Bài giảng tích phân pham kimchung

Së GD & §t nghÖ an
Tr−êng THPT §Æng thóc høa

sin4x + cos2x
∫ sin x + cos x dx
6 6

tÝch ph©n
1 ( x +1) - ( x -1)
6 6
dx
I= ∫ 8
x +1 2 ∫
= dx = ...
x 8 +1

Gi¸o viªn : Ph¹m Kim Chung
Tæ : To¸n

N¨m häc : 2007 - 2008

12

∫ bµi gi¶ng tÝch ph©n Ph¹m Kim Chung Tr−êng THPT §Æng Thóc Høa
2007

“ Thùc ra trªn mÆt ®Êt lμm g× cã ®−êng, ng−êi ta ®i l¾m th× thμnh ®−êng th«i ! ”
- Lç TÊn -
ViÕt mét cuèn tμi liÖu rÊt khã, ®Ó viÕt cho hay cho t©m ®¾c l¹i ®ßi hái c¶ mét ®¼ng cÊp thùc sù ! Còng may t«i kh«ng cã t− t−ëng lín cña
mét nhμ viÕt s¸ch, còng kh«ng hy väng ë mét ®iÒu g× ®ã lín lao v× t«i biÕt n¨ng lùc vÒ m«n To¸n lμ cã h¹n .. Khi t«i cã ý t−ëng viÕt ra nh÷ng ®iÒu
t«i gom nhÆt ®−îc t«i chØ mong sao qua tõng ngμy m×nh sÏ lÜnh héi s©u h¬n vÒ m«n To¸n s¬ cÊp..qua tõng tiÕt häc nh÷ng häc trß cña t«i bít b¨n
kho¨n, ng¬ ng¸c h¬n.. Vμ nÕu cßn ai ®äc bμi viÕt nμy nghÜa lμ ®©u ®ã t«i ®ang cã nh÷ng ng−êi thÇy, ng−êi b¹n cïng chung mét niÒm ®am mª sù
diÖu k× To¸n häc .

Thö gi¶i mét bμi to¸n khã…... nh−ng ch−a thËt hμi lßng !
1 ( x + 1) - ( x - 1) 1 ( x + 1) ⎣( x - 2x + 1) + ( 2 - 1) x ⎦ 1 (x (
- 1) ⎡ x 4 - 2x 2 + 1 + 2 + 1 x 2 ⎤) ( )
6 6 ⎡ 2 ⎤ 4 2 2 2
dx ⎣ ⎦ dx
∫ x8 + 1 =
2∫
dx =
2∫
dx +
2∫
(x + 1) - ( 2x ) (x + 1) - ( 2x ) ( x 4 + 1) - ( 2x 2 )
4 2 2 2 4 2 2 2 2 2

1 x2 + 1 ( 2 -1 ) ( x + 1) x
2 2
1 x2 - 1 ( 2 +1 ) (x 2
- 1) x 2
=
2 ∫ x 4 + 2x2 + 1 dx + 2 ∫ (x 4
)(
- 2x 2 + 1 x 4 + 2x 2 + 1 )
dx +
2 ∫ x 4 + 2x2 + 1 dx + 2 ∫ (x 4
)(
- 2x 2 + 1 x 4 + 2x 2 + 1 )
1 ⎛ 1⎞ 1 ⎛ 1⎞
1
1+ 2 2 -1 ( ⎝ )
⎜ 1+ 2 ⎟ dx
x ⎠ 1
1- 2 2 +1 ⎜ 1 - 2 ⎟ dx
⎝ x ⎠ ( )
= ∫ 2
x dx + ∫ ⎡⎛ 1 ⎞2 ⎤ ⎡⎛ ⎤ 2∫⎛
+ 2
x dx + ∫ ⎡⎛ 1 ⎞ 2 ⎤ ⎡⎛ ⎤
( )
2 2
2 ⎛ 1⎞ 1⎞
( ) ( )
2 1⎞ 2 1⎞
⎜x - ⎟ +2+ 2 ⎢ ⎜ x - ⎟ + 2 - 2 ⎥ ⎢⎜ x - ⎟ + 2 + 2 ⎥ ⎜x + ⎟ - 2- 2 ⎢⎜ x + ⎟ - 2 + 2 ⎥ ⎢⎜ x + ⎟ - 2 - 2 ⎥
⎝ x⎠ ⎢
⎣ ⎝ x⎠ ⎥⎢
⎦⎣ ⎝ x⎠ ⎥
⎦ ⎝ x⎠ ⎢
⎣ ⎝ x⎠ ⎥⎢
⎦⎣ ⎝ x⎠ ⎥
⎦
⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞
1
= ∫
d⎜x - ⎟
⎝ x⎠ 2 -1 ( d⎜x - ⎟
⎝ x⎠ ) 2 -1 d⎜x - ⎟
⎝ x⎠ 1 ( ⎝ )
d⎜x + ⎟
x⎠ 2 +1 d⎜x + ⎟
⎝ x⎠ 2 +1 d⎜x + ⎟
⎝ x⎠ ( ) ( )
4 2 ∫ ⎡⎛ 4 2 ∫ ⎡⎛ ⎤ 2∫⎛ 4 2 ∫ ⎡⎛ 4 2 ∫ ⎡⎛
+ - + + -
2
⎤ 2
⎤ ⎤
( )
2 2 2 2
2 ⎛ 1⎞ 1⎞
⎜x - ⎟ +2+ 2
⎝ x⎠
1⎞
⎢⎜ x - ⎟ + 2 - 2 ⎥
1⎞
⎢⎜ x - ⎟ + 2 + 2 ⎥ ⎜x + ⎟ - 2- 2
⎝ x⎠
1⎞
⎢⎜ x + ⎟ - 2 + 2 ⎥
1⎞
⎢⎜ x + ⎟ - 2 - 2 ⎥ ( ) ( )
⎣⎝
⎢ x⎠ ⎥
⎦ ⎣⎝
⎢ x⎠ ⎥
⎦ ⎢⎝
⎣ x⎠ ⎥
⎦ ⎢⎝
⎣ x⎠ ⎥
⎦
⎛ 1⎞ ⎛ 1⎞
⎜x + ⎟ - 2- 2 ⎜x + ⎟ - 2+ 2
ln ⎝
x⎠
ln ⎝
2+ 2 2- 2 2- 2 2+ 2 x⎠ 1
= u+ v+ + +C ( Víi x - = 2 + 2 tgu = 2 - 2 tgv )
8 8 16 ⎛ 1⎞ 16 ⎛ 1⎞ x
⎜x + ⎟+ 2- 2 ⎜x + ⎟+ 2+ 2
⎝ x⎠ ⎝ x⎠

