TP1: TÍCH PHÂN HÀM SỐ HỮU TỈ
Dạng 1: Tách phân thức
2

Câu 1.

I 

1

x2

dx
x2  7x  12



1
2

 I   1

2
16
9 

dx =  x  16ln x  4  9ln x  3  1 = 1 25ln2  16ln3 .
x  4 x  3

2



Câu 4.

I 

2
1
3
1
3
 ln( x2  1)    ln2  ln5 
2
2
2
8
2x
1
1

2

3x2  1
x  2x  5x  6
3

2

dx

2 4 13 7 14


I 

( x  1)2
(2x  1)4

dx

 7x  199

101
0  2x  1

 7x  1 
 I  

 2x  1 
0
1

99

1  x 1 
 Ta có: f ( x)  .

3  2x  1 

I 

0 (x

 2x  1 

dx

1 1  7x  1 
 


9 100  2x  1 
1

2

dx

1  100 
1

2  1
0 900

 Đặt t  x2  4  I 

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1
8




 Đặt t  1  x3  dt  3x2dx  dx 
4

Câu 10. I 

3



1

1
x( x  1)
4

2

2

1  x7

Câu 12. I  

x(1  x7 )

1

Câu 13. I 





1

3

  t  t 2  1 dt  4 ln 2

1





1 32
dt
. Đặt t  x  I  
 I   5 10
2
2
5 1 t (t  1)2
1 x .( x  1)
2

x.( x10  1)2

1

3x2

1 128 1  t
dt
7 1 t(1  t )

dx
x (1  x2 )
6

1
 I 
t

3
3



1

t6

dt 
t2  1

 4 2
117  41 3 
1 

 t  t  1  2  dt =
135


3
2 1002
1 x (1  x )

Cách 2: Ta có: I 

.dx

1

.dx  

1002


1 3 1
x  2  1
x


1000

2

1  x2

4
1 1 x



dx

Trang 2
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2
x3

dx .


1 x

 Ta có:

1

2



1  x4

3
2

1

x2 . Đặt t  x  1  dt   1  1  dx

2
2
t

2
2
2
2
2
t

2


1
1


1
dt

2

Câu 16. I  

1

1  x2

1 1

 x2 
2
x2
du
5
5
; tan u  2  u1  arctan2; tan u   u2  arctan
Đặt t  2 tan u  dt  2
2
2
2
cos u

1 x

 Ta có:

2

u


2 2
2
2
5
du 
(u2  u1) 
 arctan  arctan2 


2
1
4
x
 Ta có: I  
dx . Đặt t  x   I  ln
1
x
5
1
x
x
2

dx

dx

x4  1 ( x4  x2  1)  x2
x4  x2  1
x2
1
x2





x6  1
x6  1

3



0

x2

x

dx

2

( x  1)( x  1)
2

1

Câu 20. I  

0

2

xdx
x 4  x2  1

.



