200 bài tập tích phân môn toán 12 ôn thi THPT - Pdf 40

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Bi tp Nguyờn hm - Tớch phõn cú li gii

.

TP1: TCH PHN HM S HU T
Dng 1: Tỏch phõn thc
2

Cõu 1.

x2

I =

2
1 x 7 x + 12

2



1

I = 1 +

2
16
9


x
+
+
x x3 x2 + 1

2
1
3
1
3
+ ln( x 2 + 1) = ln 2 + ln 5 +
2
2
2
8
2x
1
1

2

3x 2 + 1
3

2

x 2 x 5x + 6

4


= ( x + 1)2 ( x + 1)3 I = ( x + 1)2 ( x + 1)3 dx =
3
3
0
8
( x + 1)
( x + 1)

Dng 2: i bin s

Cõu 5.

I =
1

Cõu 6.

I =

( x 1)2
(2 x + 1)4

dx

( 7 x 1)99

101
0 ( 2 x + 1)

1


x7

0 (1 +

x 2 )5

99

7x 1
1 1 7x 1
=
d


( 2 x + 1)2 9 0 2 x + 1 2 x + 1
dx

100

Cõu 7.

3
x 1
1 x 1
.
I =
+C
9 2x + 1
2x + 1

1 1
dt = .

2 1 t5
4 25


Bi tp Nguyờn hm - Tớch phõn cú li gii

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1

Cõu 9.

I = x 5 (1 x 3 )6 dx
0

t t = 1 x 3 dt = 3x 2dx dx =
4

Cõu 10. I =

3



1

2

3 7 8 168
1
2

3

1

t



1

3

t t 2 + 1 dt = 4 ln 2

1





1 32
dt
I = 5 10
. t t = x I =
2
2

5

1 128 1 t
I = 7
dx . t t = x I =
dt
7
7 1 t (1 + t )
1 x .(1 + x )

dx

(1 x 7 ).x 6

7

dx
6

x (1 + x 2 )

1

t : x =

1
I =
t

3

3

2

x 2001

Cõu 14. I =

1 (1 +

2

x 2 )1002

2

x 2004

I =

.dx

3
2 1002
1 x (1 + x )

Cỏch 2: Ta cú: I =

.dx =


1
x2

+ 1 dt =

2
x3

dx .

11
x 2000 .2 xdx
. t t = 1 + x 2 dt = 2 xdx

2
2000
2
2
2 0 (1 + x )
(1 + x )

1 2 (t 1)1000
1 2 1
I = 1000 2 dt = 1
21 t
2 1 t
t
Cõu 15. I =

.dx . t t =


1
t 2
1
I= 2
=

dt
=
.ln
=
ln


2


2 2
2 2 1t 2 t + 2
t + 2 1 2 2 2 + 1
1 t 2
dt

1

1

1

Bieõn soaùn: Thay Tran Sú Tuứng - Trang 2


1
1
dt
= x
. t t = x + dt = 1 dx I =
.
2
2
1
x
2
1+ x
x
t
+
2


2
x +
x2
du
5
5
t t = 2 tan u dt = 2
; tan u = 2 u1 = arctan 2; tan u = u2 = arctan
2
2
2


1

2

dx

dx

x6 + 1
x4 + 1

1

1
2
1
4
x
Ta cú: I =
dx . t t = x + I = ln
1
x
5
1
+x
x

2


x2
x6 + 1

=

1
x2 + 1

+

x2
x6 + 1

1 1 d (x3 )
1
I = 2 dx + 3 2 dx = + . =
3 0 (x ) + 1
4 3 4 3
0 x +1

Cõu 19.