(NÕu dïng kÕt qu¶ nμy ®Ó suy ng−îc cã t×m ®−îc lêi gi¶i hay h¬n ?.. )

_____________________________ Th¸ng 12 – n¨m 2007 __________________________________ Trang 1

12

2007

PhÇn lý thuyÕt

§Þnh nghÜa : Gi¶ sö f(x) lμ mét hμm sè liªn tôc trªn mét kho¶ng K, a vμ b lμ hai phÇn tö bÊt k× cña K, F(x) lμ
mét nguyªn hμm cña f(x) trªn K . HiÖu sè F(b) - F(a) ®−îc gäi lμ tÝch ph©n tõ a ®Õn b cña f(x) vμ ®−îc kÝ hiÖu lμ
b
b
∫ f ( x )dx . Ta dïng kÝ hiÖu F ( x ) a ®Ó chØ hiÖu sè : F(b) – F(a)
a

b
b
C«ng thøc Newton – Laipnit : ∫ f ( x )dx =
a
F (x)
a
= F(b) – F(a)

1
x3 1 1 3
= (1 − 03 ) =
1
∫ x dx =
2
VÝ dô :
0
3 0 3 3
b
Chó ý : TÝch ph©n ∫ f ( x )dx
a
chØ phô thuéc vμ f, a vμ b mμ kh«ng phô thuéc vμo kÝ hiÖu biÕn sè tÝch ph©n . V× vËy ta
b b b
cã thÓ viÕt : F(b) – F(a) = ∫ f ( x )dx =
a
∫ f ( t )dt = ∫ f ( u )du ...
a a

C¸c tÝnh chÊt cña tÝch ph©n .
a
1. ∫ f ( x )dx = 0
a
b a
2. ∫ f ( x )dx = - ∫ f ( x )dx
a b
b b b
3. ∫ ⎡αf ( x ) ± βg ( x )⎤dx = α ∫ f ( x )dx ± β∫ g ( x )dx
a
⎣ ⎦
a a
e e e
⎛ 3⎞ e e
VD : ∫ ⎜ 2x + ⎟dx = 2∫ xdx + 3∫ dx = x 2 + 3 ln x = ( e2 − 1) + 3 (1 − 0 ) = e2 + 2
1
1⎝
x⎠ 1 1
x 1 1
c b c
4. ∫ f ( x )dx = ∫ f ( x )dx + ∫ f ( x )dx
a a b
1 0 1 0 1
x2 0 x2 1
VD :
−1
∫ x dx = ∫
−1
x dx + ∫ x dx = − ∫ xdx + ∫ xdx = −
0 −1 0
+
2 −1 2 0
=1

b
5. f(x) ≥ 0 trªn ®o¹n [a ; b] ⇒ ∫ f ( x )dx ≥
a
0
b b
6. f(x) ≥ g(x) trªn ®o¹n [a ; b] ⇒ ∫ f ( x )dx ≥ ∫ g ( x )dx
a a
π π
2 2
VD : Chøng minh r»ng : ∫ sin2xdx ≤ 2∫ sinxdx
0 0
b b b
7. m ≤ f(x) ≤ M trªn ®o¹n [a ; b] ⇒ m(b – a) = m ∫ dx ≤ ∫ f ( x )dx ≤ M ∫ dx = M(b – a)
a a a
2
⎛ 1⎞ 5
VD : Chøng minh r»ng : 2 ≤ ∫ ⎜ x + ⎟dx ≤
1⎝
x⎠ 2
1 5
HD . Kh¶o s¸t hμm sè y = x + trªn ®o¹n [1; 2] ta cã : max y = ; min y = 2
x [1;2] 2 [1;2]

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2007

2 2
⎛ 1⎞ 5
2
2 2⎛ 1⎞ 5 2 2
⎛ 1⎞ 5
Do ®ã : 2∫ dx ≤ ∫ ⎜ x + ⎟dx ≤ ∫ dx ⇒ 2x ≤ ∫ ⎜ x + ⎟dx ≤ x ⇒ 2 ≤ ∫ ⎜ x + ⎟dx ≤
1 1 ⎝ x⎠ 21 1 1⎝ x⎠ 2 1 1 ⎝ x⎠ 2

PhÇn ph−¬ng ph¸p

Ph−¬ng ph¸p ®æi biÕn sè : t = v(x) .
1
x
VD . TÝnh tÝch ph©n : I = ∫ 2 dx
0 x +1

§Æt : t = x 2 + 1 . Khi x= 0 th× t=1, khi x=1 th× t=2 .
dt
Ta cã : dt = 2xdx ⇒ = xdx . Do ®ã :
2
1 2
x 1 dt 1 2 1
I=∫ 2 dx = ∫ = ln t = ln 2
0 x +1
21 t 2 1 2
b b
Quy tr×nh gi¶i to¸n . ∫ f ( x )dx = ∫ g ( v ( x ) )v' ( x ) dx
a a

B−íc 1 . §Æt t = v(x) , v(x) cã ®¹o hμm liªn tôc, ®æi cËn .
B−íc 2 . BiÓu thÞ f(x)dx theo t vμ dt : f(x)dx = g(t)dt
v ( b)

B−íc 3 . TÝnh ∫
v(a)
g ( t )dt .

Bμi tËp rÌn luyÖn ph−¬ng ph¸p :
TÝnh c¸c tÝch ph©n sau :
e2 2 1 3
dx dx x 2 dx xdx
1. ∫
e
x ln x
2. ∫ ( 2x − 1)
1
2
3. ∫ x3 + 1
0
4. ∫x 4
−1
2
π
2 1 4
dx dx dx
5. ∫ sin3 x
π
6. ∫ ( 2x + 1)
0 x +1
7. ∫ x (1 +
1 x )
4

Ph−¬ng ph¸p ®æi biÕn sè : x = u(t) .
1

sinx VD . TÝnh tÝch ph©n : ∫
0
1 − x 2 dx

⎛ ⎡ π π⎤⎞ π
§Æt x = sint ⎜ t ∈ ⎢ − ; ⎥ ⎟ . Khi x=0 th× t=0, khi x=1 th× t=
⎝ ⎣ 2 2⎦ ⎠ 2
⎡ π⎤
VËy víi x = sint th× x ∈ ⎡0;1⎤ ⇒ t ∈ ⎢0; ⎥ vμ dx = costdt .
⎣ ⎦
⎣ 2⎦
O cosx π π π
1 2 2 2
Do ®ã : ∫ 1 − x dx = 2
∫ 1 − sin t cos tdt =
2
∫ cos t cos tdt = ∫ cos
2
tdt =
0 0 0 0
π
π
2
1 + cos 2t 1⎛ 1 ⎞ π
=∫ dt = ⎜ t + sin 2t ⎟ 2 =
0
2 2⎝ 2 ⎠0 4
b
Quy tr×nh gi¶i to¸n . ∫ f ( x )dx
a