Trang 3
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dt
2
 1  3 

t    
 2  2 

2




6 3


Câu 21. I 

1 5
2



x2  1
x 4  x2  1

1

x2 



1

dt

 I 

1

2

0t

4

du

. Đặt t  tan u  dt 

2

cos u

 I   du 
0



2

3

1
 I  (9x2  1) 2  x3  C
27
Câu 23. I  





x2  x
1 x x

x2  x
1 x x

dx

x2

dx  

1 x x

x

dx  


4
9



4

Câu 24. I  

0 1

dx =

1 x x

2x  1
2x  1



3

dx



1 x

x



6

dx

Câu 25. I  

2 2x  1 

4x  1

1

Câu 26. I   x3 1  x2 dx

3 1
2 12

 Đặt t  4x  1 . I  ln 

1

 Đặt: t  1  x2  I    t 2  t 4  dt 
0

0

1


 Đặt t  x  dx  2t.dt . I = 2
3

x3

Câu 28. I  

dx
3
x

1

x

3
0
2

2
2
1
3
dt   (2t  6)dt  6
dt  3  6ln
2
2
t 1
1 t  3t  2
1


Câu 30. I  

1

x2  1
x 3x  1

1

2

3

dx
2

 t2  1

 1
4 3 
2tdt
2tdt
 Đặt t  3x  1  dx 
 I  2 
.
3
3
t 1
2

t

1)
dt

2


2
92
2 t 1

dx

x  1  t  x  t 2  1  dx  2tdt

 Đặt
2

2(t 2  1)2  (t 2  1)  1
 I 
2tdt
t
1

2

 4t 5

54


1
4

.2tdt 2 

1

x 1

Câu 33. I  

1 

0

2

2
 t3
 1
1
16  11 2
 t   dt  2   2t   
t 1
3
 t
3

2

t
t
4

=

Câu 34. I 

8



3

1  t2
2
1
  3t  4ln t   = 2ln2 
2  2
t 
4
x 1

dx
x2  1







1

 I   ( x  1)3 2x  x2 dx   ( x2  2x  1) 2x  x2 ( x  1)dx . Đặt t  2x  x2  I  
0

0
2

2x  3x  x
3

Câu 36. I  

x2  x  1

0

2

 I 

2

( x2  x)(2x  1)
x2  x  1

0

2


Câu 38. I 

1



4
.
3

11 

dx
x  1  x2

Trang 6
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2
.
15


 Ta có: I 
+ I1 

+ I2 

1

1 11 
1
1
  1 dx   ln x  x |1 1

2 1 x 
2
1

1  x2
dx . Đặt t  1 x2  t 2  1 x2  2tdt  2xdx  I2=
2x



1

2



t 2dt

2
2 2(t  1)

0

Vậy: I  1 .



1

I=

0

t (tdt )



4  t2

3

Câu 41. I 

2



0

( x2  1) x2  5
x 2

1

x  x2


x2  x  1


 2
2  5
2t
1 
dt  5  1  

dt  5 3  1  ln  

2
2
3  12
t t  1 t  1

t(t  1)
1 
3

t3  2
2

dx
1 3



1


0

t2

x

27



4  x2  t 2  4  x2  tdt   xdx

xdx . Đặt t =


t 2 
 
dt   (1 
)dt   t  ln

2
t2 
t2  4

3t 4
3

2 5
2


1
2dt
 ln(2t  1)
1
2t  1

3

 ln

dx

42 36 

4

 Đặt 2  1 x  t  I    2t  16   2  dt  12  42ln
t
3
t 
3

Trang 7
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3 2 3
3


3




1

M

2 2

1

3



N

2011



3

1
2 2
2011
x2
dx  
dx  M  N
3


0 (1 

x2

1  M  

3
2



0

2 2

 2011
2011x dx   

 2x2  1
3

t 3dt  



213 7
128

14077


1
t3

2 2



3

 du 



t2

1

t 4.(t 3  1) 3

2

1

3dt

dt 

3



3

14077 213 7
.

16
128
1

3

dx

x4

1

2
2 2
2
 2 (t  1)2 dt  (t  1)3 
1
3
3
1

2t(t 2  1)2 dt

 Đặt t  x  1  I  


1

t 2.(t 3  1) 3

2



1



t 
t4

2 1  3 



1

du 



2
3

dt

 
3 1 
 
 3 0

1
1 2
 u3
0



1
3

2


3

 I 

3 4

(t 2  1)2

dt =

t2  2


ln 

3
4  4  2 

Dạng 2: Đổi biến số dạng 2
1


 dx

2
x
ln
1

x


 1 x



0
1 x

Câu 49. I   

1



5

 x2 ) 4  x2 dx

2

I=

2

 (x

5

 x ) 4  x dx =
2

2

2

x

5

4  x dx +
2

2


+ Tính B =

x

2

Vậy: I  2 .
2

Câu 51. I  

3 



4  x2 dx
2x4

1
2

 Ta có: I  

2

3

4
1 2x


3 2 4
7
x dx  .

21
16

dx . Đặt x  2sin t  dx  2costdt .

