nguy¶n h m :
Z
dx
p
x2 + a
Líi gi£i.
dx
p
x2 + a
=
d(
p
x2 + a)
x
=
p
x2 + a)
dx + d(
x +
p
x2 + a
=
d(x +
p
x2 + a)
x +
p
x2 + a
= d
ln

13. x +

15. p
x2 + a
Vªy,
Z
dx
p
x2 + a
=
Z
d
ln

18. x +

21. p
x2 + a
= ln

24. x +

27. p
x2 + a
+ C
Th½ dö 5. T¼m nguy¶n h m :
Z p
x2 + adx
Líi gi£i.
Ta câ: p
x2 + adx = d
x
xd(
p
x2 + a
p
x2 + a)
= d
x
p
x2 + a
x2
p
x2 + a
dx
= d
x
p
x2 + a
p
x2 + adx +
a
p
x2 + a
dx
p
x2 + adx = d
) 2
x
+
p
x2 + a
adx
p
x2 + a
Chó þ:
dx
p
x2 + a
=
d(
p
x2 + a)
x
=
p
x2 + a)
dx + d(
x +
p
x2 + a
=
d(x +
p
x2 + a)
x +
p
x2 + a
= d
ln

29. x +

31. p
x2 + a
Do â:
p
x2 + adx = d
2
x
+ ad
p
x2 + a
ln

33. x +

35. p
x2 + a
)
p
x2 + adx = d
x
p
x2 + a + a ln

37. x +

39. p
x2 + a
2
!
Vªy,
Z p
x2 + adx =
x
p
x2 + a + a ln

41. x +

43. p
x2 + a
2
+ C.
Th½ dö 6. T¼m nguy¶n h m :
Z
x2dx
p
x2 + 1
Líi gi£i.
Ta câ:
x2dx
p
x2 + 1
= xd
p
x2 + 1
= d
x
p
x2 + 1
p
x2 + 1dx
= d
x
p
x2 + 1
d
x
p
x2 + 1 + ln

45. x +
p
x2 + 1

47. 2
!
= d
x
p
x2 + 1
x
p
x2 + 1 + ln

49. x +
p
x2 + 1

51. 2
!
Vªy: Z
dx
p
x2 + 1
=
Z
d
x
p
x2 + 1
x
p
x2 + 1 + ln

53. x +
p
x2 + 1

55. 2
!
= x
p
x2 + 1
x
p
x2 + 1 + ln

57. x +
p
x2 + 1

59. 2
+ C
3

60. Th½ dö 7. T½nh nguy¶n h m:
Z
dx
x2
p
x2 + 9
Líi gi£i.
dx
x2
p
x2 + 9
=
d
p
x2 + 9
x3 =
1
x2
d
p
x2 + 9
x
!
+
p
x2 + 9dx
x2
!
,
dx
p
x2 + 9
= d
p
x2 + 9
x
!
+
p
x2 + 9dx
x2
,
dx
p
x2 + 9
= d
p
x2 + 9
x
!
+
9dx
x2
p
x2 + 9
+
dx
p
x2 + 9
,
dx
x2
p
x2 + 9
=
1
9
d
p
x2 + 9
x
!
= d
p
x2 + 9
9x
!
Vªy,
Z
dx
x2
p
x2 + 9
=
Z
d
p
x2 + 9
9x
!
=
p
x2 + 9
9x
+ C
Th½ dö 8. T½nh nguy¶n h m:
Z
x2
p
x2 + 9dx
Líi gi£i.
Ta câ:
x2
p
x2 + 9dx =
p
x2 + 9d
x3
3
= d
x3
p
x2 + 9
3
!
x3
3
d
p
x2 + 9
= d
x3
p
x2 + 9
3
!
x4
p
x2 + 9
3
dx
= d
x3
p
x2 + 9
3
!
x4 81
3
p
x2 + 9
dx
81
p
x2 + 9
3
dx
= d
x3
p
x2 + 9
3
!
p
x2 + 9(x2 9)
3
dx
81
p
x2 + 9
3
dx
, 3x2
p
x2 + 9dx = 3d
x3
p
x2 + 9
3
!
p
x2 + 9(x2 9)dx
81
p
x2 + 9
dx
, 4x2
p
x2 + 9dx = d
x3
p
x2 + 9
+ 9
p
x2 + 9dx
81
p
x2 + 9
dx
, 4x2
p
x2 + 9dx = d
x3
p
x2 + 9
+ 9d
x
p
x2 + 9 + 9 ln