B−íc 1 . §Æt x = u(t), t ∈ ⎡α; β⎤ sao cho u(t) cã ®¹o hμm liªn tôc trªn ®o¹n ⎡α; β⎤ , f(u(t)) ®−îc x¸c ®Þnh trªn ®o¹n
⎣ ⎦ ⎣ ⎦
⎣α; β⎦ vμ u ( α ) = a; u ( β ) = b .
⎡ ⎤

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2007

B−íc 2 . BiÓu thÞ f(x)dx theo t vμ dt : f(x)dx = g(t)dt
β

B−íc 3 . TÝnh ∫ g ( t )dt
α
.

1
1 2 1
dx dx dx
1. ∫ 2. ∫ 3. ∫x
0
1 + x2 0 1− x 2
0
2
+ x +1
5

5+x
1 1 2

∫ 1 − x 2 dx ∫ 1 + x 2 dx ∫
2 3
4. x 5. x 6. dx ( §Æt x=5cos2t)
0 0 0 5−x
Ph−¬ng ph¸p ®æi biÕn sè : u(x) = g(x,t)
1
VD1 . TÝnh tÝch ph©n : I = ∫
0
1 + x 2 dx

t2 − 1
Çch (1) §Æt 1 + x 2 = x - t ⇒ 1 = -2xt + t 2 ⇒ x =
2t
t2 + 1
Khi x =0 th× t= -1, khi x=1 th× t= 1 − 2 vμ dx = dt . Do ®ã :
2t 2
1⎛ 1 ⎞
1− 2 1− 2 4 1− 2 1− 2 1− 2
− t2 − 1 t2 + 1 1 t + 2t 2 + 1 1
I= ∫ . 2 dt = − ∫1 dt = − ⎜ ∫ tdt + 2 ∫ dt + ∫ 3 dt ⎟ =
2t 2t 4 − t 3 ⎜
4 ⎝ −1 t t ⎟
−1 −1 −1 ⎠
t2 1 − 2 1 1− 2 1 1− 2
= −
8 −1
− ln t
2 −1
+ 2
8t −1
1
= − ln
2
( 2 −1 + ) 2
2

⎡ π⎤ π
Çch (2) : §Æt x=tgt , do x ∈ ⎡0;1⎤ nªn ta cã thÓ chän t ∈ ⎢0; ⎥ . Khi x=0 th× t=0, khi x=1 th× t =
⎣ ⎦
⎣ 4⎦ 4
1
vμ dx= dt . Do ®ã :
cos2 t
π π π π π
1 4
1 4
1 1 4
1 4
cos t 4
d ( sin t )
∫ 1 + x 2 dx = ∫ 1 + tg 2 t dt = ∫ dt = ∫ dt = ∫ dt = ∫ =
(1 − sin t )
2
0 0 cos 2 t 0
2
cos t cos t 3
0 cos t
4
0 cos t 0
2

π 2 π 2

1 4 ⎡ (1 − sin t ) + (1 + sin t ) ⎤ 14⎡ 1 1 ⎤
= ∫⎢ ⎥ d ( sin t ) = ∫ ⎢ + ⎥ d ( sin t ) =
4 0 ⎣ (1 − sin t )(1 + sin t ) ⎦ 4 0 ⎣ (1 − sin t ) (1 + sin t ) ⎦
π 2 π π π

1 ⎡ 4
1 1 ⎤ 1 4 d (1 − sin t ) 1 4 d ( sin t ) 1 4 d (1 + sin t )
= ∫⎢ + ⎥ d ( sin t ) = − ∫ + ∫ + ∫ =
4 0 ⎣ (1 − sin t ) (1 + sin t ) ⎦ 4 0 (1 − sin t ) 2
2 0 (1 − sin t )(1 + sin t ) 4 0 (1 + sin t )2
π π π π
1 1 + sin t 1 1 + sin t
1 ⎡ 1
= .⎢ −
1 ⎤
⎥ 4 + ln 4=
1 sin t
4 ⎣1 − sin t 1 + sin t ⎦ 0 4 1 − sin t 0 2 cos t 0 4 1 − sin t 0
2 4 + ln
1
4 = − ln 2 − 1 +
2 2
2
. ( )
B×nh luËn : Bμi to¸n nμy cßn gi¶i ®−îc b»ng ph−¬ng ph¸p tÝch ph©n tõng phÇn . Cßn víi 2 çch gi¶I trªn râ rμng
khi b¾t gÆp çch 1) ta nghÜ r»ng nã sÏ chøa ®ùng nh÷ng phÐp tÝnh to¸n phøc t¹p cßn çch 2) sÏ chøa nh÷ng phÐp
tÝnh to¸n ®¬n gi¶n h¬n. Nh−ng ng−îc l¹i sù suy ®o¸n - çch 2) l¹i chøa nh÷ng phÐp tÝnh to¸n dμi dßng vμ nÕu qu¶
thËt kh«ng kh¸ tÝch ph©n th× ch−a h¼n ®· lμ ®−îc hoÆc lμm ®−îc mμ l¹i dμi dßng h¬n .

1
1
VD2 . TÝnh tÝch ph©n : I = ∫
0 1 + x2
dx

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2007

t2 − 1
Çch (1) §Æt 1 + x 2 = x - t ⇒ 1 = -2xt + t 2 ⇒ x =
2t
t2 + 1
Khi x =0 th× t= -1, khi x=1 th× t= 1 − 2 vμ dx = dt . Do ®ã :
2t 2
1− 2 1− 2
−2t t 2 + 1 1
I= ∫ 2 . dt = − ∫ dt =
−1
t + 1 2t 2
−1
t
1− 2
= − ln t
−1
= − ln ( 2 −1 )
⎡ π⎤ π
Çch (2) : §Æt x=tgt , do x ∈ ⎡0;1⎤ nªn ta cã thÓ chän t ∈ ⎢0; ⎥ . Khi x=0 th× t=0, khi x=1 th× t =
⎣ ⎦
⎣ 4⎦ 4
1
vμ dx= dt .
cos2 t
π π π π
1 4 4 4 4
1 1 1 cos t 1 cos t
Do ®ã : ∫
0 1 + x2
dx = ∫
0 1 + tg t
2
2 cos t
dt = ∫
0
2
cos t
dt = ∫
0 cost
dt = ∫
0 cos2 t
dt =

π
π
d ( sin t ) 1 1 − sin t
( )
4
= ∫ (1 − sin t )
0
2
= ln 4 = − ln
2 1 + sin t 0
2 −1 .

