# 200 bài tập tích phân môn toán 12 ôn thi THPT

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Bi tp Nguyờn hm - Tớch phõn cú li gii

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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

dx = ( x + 16 ln x 4 9 ln x 3 ) 1 = 1 + 25ln 2 16 ln 3 .
x 4 x 3

2

Cõu 2.

I =

1

Ta cú:

dx
5

x + x3
1

5

I =

I =

1 1

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

Cõu 4.

=

x 3 ( x 2 + 1)

I = ln x
Cõu 3.

dx

dx

2 4 13 7 14
I = ln + ln + ln 2
3 3 15 6 5

xdx

1

0

( x + 1)3
x
x + 11
1
1
Ta cú:
=
= ( 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

7x 1
I =

2x + 1
0

99

1 x 1
Ta cú: f ( x ) = .

3 2x + 1

I =

Cõu 8.

I =

0 (x

1

5x
2

2

+ 4)

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

dx

1 1 7x 1
=

9 100 2 x + 1
1

2

1 100
1
=
2 1
0 900
1
8

dx

t t = x 2 + 4 I =

dx

t t = 1 + x 2 dt = 2 xdx I =
Bieõn soaùn: Thay Tran Sú Tuứng - Trang 1

1 2 (t 1)3
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

2

2

1 x7

Cõu 12. I =

x (1 + x 7 )

1

3

11 6
1 t 7 t8
1
(1

)
=
t
t
dt
=

30
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 t (t + 1)2
1 x .( x + 1)

x.( x10 + 1)2

1

Cõu 13. I =

2

dx

Cõu 11. I =

3x 2

I =

t t = x 2 I =

dx
x ( x 4 + 1)

1

dt

x 4 .dx

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

1

t6

dt =
t2 + 1

1

4 2
117 41 3
1
+
t t +1 2
dt =
135
12
1
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

1
1002

1

x 3 2 + 1
x

1000

2

1 + x2

4
11+ x

Ta cú:
3
2

1+ x

2

1+ x4

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
1
d 1 =
t 2002.21001

dx
1+

=

1

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

2
1
x
2
x

x +
2
x
3
2

3
2 1

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

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.

2

1 x2

Cõu 16. I =

11+

x4

1 x

Bi tp Nguyờn hm - Tớch phõn cú li gii

dx
1

2

5
2

1

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
cos u

Ta cú:

2

4

u2

2
I=
2
2

Cõu 17. I =

u1

1 x

Cõu 18. I =

0

x4 + 1

Ta cú:

x6 + 1

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

3
1x+x

1

2
2
5
(u2 u1 ) =
arctan arctan 2
2
2
2

du =

=

( x 4 x 2 + 1) + x 2
x6 + 1

=

x4 x2 + 1
( x 2 + 1)( x 4 x 2 + 1)

+

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

x4 1

0

I=

3
3

0

x

2

( x 1)( x + 1)
1
0

xdx
4

2

x + x +1

1+ 5
2

Ta cú:

1
0t

dx =

x2 + 1

2

x +1
4

2

x x +1

0

1
1
1

+
2
dx = ln(2 3) +
2
4
12
x 1 x +1
1 1 dt
11
=
2 0 t 2 + t + 1 2 0

dt
2
1 3

t + +
2 2

dx
1+

=

1
2

3
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

. t t = x

1
1
dt = 1 + dx
x
x2

dt
2

+1

. t t = tan u dt =

du
2

cos u

4

I = du =
0

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 =
dx = x (3x 9 x 2 1)dx = 3x 2dx x 9 x 2 1dx
2
3x + 9 x 1
2

3

+ I1 = 3x dx = x + C1

3

1
1
+ I 2 = x 9 x 1dx = 9 x 2 1 d (9 x 2 1) = (9 x 2 1) 2 + C2
18
27
2

3

I=

1
(9 x 2 1) 2 + x 3 + C
27
x2 + x

Cõu 23. 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 =

4
9

(

4

dx =

1+ x x

2x + 1

Cõu 24. I =

01+

2x + 1

6
2 2x

)

3

+ 1 + 4x + 1

01+

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

1+ x x

t t = 2 x + 1 . I =

dx

1

Cõu 27. I =

(

+C

Cõu 26. I = x 3 1 x 2 dx
1

dx .

2 d (1 + x x )
4
=
1 + x x + C2

3
3
1+ x x

dx

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
11
t t = x dx = 2t.dt . I = 2
dt = 2 t 2 t + 2
4 ln 2 .
dt =
t +1
3
1+ t

0
0
3

Cõu 28. I =

x 3

dx
3
x
+
1
+
x
+
3
0
Bieõn soaùn: Thay Tran Sú Tuứng - Trang 4

4 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

2

1
3
dt = 3 + 6 ln
t +1
2
1

dt = (2t 6)dt + 6
1

x + 1dx

1

1

1

t7 t4
9
t t = x + 1 t = x + 1 dx = 3t dt I = 3(t 1)dt = 3 =
28
7 4 0
0
3

3

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
2
.t
3
4
4
21 3
t 1
100
9
= t t + ln
=
+ ln .
93
5
t + 1 2 27
2
3