Trang 9
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





1 cos tdt 1
12
3
2  1 

cot
t
dt




x2dx

0

4  x6

Câu 52. I  

6

6

 Đặt t  x3  dt  3x2dx  I 

1 1 dt
.
3 0 4  t 2


 
16

Đặt t  2sin u, u  0;   dt  2cosudu  I   dt  .
 2
30
18

2


. Đặt x  1  2cost .
2
3



 I 

2



2
3
1
2

Câu 55.



t
2

 Đặt x  2cost  dx  2sin tdt  I  4  sin2 dt    2 .

(1  2cost ) 2sin t
2

4  (2cost )2

Câu 56. I 

3



x2  1dx

2

Trang 10
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
12



3 1

8 8



x

dx
u  x2  1 du 
2
 Đặt 

2

x  1dx 
2

x
x2  1
3



2

dx  5 2 

dx
x2  1

 2
  x 1 

2

3


 dx
x2  1 
1


dx    sin x  cos x  4(sin x  cos x dx
sin x  cos x
 3cos x  5sin x  C .
cot x  tan x  2tan2x
Câu 58. I  
dx
sin4x
2cot 2x  2tan2x
2cot 4x
cos4x
1
 Ta có: I  
dx  
dx  2
dx  
C
2
sin4x
sin4x
2sin4x
sin 4x


cos2  x  
8

Câu 59. I  
dx
sin2x  cos2x  2




  


sin  x    cos x    

4

8
8  

 


 I 





cos 2x  


1 
dx

4  dx  1




 2



dx
3sin x  cos x

3

1
dx
1
dx
= I 
.
=

4

4
3
2 x  
1  cos x  
2sin   
3
3
3

2 6


cos

6







6

6

1
2 0

1
sin x  sin



3

dx 

6



x 
x 


cos  
sin   
6
6
1
 2 6 dx  1
 2 6  dx  ln sin  x   
 



20
x 
20
x 
2 6
sin   
cos  
2 6
2 6



0

0

3
.
 cos4x  cos8x  I 
128
64 16
64


2

 cos2x(sin

Câu 63. I 

4

x  cos4 x)dx

0





2



0



x  1) cos2 x.dx

0

A =





2

2

5
 cos xdx 

0

0


2
 cos x.dx 

0

Vậy I =


– .
15 4


Câu 65. I 

2

 cos

2

x cos 2 xdx

0





2

2

 I   cos2 x cos2xdx 
0



1




3
2 4sin x dx
0 1  cos x

Câu 66. I  



4sin3 x 4sin3 x(1  cos x)

 4sin x  4sin x cos x  4sin x  2sin2x
1  cos x
sin2 x


 I   2 (4sin x  2sin2x)dx  2
0
2

Câu 67. I 



1  sin xdx

0


2
2
x 
x  
 2   sin   dx   sin    dx   4 2
2 4
2 4 
0
3


2




Câu 68. I 

4



0

dx
cos6 x

4

 Ta có: I   (1  2tan2 x  tan4 x)d(tan x) 

3
1
3
1
C
Đặt t  tan x . I    t 3  3t   t 3  dt  tan4 x  tan2 x  3ln tan x 
t
4
2


2tan2 x
2t
Chú ý: sin2x 
.
1 t2

Câu 69. I  

Trang 14
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Câu 71. I  

 I 

dx
sin x.cos3 x
dx

Câu 72. I  

2011

sin2011 x  sin2009 x
sin5 x

cot xdx

1

2011 1 

sin2 x cot xdx 

sin4 x

 Ta có: I  

Đặt t  cot x  I  

2
2011
t
(1  t 2 )tdt

4024

2011



Câu 73. I 

2

sin2x.cos x
dx
1

cos
x
0




2
(t  1)2
sin x.cos2 x
dt  2ln2  1
dx . Đặt t  1 cos x  I  2
t
1

cos
x
1
0

2

dx . Đặt t  cosx

cos
x
0
3

1
2

1  u2
3
du  ln2 
u
8
1

 I  
Câu 75. I 



 sin

2

x(2  1  cos2x )dx








2

2

2

+ H   2sin xdx   (1  cos2x)dx   
2




2















4


3

 I  4. 


dx
2

. Đặt t  tan x  dt 

2

sin 2x.cos x

dx
cos2 x

.