2 2 0
x2
1. ∫
1
x 2 − 1dx 2. ∫
1 x2 − 1
dx 3. ∫
−1
x 2 + 2x + 2dx

1
2 −1 1
dx dx xdx
4. ∫1+
0 x − 4x + 3
2
5. ∫ 1+
−2 1 − 2x − x 2
6. ∫x+
0 x2 − 1

Chó ý : Khi ®øng tr−íc mét bμi to¸n tÝch ph©n, kh«ng ph¶i bμi to¸n nμo còng xuÊt hiÖn nh©n tö ®Ó chóng ta sö dông
ph−¬ng ph¸p ®æi biÕn sè . Cã nhiÒu bμi to¸n ph¶i qua 1 hay nhiÒu phÐp biÕn ®æi míi xuÊt hiÖn nh©n tö ®Ó ®Æt Èn phô (
sÏ nãi ®Õn ë phÇn Ph©n Lo¹i Çc d¹ng To¸n )

Ph−¬ng ph¸p tÝch ph©n tõng phÇn .

NÕu u(x) vμ v(x) lμ hai hμm sè cã ®¹o hμm liªn tôc trªn ®o¹n [a; b] th× :
b
b b
∫ u ( x )v' ( x ) dx = ( u ( x ) .v ( x ) ) - ∫ v ( x )u' ( x ) dx
a
a a
hay
b
b b
∫ u ( x )dv = ( u ( x ) .v ( x ) )
a
a ∫
- v ( x )du
a

π
2
VD1. TÝnh ∫ x cos xdx
0

⎧u=x ⎧du = dx
§Æt ⎨ , ta cã : ⎨
⎩dv = cos xdx ⎩ v = sin x

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2007

π π
π 2 π
2
π π
∫ x cos xdx = ( x sin x ) 2 − ∫ sin xdx = 2 + cosx 2 = 2 − 1
0 0 0 0
⎧u = cosx
NhËn xÐt : Mét c©u hái ®Æt ra lμ ®Æt ⎨ cã ®−îc kh«ng ?
⎩ dv = xdx
π π π
π
2
⎛ x2 ⎞ 12 2

∫ x cos xdx = ⎜ cosx ⎟ 2 + ∫ x 2 sin xdx , râ rμng tÝch ph©n ∫x
2
Ta h·y thö : sin xdx cßn phøc t¹p h¬n tÝch
0 ⎝ 2 ⎠ 0 20 0

ph©n cÇn tÝnh . VËy viÖc lùa chän u vμ dv quyÕt ®Þnh rÊt lín trong viÖc sö dông ph−¬ng ph¸p tÝch ph©n tõng phÇn . Ta
h·y xÐt mét VD n÷a ®Ó ®i t×m c©u tr¶ lêi võa ý nhÊt !
2
ln x
VD2. TÝnh ∫ 5 dx
1
x
⎧ 1
⎪u = 5
Ta thö ®Æt : ⎨ x râ rμng ®Ó tÝnh v= ∫ ln xdx lμ mét viÖc khã kh¨n !
⎪ dv = ln xdx
⎩
⎧ 1
⎧ u = ln x ⎪ du = x
⎪ ⎪
Gi¶i . §Æt ⎨ 1 ta cã : ⎨
⎪dv = x 5 dx ⎪ v = 1 dx = − 1
⎩ ⎪
⎩ ∫ x5 4x 4
2 2
ln x ⎛ ln x ⎞ 2 1 dx ln 2 1 ⎛ 1 ⎞ 2 15 ln 2
Do ®ã : ∫ 5 dx = ⎜ − 4 ⎟ + ∫ 5 = − + ⎜− ⎟1= −
1
x ⎝ 4x ⎠ 1 4 1 x 64 4 ⎝ 4x 4 ⎠ 256 64
NhËn xÐt : Tõ 2 VD trªn ta cã thÓ rót ra mét nhËn xÐt ( víi nh÷ng tÝch ph©n ®¬n gi¶n ) : ViÖc lùa chän u vμ dv
ph¶i tho¶ m·n :
1 du ®¬n gi¶n, v dÔ tÝnh .
2 TÝch ph©n sau ( ∫ vdu ) ph¶i ®¬n gi¶n h¬n tÝch ph©n cÇn tÝnh ( ∫ udv ) .
π π
1 1 2 6 1

∫ xe dx ∫ ∫ ( x − 1)cosxdx ∫ ( 2 − x ) sin 3xdx ∫x e
x 3x 2 −x
1. 2 . xe dx 3. 4. 5. dx
0 0 0 0 0
π π
2 2 e 5 e

∫ ∫ ∫ 9. 2x ln ( x − 1)dx
∫ ∫ ( ln x ) dx
2 2
6 . x sin xdx 7. e x cosxdx 8. ln xdx 10.
0 0 1 2 1

Mçi d¹ng to¸n chøa ®ùng nh÷ng ®Æc thï riªng cña nã !

PhÇn ph©n lo¹i c¸c d¹ng to¸n

TÝch ph©n cña c¸c hμm h÷u tû

P (x)
A. D¹ng : I = ∫ dx ( a ≠ 0)
ax + b

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12

2007

α α
C«ng thøc cÇn l−u ý : I = ∫ dx = ln ax + b + C
ax + b a

TÝnh I1 = x + 1 dx
∫ x −1
TÝnh I2 = x − 5 dx
2

∫ x +1
x3
TÝnh I3 =
∫ 2x + 3 dx
Ph−¬ng ph¸p : Thùc hiÖn phÐp chia ®a thøc P(x) cho nhÞ thøc : ax+b, ®−a tÝch ph©n vÒ d¹ng :
α
I = ∫ Q ( x ) dx + ∫ dx ( Trong ®ã Q(x) lμ hμm ®a thøc viÕt d−íi d¹ng khai triÓn )
ax + b