62. x +
p
x2 + 9

64. 2
!
81d
ln

66. x +
p
x2 + 9

68. , x2
p
x2 + 9dx = d
x3
p
x2 + 9
4
+
9x
p
x2 + 9 + 81 ln

70. x +
p
x2 + 9

72. 8
81 ln

74. x +
p
x2 + 9

76. 4
!
, x2
p
x2 + 9dx = d
x3
p
x2 + 9
4
+
9x
p
x2 + 9 81 ln

78. x +
p
x2 + 9

80. 8
!
Vªy,
Z
x2
p
x2 + 9dx =
x3
p
x2 + 9
4
+
9x
p
x2 + 9 81 ln

82. x +
p
x2 + 9

84. 8
+ C
4

85. Th½ dö 9. T¼m nguy¶n h m :
Z
ln x
ln x + 2
2
dx
Líi gi£i.
Ta câ:
ln x
ln x + 2
2
dx = dx 4
(ln x + 2)dx
(ln x + 2)2 +
4dx
(ln x + 2)2
= dx 4
d
x
ln x + 2
+
xd (ln x + 2)
(ln x + 2)2
+
4dx
(ln x + 2)2
= dx 4d
x
ln x + 2
4dx
(ln x + 2)2 +
4dx
(ln x + 2)2
= d
x
4x
ln x + 2
Z
Vªy,
ln x
ln x + 2
2
dx =
Z
d
x
4x
ln x + 2
= x
4x
ln x + 2
+ C
Th½ dö 10. T¼m nguy¶n h m :
Z
x ln(x + 1)
(x + 1)2 dx
Líi gi£i.
Ta câ:
x ln(x + 1)
(x + 1)2 dx =
ln(x + 1)dx
x + 1
ln(x + 1)dx
(x + 1)2
=
ln(x + 1)dx
x + 1
ln(x + 1)d(x + 1)
(x + 1)2
= d
ln2(x + 1)
2
(x + 1)d (ln(x + 1))
(x + 1)2
d
ln(x + 1)
x + 1
= d
ln2(x + 1)
2
d (ln(x + 1))
x + 1
+ d
ln(x + 1)
x + 1
= d
ln2(x + 1)
2
dx
(x + 1)2 + d
ln(x + 1)
x + 1
= d
ln2(x + 1)
2
+ d
1
x + 1
+ d
ln(x + 1)
x + 1
= d
ln2(x + 1)
2
+
1
x + 1
+
ln(x + 1)
x + 1
Z
Vªy,
x ln(x + 1)
(x + 1)2 dx =
ln2(x + 1)
2
+
1
x + 1
+
ln(x + 1)
x + 1
+ C
Th½ dö 11. T¼m nguy¶n h m :
Z
(x + 1)2e
x21
x dx
5

86. Líi gi£i.
(x + 1)2e
x21
x dx = x2(x+1)2
x2+1 d
e
x21
x
=
(x + 1)2
x2 + 1
d
x2e
x21
x
e
x21
x d (x2)
=
(x + 1)2
x2 + 1
d
x2e
x21
x
2xe
x21
x dx
, (x2 + 1)e
x21
x dx = d
x2e
x21
x
2xe
x21
x dx
, (x + 1)2e
x21
x dx = d
x2e
x21
x
Vªy, I =
Z 2
1
d
x2e
x21
x

89. 2
1
= x2e
x21
x + C
ang cªp nhªt . . .
Hçng Ngü, ng y 15 th¡ng 3 n«m 2014
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