1

3
3

I=

x2


1+ 5
2



Ta cú:

1
0t

dx =

x2 + 1

2

x +1
4

2

x x +1



0

1
1

3

t t = x 2 I =

.

x 4 x2 + 1

1

I =

2

2

Cõu 20. I =

Cõu 21. I =

dx

x2 +

1
x2
1
x2

1

4

Bieõn soaùn: Thay Tran Sú Tuứng - Trang 3

2

=


6 3


Bi tp Nguyờn hm - Tớch phõn cú li gii

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TP2: TCH PHN HM S Vễ T
Dng 1: i bin s dng 1

x

Cõu 22. I =

dx

2

3x + 9 x 1
x
I =




x + x
1+ x x

x2

dx =

x2

+ I1 =



1+ x x

2

dx

1+ x x

1+ x x

x
1+ x x

Vy: I =


x

)

3



4
1 + x x + C1
3

3

t2
1 + t dt =2 + ln 2 .
1
3 1
2 12

t t = 4 x + 1 . I = ln
1

t: t = 1 x 2 I = ( t 2 t 4 ) dt =

0

1+ x


Cõu 25. I =

1+ x x

dx . t t= 1 + x x t 2 1 = x x x 3 = (t 2 1)2 x 2dx =

4 2
4 3 4
4
3 (t 1)dt = 9 t 3 t + C = 9

+ I2 =

x

dx +

0

2
.
15

dx
1 3

1

t +t
2

t(t 1)dt
3


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Bi tp Nguyờn hm - Tớch phõn cú li gii
2

t t = x + 1 2tdu = dx I =

2t 3 8t

1t

0

x.

Cõu 29. I =

3

2

+ 3t + 2

2


5

x2 + 1

1

x 3x + 1

Cõu 30. I =

2

3

dx
2

t2 1

+1
4 3
2tdt
2tdt
t t = 3x + 1 dx =
I = 2
.
3
3
t 1

(
t

1)
dt
+
2


2
92
2 t 1

dx

x + 1 = t x = t 2 1 dx = 2tdt

t
2

2(t 2 1)2 + (t 2 1) 1
I =
2tdt
t
1
1

2

2


Cõu 33. I =

0

.2tdt =2

2



1

x +1
2
1 + 2x )

(1 +

2

2
t3
1
1
16 11 2
t
dt

=




22
22
2 2
t t2
t2
t2

=

Cõu 34. I =

8



3

1 t2
2
1
3t + 4 ln t + = 2 ln 2
2 2
t
4
x 1
2



8
3

= 1 + ln

(

3 + 2 ) ln ( 8 + 3)

1

Cõu 35. I = ( x 1)3 2 x x 2 dx
0

1

3

I = ( x 1)
0

2

Cõu 36. I =

2 x x dx = ( x 2 2 x + 1) 2 x x 2 ( x 1)dx . t t = 2 x x 2 I =
0

2 x3 3x 2 + x


0

x 3dx
3

4 + x2

3

t t = 4 + x 2 x 2 = t 3 4 2 xdx = 3t 2dt I =


3 2 4
38
(t 4t )dt = + 4 3 2

23
25

4

Cõu 38. I =

1

dx




1 11
1
1
+ 1 dx = ln x + x |1 = 1

2 1 x
2
1



1

1 + x2
dx . t t = 1 + x 2 t 2 = 1 + x 2 2tdt = 2 xdx I2=
2x

Vy: I = 1 .

2



t 2dt

2
2 2(t 1)