P (x)
B. D¹ng : I = ∫ 2
dx ( a ≠ 0)
ax + bx + c
1. Tam thøc : f ( x ) = ax 2 + bx + c cã hai nghiÖm ph©n biÖt .

u' ( x )
C«ng thøc cÇn l−u ý : I = ∫ u ( x ) dx = ln u ( x ) + C

2
☺ TÝnh I = ∫x 2
−4
dx

C¸ch 1. ( ph−¬ng ph¸p hÖ sè bÊt ®Þnh )
⎧ 1
2 A B ⎧A + B = 0 ⎪A = 2
⎪
= + ⇒ 2 ≡ (A + B) x + 2(A − B) ⇒ ⎨ ⇔ ⎨
x −4 x −2 x +2 ⎩A − B = 1
2
⎪B = − 1
⎪
⎩ 2
2 1 1 1 1 1 x −2
Do ®ã : I = ∫x 2
−4
dx = ∫
2 x −2
dx - ∫
2 x+2
dx = ln
2 x+2
+C

C¸ch 2. ( ph−¬ng ph¸p nh¶y tÇng lÇu )
2 1 ⎡ 2x 2x − 4 ⎤ 1
Ta cã : I = ∫ 2 dx = ⎢ ∫ 2 dx − ∫ 2 dx ⎥ = ln x 2 − 4 − ln x + 2 + C
x −4 2⎣ x −4 x −4 ⎦ 2
α
< Tæng qu¸t >TÝnh I = ∫ 2 dx
x − a2
2x
TÝnh I = ∫ dx
9 − x2
3x + 2
TÝnh I = ∫ 2 dx
x −1
x2
TÝnh I = ∫ 2 dx
x − 5x + 6

3x 3
TÝnh I = ∫ x 2 − 3x + 2 dx

Ph−¬ng ph¸p :
Khi bËc cña ®a thøc P(x) <2 ta sö dông ph−¬ng ph¸p hÖ sè bÊt ®Þnh hoÆc ph−¬ng ph¸p nh¶y
tÇng lÇu.
Khi bËc cña ®a thøc P(x) ≥ 2 ta sö dông phÐp chia ®a thøc ®Ó ®−a tö sè vÒ ®a thøc cã bËc < 2 .

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12

2007

2. Tam thøc : f ( x ) = ax 2 + bx + c = ( αx + β )2 cã nghiÖm kÐp .

u' ( x ) 1
C«ng thøc cÇn l−u ý : I = ∫u 2
(x)
dx = −
u(x)
+C

1 d ( x − 2) 1
TÝnh I = ∫ dx = ∫ =− +C
x 2 − 4x + 4 ( x − 2) x −2
2

4x
TÝnh I = ∫ 4x 2
− 4x + 1
dx .

⎧ dt
⎪ dx=
§Æt : 2x – 1 = t ⇒ ⎨ 2 , lóc ®ã ta cã :
⎪2x = t + 1
⎩
t +1 dt dt 2
I = 2∫ 2 dx = 2∫ + 2∫ 2 = 2 ln t − + C
t t t t
x −3
2
TÝnh I = ∫ 2 dx
x − 4x + 4
x3
TÝnh I = ∫ 2 dx
x + 2x + 1
t −β
Ph−¬ng ph¸p : §Ó tr¸nh phøc t¹p khi biÕn ®æi ta th−êng ®Æt : αx + β = t ⇒ x = vμ thay vμo biÓu thøc
α
trªn tö sè .
3. Tam thøc : f ( x ) = ax 2 + bx + c v« nghiÖm .
1
TÝnh I = ∫x 2
+1
dx
1
§Æt : x = tgα ⇒ dx = dα , ta cã :
cos2 α
1
I= ∫ cos α ( tg α + 1) dα = ∫ dα = α + C
2 2
, víi ( tgα = x )

1 a
< Tæng qu¸t > TÝnh I = ∫ dx . HD §Æt x = atgα ⇒ dx = dα , ta cã :
x + a2
2
cos 2 α
dα α
I= ∫
a
= +C
a
2
TÝnh I = ∫ x 2 + 2x + 2 dx
2x + 1
TÝnh I = ∫ x 2 + 2x + 5 dx
x2
TÝnh I = ∫ x 2 + 4 dx
x3
TÝnh I = ∫ x 2 + 9 dx

P (x)
C. D¹ng : I = ∫ dx (a ≠ 0)
ax + bx 2 + cx + d
3

1. §a thøc : f ( x ) = ax 3 + bx 2 + cx + d cã mét nghiÖm béi ba.
0974.337.449 ___________________________ Th¸ng 12 – n¨m 2007 ___________________ Trang 8

12

2007

1 1
C«ng thøc cÇn l−u ý : I = ∫x dx = − +C ( n ≠ 1)
n
( n − 1) x n −1

1
☺ TÝnh I = ∫ dx
( x − 1)
3

( x − 1)
−2
1 1
dx = ∫ ( x − 1) d ( x − 1) =
−3
NÕu x > 1 , ta cã : I = ∫ +C= − +C .
( x − 1) −2 2 ( x − 1)
3 2

(1 − x )
−2
1 1
dx = ∫ (1 − x ) d (1 − x ) =
−3
NÕu x < 1 , ta cã : I = − ∫ +C = − +C
(1 − x ) −2 2 ( x − 1)
3 2

1 1
VËy : I = ∫ dx = − +C
( x − 1) 2 ( x − 1)
3 2

1
Chó ý : = x − m , víi x > 0
xm
x
TÝnh I = ∫ dx
( x − 1)
3

t +1 ⎛1 1 ⎞ 1 1
§Æt : x – 1 = t ta cã : I = ∫ dt = ∫ ⎜ 2 + 3 ⎟dt = − − 2 + C
t3 ⎝t t ⎠ t 2t
x2 − 4
TÝnh I = ∫ dx
( x − 1)
3

x3
TÝnh I = ∫ dx
( x − 1)
3

x4
TÝnh I = ∫ dx
( x + 1)
3

2. §a thøc : f ( x ) = ax 3 + bx 2 + cx + d cã hai nghiÖm .
1
( x − 1)( x + 1)
2

1 dt
§Æt : x + 1 = t , ta cã : I = ∫ t ( t − 2 ) dt = ∫ t
2 3
− 2t 2
C¸ch 1 < Ph−¬ng ph¸p nh¶y tÇng lÇu >
1 3t 2 − 4t 1 ⎛ 3t 2 − 4t − 4 ⎞ 3t 2 − 4t 1 ⎛ 3t + 2 ⎞ 3t 2 − 4t 1 ⎛ 3 2 ⎞
Ta cã : 3 = 3 − ⎜ ⎟= − ⎜ ⎟= − ⎜ + ⎟
t − 2t 2 t − 2t 2 4 ⎝ t 3 − 2t 2 ⎠ t 3 − 2t 2 4 ⎝ t 2 ⎠ t 3 − 2t 2 4 ⎝ t t2 ⎠
3t 2 − 4t 1 ⎛3 2 ⎞ 3 1
∫ t3 − 2t2 dt − 4 ∫ ⎜ t + t2 ⎟dt = ln t − 2t − 4 ln t + 2t + C .
Do ®ã : I = 3 2