4

I 

3



t


 
 2

3 1
3
 t
t




Câu 77. I 

2

sin 2 x

  2  sin x  dx
2

0



 Ta có: I 




2
3 2
dt  2    dt  2  ln t    2ln 
t t2 
t 2
2 3

2
3



Câu 78. I 

6

sin x

 cos2x dx

0



I

6



6

t 

1
2 2

ln

3
2
2t  2
2t  2

1

1

=
3
2

2 2

Trang 16
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ln

3 2 2

2

1
dx
2

Câu 80. I   sin x  sin2 x 



 Đặt t  cosx . I 

3
(  2)
16

6


Câu 81. I 

4

sin4x



sin x  cos x
6


3
4
3
t


1

1
1
4



2
.
3



Câu 82. I 

2

sin x



 sin x 


6 2
6







sin  x   dx
2
6
3
3
1 2
dx

=
I=



6
16 0


  16 0

cos3  x  
cos2  x  

4

sin x

4

sin x



cos2 x



cos2 x





1  cos2 x .dx 





3





7
 3  1.
12

3

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0






3

sin x
cos2 x

sin x dx 

4



sin x


dx = 
dx = 
20

20



0 sin x  3 cos x
sin  x  
1  cos2  x  
3
3








6

6

6

1
2




2

3



sin x  3 cos x dx = I 

0





sin x  3 cos x dx 

2

 sin x 

3 cos x dx  3  3



0

3



2

cos xdx

 (sin t  cost )3  (sin x  cos x)3


0





12
dx
1
 4
1
 
  cot( x  )  1  I 
 2I  
2

20 2
2
4 0
2
0 (sin x  cos x)
sin ( x  )



2

sin xdx

 sin x  cos x 

3

I2 

;

2



0

cos xdx

sin x  cos x 

3

.

 t . Ta chứng minh được I1 = I2






1
 
tan( x  ) 2  1
2
4 0

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 I1  I 2 

1
 I  7I 1 – 5I 2  1.
2



Câu 88. I 

3sin x  2cos x

2

 (sin x  cos x)3 dx


2

3sin x  2cos x

3cos x  2sin x

 (cos x  sin x)3 dx

0





2

3cos x  2sin x

2

1

 (sin x  cos x)3 dx   (cosx  sin x)3 dx   (sin x  cosx)2 dx  1 

0



dt 





dt   

sin t

dt  I
2
1

cos
t
0

  
2
   I 
2
 4 4
8
0 1  cos t
d(cost )

dt   



Câu 90. I 



 cos3 x  sin3 xdx

0

2



 2I 

2





cos x sin x  sin x cos x
4

4

sin3 x  cos3 x

0

dx 

2


0

 Đặt x 


2


(cos x)  dx


 t  dx  dt


2

2



2


1
 tan2(sin t)  dt   
 tan2(sin x)  dx
2
2




1
1
Do đó: 2I   

 tan2 (cos x)  tan2(sin x)  dx = 2  dt  
2
2

 cos (sin x) cos (cos x)
0
0

 I


2

.



Câu 92. I 

4



cos x  sin x
3  sin2x

4





12

  dt 

6

6



Câu 93. I 

3

sin x

0

cos x 3  sin2 x



 Đặt t  3  sin2 x =



3

sin x.cos x
cos2 x 3  sin2 x

sin2 x

+ Tính I 1  

3

2
3

+ Tính I 2= 

3





dx =

3

dt
4  t2


 ln
ln 15  4  ln  3  2 .