=0


dx

4 x2
dx
x

Ta cú: I =
I=

1

1
3 1
1
Ta cú: I = 2 1 . 3 dx . t t = 2 1 I = 6 .
x
x
1 x
3

2

0

1

t(tdt )
4 t2

x


3
3

0


2 3

= 3 + ln


2
+
3
3



Bieõn soaùn: Thay Tran Sú Tuứng - Trang 6


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Cõu 41. I =

2 5



0

dx

t t = x + x + x + 1 I =

1+ 3

1+
2dt
= ln(2t + 1)
1
2t + 1



1

3

x2
1 + x )2 (2 + 1 + x )2

0 (1 +

4

4


2t (t 2 1)2 dt

t t = x + 1 I =

t(t + 1)2

1

Cõu 46. I =

3

2 2



x x 3 + 2011x
x4

1

Ta cú: I =

3

2 2



1

1

1

x2
x3

dx +

2 2



dx =



1

2
2
2
= 2 (t 1)2 dt = (t 1)3 =
1
3
3
1

3




7
2



t 3dt =

0

2 2

2011
2011x dx =

2 x2 1
3

3

=

213 7
128

14077
16

3


2
2 5
2t
1
dt = 5 1 +

dt = 5 3 1 + ln

3 12
t t2 + 1 t2 + 1

t(t 2 + 1)
1

2

Cõu 44. I =

2

t3 2

1

Cõu 43. I =

3t

dt

dt



1

2

t .(t

3

2
1) 3

Bieõn soaùn: Thay Tran Sú Tuứng - Trang 7


Bi tp Nguyờn hm - Tớch phõn cú li gii

3

=

2



1


3
2 1 3
= t
t4
1

dt




1
t 2 . t 3 1 3
t

t u = 1

2

2
3


1
t 4 1 3
t
1
2

I=

1
2


=
3 1

3 0

1
1 2
= u3

=

0

1
3

2

x4
dx

1 2
x x x +1





1
2

2

dt =

19
2 4+ 2
+
ln

3
4 4 2

Dng 2: i bin s dng 2
1



2
x
ln
1
+
x
( ) dx
1+ x


dv = 2 xdx

0

Cõu 50. I =

2

(x

5

K=

1
2

+ x 2 ) 4 x 2 dx

2

I=

2

(x

5

2


2

4 x 2 dx . t x = 2sin t . Tớnh c: B = 2 .

2
2

+ Tớnh B =

2

2
2

+ Tớnh A =

2

2

Vy: I = 2 .

Bieõn soaùn: Thay Tran Sú Tuứng - Trang 8


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1 2x
2

+ Tớnh I 2 =

1



I2 =

4

2x4

1

4 x2
2x4

dx .

3 2 4
7
x dx = .

21
16


=

cot 2 t.d (cot t ) =
2



4
8 sin t
8
8
8
sin t
6

Vy: I =

1(
7 2 3) .
16

1

x 2dx

0

4 x6

Cõu 52. I =


Cõu 54. I =

0

Ta cú: I =

0

I =

2



2
3
1
2

Cõu 55.



0

t
2

0


+

3 3
4
2

2



1 2 x 1 x 2 dx

6

t x = sin t I = (cos t sin t )cos tdt =
0

Bieõn soaùn: Thay Tran Sú Tuứng - Trang 9


12

+

3 1

8 8




3



2

I=

3
2



3



x.

2

x 2 1dx

x
x2 1
3

dx = 5 2



Chỳ ý: Khụng c dựng phộp i bin x =

3

2

1
vỡ 2;3 [ 1;1]
cos t

TP3: TCH PHN HM S LNG GIC
Dng 1: Bin i lng giỏc
Cõu 57. I =

8cos2 x sin 2 x 3
dx
sin x cos x

(sin x cos x )2 + 4 cos 2 x
I =
dx = ( sin x cos x 4(sin x + cos x ) dx
sin x cos x
= 3cos x 5sin x + C .
cot x tan x 2 tan 2 x
dx
Cõu 58. I =
sin 4 x
2 cot 2 x 2 tan 2 x
2 cot 4 x


4


cos 2 x +



1
dx


4

dx +
=

2
2 2 1 + sin 2 x +













=

2 2
3
2 2 1 + sin 2 x +
sin x +





4
8




1

3
=
ln 1 + sin 2 x + cot x +
+ C
4
8


4 2


3
2 x
1 cos x +
2sin +
3
3
3

2 6


Cõu 61. I =

6

1

2 sin x

3

0

dx



1
Ta cú: I =
2





dx =

1
2

6





dx

0 sin x sin
3
3
x x
cos +
2 6 2 6

dx




x


2 6
20
x
20
x

sin
cos +
2 6
2 6




6
0

x
ln cos +
2 6


6
0



Cõu 62. I =



x + cos4 x )dx

0





2



0



1



1 2



0

1




5
cos xdx =

2
(1 sin x )