⎝ ⎠
C¸ch 2 < Ph−¬ng ph¸p hÖ sè bÊt ®Þnh >
⎧ 1
⎪B = − 2
⎧ −2B = 1 ⎪
1 At + B C ⎪ ⎪ 1
= + ⇒ 1 ≡ ( A + C ) t + ( −2A + B ) t − 2B ⇒ ⎨−2A + B = 0 ⇒ ⎨ A = −
2

t − 2t
3 2
t 2
t−2 ⎪ A+C =0 ⎪ 4
⎩ ⎪ 1
⎪ C=4
⎩
1 1 ⎡t + 2 1 ⎤ 1 ⎡1 2 1 ⎤ 1⎡ 2 ⎤
Do ®ã : ∫t 3
− 2t 2
dt = − ∫ ⎢ 2 −
4⎣ t ⎥ dt = − ∫ 4 ⎢ t + t 2 − t − 2 ⎥ dt = − 4 ⎢ln t − t − ln t − 2 ⎥ + C
t − 2⎦ ⎣ ⎦ ⎣ ⎦
0974.337.449 ___________________________ Th¸ng 12 – n¨m 2007 ___________________ Trang 9

12

2007

Ph−¬ng ph¸p “nh¶y tÇng lÇu” ®Æc biÖt cã hiÖu qu¶ khi tö sè cña ph©n thøc lμ
mét h»ng sè .
Ph−¬ng ph¸p “hÖ sè bÊt ®Þnh” : bËc cña ®a thøc trªn tö sè lu«n nhá h¬n bËc
mÉu sè 1 bËc .

2x + 1
TÝnh I = ∫ x ( x − 2 ) dx
2

§Ó sö dông ph−¬ng ph¸p nh¶y tÇng lÇu ta sÏ ph©n tÝch nh− sau :
2x + 1 2 1
= +
x2 ( x − 2) x ( x − 2) x2 ( x − 2)
x2
TÝnh I = ∫ dx
( x − 1) ( x + 2 )
2

x2 Ax + B C
Sö dông ph−¬ng ph¸p hÖ sè bÊt ®Þnh : = +
( x − 1) ( x + 2 ) ( x − 1) x+2
2 2

≡ ( x + 2 )( Ax + B ) + C ( x − 1)
2
Do ®ã : x 2
4
Cho : x=-2, suy ra : C =
9
2
x=0 , suy ra : B = −
9
5
x=1, suy ra : A =
9
Ph−¬ng ph¸p trªn gäi lμ ph−¬ng ph¸p “g¸n trùc tiÕp gi¸ trÞ cña biÕn sè” ®Ó t×m A, B, C.
x3 − 1
TÝnh I = ∫ 3 dx
x + 2x 2 + x

3. §a thøc : f ( x ) = ax 3 + bx 2 + cx + d cã ba nghiÖm ph©n biÖt .

1
☺ TÝnh I = ∫ x (x 2
− 1)
dx

1 ⎡ 3x 2 − 1 3x 2 − 3 ⎤ 1 ⎡ 3x 2 − 1 3 ⎤
1
C¸ch 1. Ta cã : ⎢ 3 − = ⎥= ⎢ 3 − ⎥
x ( x − 1) 2 ⎢ x − x x ( x 2 − 1) ⎥ 2 ⎣ x − x x ⎦
⎣
2
⎦
1 ⎡ 3x 2 − 1 3 ⎤ 1 3
Do ®ã : I = ∫ ⎢ 3 − ⎥ dx = ln x 3 − x − ln x + C
2⎣ x − x x⎦ 2 2

⇒ 1 ≡ A ( x 2 − 1) + Bx ( x + 1) + Cx ( x − 1)
1 A B C
C¸ch 2 . Ta cã : = + +
x ( x 2 − 1) x x − 1 x + 1
Cho x=0, suy ra A = -1 .
1
x=1, suy ra B =
2
1
x=-1, suy ra C =
2
1
Do ®ã : I = − ln x + ln x − 1 + C
2

2

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2007

x +1
TÝnh I = ∫ x (x 2
− 4)
dx

x2
TÝnh I = ∫ dx
( x 2 − 1) ( x + 2 )
x3
TÝnh I = ∫ (x 2
− 1) ( x − 2 )
dx

dx
TÝnh I = ∫ ( 2x + 1) ( 4x 2
+ 4x + 5 )
dt
§Æt : 2x + 1 =t ⇒ dx = , ta cã :
2

1 dt 1 ⎡ 3t 2 − 6 3t 2 − 18 ⎤ 1
I=
2 ∫ t ( t2 − 6 ) 24 ⎢ t − 6t
= ⎢∫ 3 dt − ∫ 2 dt ⎥ =
t ( t − 6 ) ⎥ 24
ln t 3 − 6t − 3 ln t + C
⎣ ⎦
4. §a thøc : f ( x ) = ax 3 + bx 2 + cx + d cã mét nghiÖm (kh¸c béi ba)
1
☺ TÝnh I = ∫ 3 dx
x −1
§Æt x – 1 = t ⇒ dx = dt , ta cã :
dt 1 ⎡ t 2 + 3t + 3 t 2 + 3t ⎤ 1 ⎡ dt t+3 ⎤
I= ∫ 2 = ⎢∫ 2 dt − ∫ 2 dt ⎥ = ⎢ ∫ − ∫ 2 dt =
t ( t + 3t + 3 ) 3 ⎢ t ( t + 3t + 3 ) t ( t + 3t + 3 ) ⎥ 3⎣ t t + 3t + 3 ⎥⎦
⎣ ⎦
⎡ ⎤
⎢ ⎥
1 dt 1 2t + 3 3 dt ⎥ = 1 ln t − 1 ln t 2 + 3t + 3 − α 3 + C ( Víi x = 3 tgα )
= ⎢∫ − ∫ 2 dt − ∫
3 ⎢ t 2 t + 3t + 3 2 ⎛ 3⎞
2
3⎥ 3 2 2
⎢ ⎜t + ⎟ + ⎥
⎢
⎣ ⎝ 2⎠ 4⎥ ⎦
1
TÝnh I = ∫ dx
x ( x 2 + 1)
1
TÝnh I = ∫ x (x 2
+ 2x + 2 )
dx

x2
TÝnh I = ∫ x 3 + 1 dx
x3
TÝnh I = ∫ 3 dx
x −8
1
TÝnh I = ∫ 3 dx
x − 3x 2 + 3x − 2