4
2
15

4
3

2





2
3

x
2
3

Vậy: I 

0

sin3 x  sin2 x

3

u  x


du  dx
dx  
I1 

dx . Đặt 
dv 
v   cot x
3
sin2 x

sin2 x


x

2

dx
 3
1  sin x 
3

2

dx
dx
 3

I

cos2 x  4sin2 x

0



I

2



dx

2
udu
22
2
3
dx . Đặt u  3sin x  1  I  
  du 
u
31
3
1
3sin2 x  1

2sin x cos x

dx   
dx . Đặt t  tan x  dt 
dx  (tan2 x  1)dx
2
2
cos2x
cos x
0
0 (tan x  1)


6

1

 I 

3



0

1

1 3 1 3
.


2


dx . Đặt 1 cot x  t 

1
sin2 x

dx  dt

6

 I 2

3 1



3 1

t 1
dt  2  t  ln t 
t

3 1
3 1
3

 2

 2
 ln 3 

. Đặt t  tan x  dx 

dt
1 t2

4

3
3
3 (1  t 2)2 dt
3 1
1
t
8 34
  (  2  t 2)dt  (  2t  )

 I 
2
t
3
3
t2
1
1 t
1

Trang 21
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
 dt  ln3  ln2
3 0  t  2 2t  1 
2
3
0 2t  5t  2
1

t


4

sin2 xdx



cos4 x(tan2 x  2tan x  5)



Câu 100. I 



4

 Đặt t  tan x  dx 
1



ln

3
 2

3 1 t 2  2t  5
1 t  2t  5

 tan u  I 1 

1
2

0




du 



2 3
.
. Vậy I  2  ln 
8
3 8



 3sin x  4sin3 x dx   4cos2 x  1 dx





6

6

Đặt t  cosx  dt   sin xdx  I  

0



3
2



Câu 102. I  2

sin x  cos x
1  sin2x

4

dt
4t  1

4 2


 I  2
4

I 

1

sin x  cos x
dx . Đặt t  sin x  cos x  dt  (cos x  sin x)dx
sin x  cos x

21

t

dt  ln t

2
1

1
 ln2
2

Trang 22
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4

tan xdx

0

cos x 1  cos2 x





 Ta có: I 

4

tan xdx

0

cos2 x tan2 x  2



3

tdt
 I 


1
dt   .
3
32
2 t

 Đặt t  cos x  sin x  3  I  

dx

0



Câu 106. I 

4

sin4x

0

cos x. tan x  1



2

4




Câu 107. I 

4

sin4 x

 1  cos2 xdx

0



 Ta có: I 

4



2sin2x(2cos x  1)
2

1  cos x
2

0




.

2
2
(tan
x

1)
(
t

1)
2
0
0
6

 Ta có: I   

2

Trang 23
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

Câu 109. I 

tan 3 x





Câu 110. I 

2

cos x



7  cos2 x

0

I

dx

2

1

cos x dx



2





 Ta có:

3



sin x

4

4

3

dx 



1

.

1

2
tan x cos x


cos x  cos x  sin x
)dx
1  cos 2 x
3





 cos x(1  cos2 x)  sin x 
x.sin x
dx

x
.cos
x
.
dx

dx  J  K



2
2


1

cos




K

dx .

(  t ).sin(  t)

0



 2K  
0

1  cos (  t)
2

( x    x).sin x
1  cos2 x

Đặt x    t  dx  dt


dt  
0

(  t).sin t
1  cos t



Đặt t  cosx  K 



1

dt

,
2  1 t2

đặt t  tan u  dt  (1 tan2 u)du

1



K 





4

(1  tan u)du

2



du 


2



. u 4 


2

4

4

4

2



Câu 113. I 

2

cos x




3

dt
4  t2



1
 ln( 15  4)  ln( 3  2)
2

Dạng 3: Đổi biến số dạng 2


2

Câu 114. I   sin x  sin2 x 



1
.dx
2

6


 Đặt cos x 








2
3sin x  4cos x
3sin x
4cos x
3sin x
4cos x
dx

dx

dx

dx

dx
2
2
2
2





3  cos x
0
0

+ Tính I1  

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