2

A =

0



2

d (sin x ) =

0

8
15



2

2
cos x.dx =

2

x cos 2 xdx

0





2

I = cos2 x cos2 xdx =
0



2

1
12
(1
+
cos
2
x
)
cos2
xdx
=

1 + cos x
sin2 x


I = 2 (4sin x 2sin 2 x )dx = 2
0

Cõu 67. I =

2



1 + sin xdx

0

I=

2

2
2

x
x
x
x
x
sin + cos dx = sin + cos dx = 2 sin + dx




Cõu 68. I =

4



0

dx
6

cos x

4

Ta cú: I = (1 + 2 tan2 x + tan 4 x )d (tan x ) =
0

Dng 2: i bin s dng 1

Bieõn soaùn: Thay Tran Sú Tuứng - Trang 12

28
.
15



1
3
1
t t = tan x . I = t 3 + 3t + + t 3 dt = tan 4 x + tan2 x + 3ln tan x
+C
t
4
2


2 tan2 x
2t
.
Chỳ ý: sin 2 x =
1 + t2
dx
Cõu 71. I =
sin x.cos3 x
dx
dx
dx
2t
I =
. t t = tan x dt =
=
2
; sin 2 x =
sin x.cos x.cos2 x
sin 2 x.cos2 x
cos2 x


1

2011 1

sin 2 x cot xdx =

sin 4 x

Ta cú: I =

t t = cot x I =

2
2011
t
(1 + t 2 )tdt

4024

2011

cot 2 x

sin 4 x
4024

cot xdx
8046


0





2
sin x.cos2 x
(t 1)2
dx . t t = 1 + cos x I = 2
dt = 2 ln 2 1
1
+
cos
x
t
0
1
2

Ta cú: I = 2


Cõu 74. I =

3

sin

2


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1
2

1 u2
3
du = ln 2
u
8
1

I =


sin

Cõu 75. I =

2

x (2 1 + cos2 x )dx



2





2













2

2

2

+ K = sin2 x 2 cos2 x = 2 sin2 x cos xdx = 2 sin2 xd (sin x ) =



I =

2
3


cos2 x

.

4

I=

3



(1 + t 2 )2 dt
t2

1

3

=



1

3

1
1
t3


( 2 + sin x ) dx
2

0



Ta cú: I =



2

sin 2 x

(2 + sin x )2
0

3

I = 2

2

t2
t2

2


sin x

cos 2 x dx
0



I=

6



0



sin x
dx =
cos 2 x

6

sin x

2 cos2 x 1 dx . t t = cos x dt = sin xdx

0

Bieõn soaùn: Thay Tran Sú Tuứng - Trang 14


6

dt =

1
2 2

ln

3
2
2t 2
2t + 2

1
3
2

=

1
2 2

ln

32 2
52 6



2

t t = cos x . I =

3
( + 2)
16

6


Cõu 81. I =

4

sin 4 x



dx

sin6 x + cos6 x

0



4

I=

2
.
3



Cõu 82. I =

2

sin x



( sin x +

0

3 cos x

)

3

dx



Ta cú: sin x + 3 cos x = 2 cos x


dx

I=
=
+


16 0


6
16 0

cos3 x
cos2 x
6
6




Cõu 83. I =

sin x 1 cos2 x

4








sin x
2

cos x

1 cos2 x .dx =

sin x
2

cos x





3
2

4



4

dx +


2

cos x

sin x dx +

3

7
3 1.
12

3

Bieõn soaùn: Thay Tran Sú Tuứng - Trang 15

4



sin x

2
0 cos

x

sin x dx



dx =
20

20



0 sin x + 3 cos x
1 cos2 x +
sin x +
3
3




6





6

6

1
2




I=

2



sin x 3 cos x dx = I =

0



3

sin x 3 cos x dx +



2

sin x

3 cos x dx = 3 3



0

3


(sin t + cos t )3 (sin x + cos x )3
0



2

0



2



12
dx
1
4
1
2I =
=
= cot( x + ) = 1 I =
2

20 2
2
4 0
2


0

sin xdx

( sin x + cos x )

3

;

I2 =

2

cos xdx

0

( sin x + cos x )



3

.

t . Ta chng minh c I1 = I2






=


1
tan( x ) 2 = 1
2
4 0

1
I = 7I1 5I 2 = 1 .
2



Cõu 88. I =

2

3sin x 2 cos x

(sin x + cos x )3 dx
0

Bieõn soaùn: Thay Tran Sú Tuứng - Trang 16


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2

3cos x 2sin x

(cos x + sin x )3 dx

0



3sin x 2 cos x

(sin x + cos x )3

dx +

2

0



dt =



3cos x 2sin x

(cos x + sin x )3


0 1 + cos



2

t

2

1 + cos t



dt =

sin t

2
0 1 + cos t

dt I


2
= + I =
2
4 4
8



4

sin t cos t
cos3 t + sin3 t

dt =

2

sin 4 x cos x

cos3 x + sin3 x dx

0

2



2

2I =



4




20
2

1
4

I= .