Tãm l¹i : Ta th−êng sö dông hai phÐp biÕn ®æi :
Tö sè lμ nghiÖm cña mÉu sè .
Tö sè lμ ®¹o hμm cña mÉu sè .
vμ ph©n thøc ®−îc quy vÒ 4 d¹ng c¬ b¶n sau :
1 1 1
ax + b øng víi ∫ ax + b
↔
{ dx = ln ax + b + C
a
u' u'
u øng víi ∫ u
↔
{ dx = ln u + C

0974.337.449 ___________________________ Th¸ng 12 – n¨m 2007 ___________________ Trang 11

12

2007

u' u' 1
( n ≥ 2 ) ↔ ∫ n dx = -
{ +C
un øng víi u ( n - 1) un-1
1 1 a
↔ ∫
{ dx = + C , víi x + d = atgα
( x + d ) + a øng víi ( x + d ) + a
2 2 2 2
a

Q (x)
D. D¹ng : I = ∫ dx < P(x) lμ ®a thøc bËc cao> Vμ mét sè kÜ thuËt t×m nguyªn hμm .
P (x)
1. KÜ thuËt biÕn ®æi tö sè chøa nghiÖm cña mÉu sè .
dx
TÝnh I = ∫ x ( x − 1) ( x + 7 ) ( x + 8 )
x ( x + 7 ) − ( x − 1)( x + 8 )
HD : I = ∫ x ( x − 1) ( x + 7 ) ( x + 8 ) dx
dx
TÝnh I = ∫x 4
+ 10x 2 + 9
1 ( x + 9 ) − ( x + 1)
2 2
dx
HD : I = ∫ 2
( x + 1)( x 2 + 9 ) 8 ∫ ( x 2 + 1)( x 2 + 9 )
=

dx
TÝnh I = ∫x 6
+ 6x 4 − 13x 2 − 42
dx
HD : I = ∫ (x 2
− 3 )( x 2 + 2 )( x 2 + 7 )
dx
TÝnh I = ∫ 5x 5
+ 20x
1 ( x + 4) − x
4 4
1 dx
HD : I = ∫
5 x ( x 4 + 4 ) 20 ∫ x ( x 4 + 4 )
=

dx
TÝnh I = ∫x 7
− 10x 3
1 x − ( x − 10 )
4 4
dx
HD : I = ∫ x 3 ( x 4 − 10 ) 10 ∫ x 3 ( x 4 − 10 )
=

dx
TÝnh I = ∫ (x 2
− 2 )( 2x 2 + 1)( 3x 2 − 4 )
dx
TÝnh I = ∫x 8
− 10x 6 + 35x 4 − 50x 2 + 24
dx
TÝnh I = ∫ ( x + 1) ( x 4
+ 4x 3 + 6x 2 + 4x − 9 )
x 2 dx
TÝnh I =∫ x4 − 1
x 4 dx
TÝnh I = ∫ 4
x −1
x 4 dx
TÝnh I = ∫ 4
x +1
x 4 dx
TÝnh I = ∫ 6
x −1

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12

2007

x 6 dx
TÝnh I = ∫ x6 − 1
dx
TÝnh I = ∫ 100
3x + 5x
dx
TÝnh I = ∫
x ( 2x 50 + 7 )
2

(1 − x ) dx 2000

TÝnh I = ∫ x (1 + x ) 2000

P(x)
2. KÜ thuËt ®Æt Èn phô víi tÝch ph©n cã d¹ng : I = ∫ dx ( α ≠ 1)
( ax + b )α
x3 + x + 1
( x − 2)
30

⎧ dx = dt
§Æt x – 2 = t ⇒ ⎨ , ta cã :
⎩x = t + 2
( t + 2)
3
+t+3 t 3 + 6t 2 + 13t + 11 ⎡ 1 1 1 1 ⎤
I= ∫ t 30
dt = ∫ t 30
dt = − ⎢
⎣ 26t
26
+6
27t 27
+ 13
28t 28
+ 11
29t 29 ⎥
⎦
+ C =…

x4
TÝnh I = ∫ ( x − 3) 45
dx

3x 4 − 5x 3 + 7x − 8
TÝnh I = ∫ dx
( x + 2)
50

Chó ý : Víi lo¹i to¸n nμy trong cuèn “TÝch Ph©n – T.Ph−¬ng ” ®· sö dông ph−¬ng ph¸p khai triÓn
Taylor nh−ng t«i c¶m thÊy c¸ch lμm nμy kh«ng nhanh h¬n l¹i g©y nhiÒu phøc t¹p cho häc sinh nªn ®·
kh«ng nªu ra .

3. KÜ thuËt biÕn ®æi tö sè chøa ®¹o hμm cña mÉu sè .
xdx
TÝnh I = 4 ∫x
−1
§Æt x 2 = t ⇒ 2xdx = dt
x 3 dx
TÝnh I = ∫ 4
x +1
x −1
2
☺ TÝnh I = ∫ 4 dx
x +1
1 ⎛ 1⎞
1− 2 d⎜ x + ⎟
x −1
2
x dx = ⎝ x⎠ 1 x2 − x 2 + 1
I= ∫ 4 dx = ∫ ∫ ⎛ 1 ⎞2 = ln +C
x +1 1 x2 + x 2 + 1
x + 2 ( ) 2 2
2 2

x ⎜x + ⎟ − 2
⎝ x⎠
x +1
2
TÝnh I = ∫ 4 dx
x +1
x2
TÝnh I = ∫ 4 dx
x +1

TÝnh I = ∫ 4
( x 2 − 1) dx
x − 5x 3 − 4x 2 − 5x + 1

TÝnh I = ∫ 4
( x 2 + 1) dx
x + 2x 3 − 10x 2 − 2x + 1
0974.337.449 ___________________________ Th¸ng 12 – n¨m 2007 ___________________ Trang 13