2



0



1

cos2 (sin x ) tan

Cõu 91. I =

t x =


2

2

1

I=





2

2


1
1
Do ú: 2I =
+
tan 2 (cos x ) tan 2 (sin x ) dx = 2 dt =
2
2

cos (sin x ) cos (cos x )
0
0

I=


2





1

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du
4u

2

. t u = 2sin t I =



4

2 cos tdt



2

4 4sin t



6


I=





3

3



0

=

dx

sin x

.dx =

cos x 3 + sin2 x
15
2

1 t+2
ln
4 t2

ln
4
15

4


2
3

x

3

cos2 x 3 + sin2 x

sin3 x + sin2 x

3

I =

0

sin x.cos x

x + ( x + sin x )sin x

Cõu 94. I =
2

dx .

15
2



3

1
1


dt
t+2 t2

3+2
1 (
=
ln 15 + 4 ) ln ( 3 + 2 ) .
2
3 2

(

)

dx

dx

3

2

dx
dx
= 3
=4 2 3


x
2
1 + cos x
3 2 cos
2

4 2

+42 3.


2

Cõu 95.

I=

0




2
udu
22
2
dx . t u = 3sin 2 x + 1 I = 3
= du =
u
31
3
1
3sin2 x + 1

2



sin 2 x



tan x

4 dx
cos2 x




tan x

.
=
=
2
t +1 0
2
(t + 1)
dt



0



Cõu 97. I =

3

cot x
dx
sin x.sin x +

4
6







= 2
ln 3
3


3 +1
3 +1
3

3



Cõu 98. I =

3

dx

sin2 x.cos4 x



4



3


=
+
+
t
dt
=

+
t
+



2
2
t
3
t
1
1 t
1

3
=

8 34
3





2

+ 5t + 2








cos4 x (tan 2 x 2 tan x + 5)

4

t t = tan x dx =
Tớnh I1 =

1 1 2
1
1
2


dt = ln 3 ln 2

3 0 t + 2 2t + 1
2
3


I=

1 t

t 1
2

2

2t + 5

= tan u I1 =

1
2

= 2 + ln
0






du =

2
3



2

Cõu 101. I =

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6



I=



2

2

sin x

3sin x 4sin3 x

dx =




dt
2

4t 1

=

1
4

3
2



0

dt
t2

1
4

=

1
ln(2 3)
4



6

Cõu 103. I = 2 1 cos3 x .sin x.cos5 xdx
1

t t = 6 1 cos3 x t 6 = 1 cos3 x 6t 5dt = 3cos2 x sin xdx dx =

2t 5dt
cos2 x sin x

1

1

t 7 t13
12
I = 2 t 6 (1 t 6 )dt = 2
=
7 13 0 91
0


Cõu 104. I =

4

tan xdx

0

dt =

3 2

2



Cõu 105. I =

2

cos2 x

(cos x sin x + 3)3

0

tan x
dx
cos 2 x

4

dx

t 3
1
dt = .
3

cos x. tan x + 1


Ta cú: I =

4

sin 4 x



4

4

sin x + cos x

0

dx . t t = sin 4 x + cos4 x I = 2

2
2



dt = 2 2 .

1


Cõu 108. I =

6



0

1
2

2(2t 1)
1
dt = 2 6 ln .
t +1
3
1

dx . t t = cos2 x I =



tan( x )
4 dx
cos 2 x


1
3


6




tan 3 x

6
6
tan 3 x
Ta cú: I =
dx =
dx .
2
2
2
2
0 cos x sin x
0 cos x(1 tan x)
3
3 t3
1 1 2
t t = tan x I =
dt = ln .
2
6 2 3
0 1 t






3

Cõu 111.

4



4

dx
sin3 x.cos5 x


Ta cú:



3

1





4


1

2
tan x cos x
3

dx .