12

2007

TÝnh I = ∫
(x 2
− 2)
dx
x 4 − 3x 3 + 11x 2 − 6x + 4

TÝnh I = ∫
( x2 + 3) dx
x 4 − 2x 3 − 2x 2 + 6x + 9
dx
TÝnh I = ∫ 4
x + x2 + 1
dx
TÝnh I = ∫ 4
x − 3x 2 + 4
B×nh luËn : Lo¹t bμi to¸n nμy lμm t«i kh¸ Ên t−îng víi phÐp chia c¶ tö sè vμ mÉu sè cho
x 2 . Qu¶ thËt t«i lu«n cè g¾ng t×m tßi xem liÖu m×nh cã thÓ nghÜ ra mét ph−¬ng ph¸p nμo
kh¸c hay h¬n ch¨ng, nh−ng …” bã tay.com “ . ThÕ míi hiÓu to¸n häc : “lu«n tiÒm Èn nh÷ng vÎ
®Ñp lμm ng−êi ta söng sèt”.
x5
TÝnh I = ∫ 6 dx
x +1
x
TÝnh I = ∫ 6 dx
x −1
1 dt
§Æt x 2 = t ⇒ 2xdx = dt , ta cã : I = ∫ 3
2 t −1
x3
TÝnh I = ∫ 6 dx
x −1
x4 + 1
TÝnh I = ∫ 6 dx
x +1
1 d(x ) 1 d(x )
3 2
x3 + x
TÝnh I = ∫ 6 dx HD : I = ∫ 6 + ∫ 6
x +1 3 x +1 2 x +1
3
x2
d ( x2 )
x 1
TÝnh I = ∫ 6 dx HD : I = ∫
x +1 2 ( x 2 )3 + 1

TÝnh I = ∫
(x 2
+ 1)( x 2 + 2x − 1)
dx
x 6 − 14x 3 − 1
⎛ 1 ⎞⎛ 1 ⎞ ⎛ 1 ⎞
⎜ 1 + 2 ⎟⎜ x − + 2 ⎟ ⎜ x − + 2⎟
⎛ 1⎞
HD : I = ∫ ⎝
x ⎠⎝ x ⎠ dx = ⎝ x ⎠
⎛ 3 1 ⎞ ∫ ⎛ 1 ⎞3 ⎛ 1 ⎞ d⎜x − ⎟
⎝ x⎠
⎜ x − 3 ⎟ − 14 ⎜ x − ⎟ + 3 ⎜ x − ⎟ − 14
⎝ x ⎠ ⎝ x⎠ ⎝ x⎠
19
x
TÝnh I = ∫ dx
( 3 + x10 )
2

x10 .10x 9 x10
∫ 3 + x10 2 d ( x )
1
HD . I = ∫ dx = 10

(3 + x )10 2 10 ( )
x 99
TÝnh I = ∫ dx
( 2x − 3)
50 7

x 2n −1
TÝnh I = ∫ dx
( ax + b)
n k

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12

2007

4. KÜ thuËt chång nhÞ thøc .

C¬ së cña ph−¬ng ph¸p :
a b
( ax + b )
n ,
⎛ ax + b ⎞ c d
§Ó t×m nguyªn hμm cã d¹ng : I = ∫ ( cx + d ) m
dx , ta dùa vμo c¬ së : ⎜ ⎟ =
⎝ cx + d ⎠ ( cx + d )
2

vμ ph©n tÝch biÓu thøc d−íi dÊu tÝch ph©n vÒ d¹ng :
⎛ ax + b ⎞ dx ⎛ ax + b ⎞ ⎛ ax + b ⎞
I = k∫ f ⎜ ⎟ = k∫ f ⎜ ⎟d⎜ ⎟
⎝ cx + d ⎠ ( cx + d )2 ⎝ cx + d ⎠ ⎝ cx + d ⎠

VD . TÝnh
( 3x − 5 )
10 10 10 11
⎛ 3x − 5 ⎞ dx 1 ⎛ 3x − 5 ⎞ ⎛ 3x − 5 ⎞ 1 ⎛ 3x − 5 ⎞
I= ∫ dx = ∫ ⎜
11 ∫ ⎝ x + 2 ⎠
⎟ = ⎜ ⎟ d⎜ ⎟= ⎜ ⎟ +C
( x + 2) ⎝ x+2 ⎠ ( x + 2) x + 2 ⎠ 121 ⎝ x + 2 ⎠
12 2
⎝
(7x − 1)
99

TÝnh I = ∫ dx
( 2x + 1)
101

dx
TÝnh I = ∫
( x + 3) ( x + 5)
5 3

⎡ ( x + 3) − ( x + 5) ⎤
6
dx 1 1 1 1 dx dx
HD . I = ∫ =∫ 6 ∫ 5 ⎢ ⎥ =
⎛ x + 3 ⎞ ( x + 5) ( x + 5) x+5 ( x + 5)
5 5 6 2 2
⎛x +3⎞ 2 ⎛x +3⎞ ⎣ ⎦
( x + 5)
8
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝x +5⎠ ⎝ x +5⎠ ⎝ x +5⎠
§Ó tr¸nh sù ®å sé trong tÝnh to¸n ta cã thÓ sö dông phÐp ®Æt Èn phô nh− sau :
⎧ 1 dt
⎪ dx =
x+3 ⎪ ( x + 5)
2
2
§Æt =t⇒⎨ , nªn ta cã :
x+5 ⎪x + 5 − 2 = t ⇒ 1 = 1− t
⎪ x+5
⎩ x+5 2
⎡ ( x + 3) − ( x + 5) ⎤ 1 ( t − 1) dt
6 6
1 1 dx
6 ∫ 5 ⎢ ⎥ =∫
x+5 ( x + 5)
2
2 ⎛x+3⎞ ⎣ ⎦ 27 t5
⎜ ⎟
⎝x+5⎠

dx
TÝnh I = ∫
( 3x − 2) ( 3x + 4 )
7 3

dx
TÝnh I = ∫
( 2x − 1) ( 3x − 1)
3 4

3x − 1 1 1
§Æt =t⇒− dx = dt vμ = 2t − 3
2x − 1 ( 2x − 1) 2x − 1
2

( 2t − 3 )
5
dx dx dt
Do ®ã ta cã : I = ∫ ( 2x − 1) ( 3x − 1)
3 4
= ∫ ⎛ 3x − 1 ⎞
4
= −∫
t4
( 2x − 1)
7
⎜ ⎟
⎝ 2x − 1 ⎠

0974.337.449 ___________________________ Th¸ng 12 – n¨m 2007 ___________________ Trang 15

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