= 4 ( 8 3 1)

1

Bieõn soaùn: Thay Tran Sú Tuứng - Trang 21


6 2


Bi tp Nguyờn hm - Tớch phõn cú li gii
Cõu 112. I =



x(
0

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cos3 x + cos x + sin x
)dx



u = x
du = dx
+ Tớnh J = x.cos x.dx . t
J = 2

=
dv
cos
xdx
v = sin x


0


x.sin x

+ Tớnh K =

0 1 + cos



2

2

1 + cos ( t )



=

4


2

4

( x ).sin x

0

sin x.dx
2



dt =

x

1 + cos2 x

K=

dx








dt =

0

( x + x ).sin x

0

t x = t dx = dt

dx .

( t ).sin( t )

K=

K=

x







3 + cos 2 x

dx

6


2

Ta cú: I =


sin x cos x
sin x 3 + cos x
2

2

dx . t t = 3 + cos2 x

6

I=

15
2





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.

Bi tp Nguyờn hm - Tớch phõn cú li gii




3 1
3
34
sin t , 0 t I = cos2 tdt = + .
2

2
2 4 2
20

t cos x =


Cõu 115. I =

3sin x + 4 cos x
dx
2
x + 4 cos 2 x


2
2
2
2
2




3
+
cos
3
+
cos
3
+
cos
3
+
cos
4

sin
x
x
x
x
x
0

3 3(1 + tan 2 u )du 3
=
3(1 + tan 2 u )
6
0
6

t t = 3 tan u dt = 3(1 + tan 2 u )du I1 =

2

+ Tớnh I 2 =
0

Vy: I =

1

4 cos x
4dt1
dx . t t1 = sin x dt1 = cos xdx I 2 =
dt1 = ln 3
2
4 sin x
4 t12
0

3
6





6

tan x
cos2 x

t u = tan x du =

dx =

1
2

cos x
1

4





+1

6

tan x
2

3



3

dt = t

7
3

7
3



Cõu 117. I =

2





7

= 3

3



sin x + cos x

( sin x cos x )

t t = 2 tan u I =

1
2

2

arctan



0

+2

dx . t t = sin x cos x I =

1

2(1 + tan 2 u)

1
1
du = arctan
2

Cõu 118. I =




3

x sin x
cos2 x

dx .

S dng cụng thc tớch phõn tng phn ta cú:


I=

3







1
x
xd
=
cos x cos x


3


3

tớnh J ta t t = sin x. Khi ú J =




Vy I =


3

3
2

dx
=
cos x


3







1 + cos x .e

x

dx

x
x
1 + sin x 1 + 2sin 2 cos 2
1
x
Ta cú:
=
=
+ tan
x
x
1 + cos x
2
2 cos2
2 cos2
2
2


2


x


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.



Cõu 120. I =



4

Bi tp Nguyờn hm - Tớch phõn cú li gii

x cos 2 x

(1 + sin 2 x )

0

2

dx

u = x
du = dx


cos2 x

1
dx = +
4+
16 2
2 1 + sin 2 x 0 2 0 1 + sin 2 x

I = x. .

1



.

1

2 cos2 x

4


0

dx




1 1


dt = t 3 t 2 + 2t 2 ln t + 1 + C = e x e x e x + 2 e x 2 ln e x + 1 + C
1+ t
3
3
Cõu 122. I =

I =

( x 2 + x )e x
x + e x

( x 2 + x )e x
x

x+e
dx

Cõu 123. I =

dx

dx =

xe x .( x + 1)e x
xe + 1
x

dx . t t = x.e x + 1 I = xe x + 1 ln xe x + 1 + C .

e2 x + 9


x ln( x 2 + 1) + 2011
Ta cú: I =
dx . t t = ln( x 2 + 1) + 1
2
2
( x + 1) ln( x + 1) + 1
1 t + 2010
1
1
1
I=
dt = t + 1005ln t + C = ln( x 2 + 1) + + 1005ln(ln( x 2 + 1) + 1) + C
2
t
2
2
2

Cõu 124. I =

e

Cõu 125. J =

1

xe x + 1
x (e x + ln x )


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