Cõu 31. I =

2x2 + x 1
x +1

0

=

4
24 2
dt
(
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

4t 5

54
= 2 (2t 4 3t 2 )dt =
2t 3 =
5
1 5
1

x 2dx

Cõu 32. I = 2

0 ( x + 1)

x +1

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

2

(t 2 1)2
t3

1
4

Cõu 33. I =

0

.2tdt =2

2

1

x +1
2
1 + 2x )

(1 +

2

2
t3
1
1
16 11 2
t
dt

=
2
2t =

t 1
3
t
3

dx

t 2 2t
t t = 1 + 1 + 2 x dt =
dx = (t 1)dt v x =
2
1 + 2x
dx

Ta cú: I =

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

x +1

dx

Bieõn soaùn: Thay Tran Sú Tuứng - Trang 5

Bi tp Nguyờn hm - Tớch phõn cú li gii

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(

8

)

x
1

I=

dx = x 2 + 1 ln x + x 2 + 1

2

x2 + 1
3 x +1

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
x2 x + 1

0

1

2

2
.
15

dx

3
( x 2 x )(2 x 1)
4
dx . t t = x 2 x + 1 I = 2 (t 2 1)dt = .
3
0
x2 x + 1
1

2

I =

2

Cõu 37. I =

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

x + 1 + x2

1 1 +

1

1
1 + x 1 + x2
1 11
1 + x2
Ta cú: I =
dx =
dx = + 1 dx
dx
2
2
2x
2 1 x
2x
1 (1 + x ) (1 + x )
1
1

+ I1 =
+ I2 =

1

1 + x 1 + x2

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

Cỏch 2: t t = x + x 2 + 1 .
Cõu 39. I =

1

1
3
2

Cõu 40. I =

1

1
3 3
x

(x )
x4

4 x2

1

3

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

2

0

xdx . t t =

t2

0

4 x 2 t 2 = 4 x 2 tdt = xdx

4

t2
=
dt = (1 +
)dt = t + ln

2
2
t+2

t
4
t
4

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

5

x
( x 2 + 1) x 2 + 5

2

Cõu 42. I =

Bi tp Nguyờn hm - Tớch phõn cú li gii

27

x 2

3

x+ x

1

3

1

x2 + x + 1

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

3

= ln

3+ 2 3
3

dx
42

36

4

t 2 + 1 + x = t I = 2t 16 + 2 dt = 12 + 42 ln
t
3
t
3
3

x2

Cõu 45. I =

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

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

M=

3

2 2

2 2

2011

x3

1

I=

2

x
x3

1

N=

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

2011

x3

1

1

2 2

2

dx

1

dx . t t =

dx

1
x2

dx = M + N

1 M =

3
2

7
2

t 3dt =

0

2 2

2011
2011x dx =

2 x2 1
3

3

=

213 7
128

14077
16

3

14077 21 7

.
16
128
1

dx

Cõu 47. I =

0 (1 +

3

x 3 ). 1 + x 3
3

3

3

=

1 15
ln .
4 7

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

dx

2

t t = 6 x I = 5
1

t t = x 2 + 5 I =

dx

t t = 1 + x I =

2

t2

2
1 4 3
t .(t 1) 3

3

dt =

2

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

dt

=

2

3

1

Cõu 48. I =

2 2

3

1
t3

du =

3dt
t4

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=

2

u 3

0

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3

du =

2
3

dt

1
2 2
u 3 du

1
3 0

1
1 u3

1
2

=
3 1

3 0

1
1 2
= u3

=

0

1
3

2

x4
dx

1 2
x x x +1

t t = x 2 + 1
3

I =

2

(t 2 1)2
t2 2

3 4

dt =

t 2t 2 + 1
t2 2

2

3

3

2

2t

dt = t 2 dt +

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

0

Cõu 49. I =

1 x

1

1 x

Tớnh H =

1+ x

0

dx . t

x = cos t; t 0; H = 2
2
2

1

u = ln(1 + x )

Tớnh K = 2 x ln(1 + x )dx . t

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

+ x ) 4 x dx =

2

x

5

2

4 x dx +

2

x

2

4 x 2 dx = A + B.

2

x

5

4 x 2 dx . t t = x . Tớnh c: A = 0.

x

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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(3

2

Cõu 51. I =

Bi tp Nguyờn hm - Tớch phõn cú li gii

)

4 x 2 dx
2x4

1
2

Ta cú: I =

2

3

4
1 2x

2

+ Tớnh I1 =

3

1 2x
2

+ Tớnh I 2 =

1

I2 =

4

2x4

1

4 x2
2x4

dx .

3 2 4
7
x dx = .

21
16

dx =

dx . t x = 2sin t dx = 2 cos tdt .

2

2

4 x2

dx

6

6

2

1 cos tdt 1
12
3
2 1
=
cot
t
dt
=

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 =

t t = x 3 dt = 3 x 2 dx I =

1 1 dt
.
3 0 4 t 2

16

t t = 2sin u, u 0; dt = 2 cos udu I = dt = .
2
30
18
2

Cõu 53. I =

2 x
dx
x+2

1

x 2dx

0

Cõu 54. I =

0

Ta cú: I =

0

I =

2

2
3
1
2

Cõu 55.

0

t
2

0

3 + 2x x2
1

2

t x = 2 cos t dx = 2sin tdt I = 4 sin2 dt = 2 .

x 2dx
22 ( x 1)2

. t x 1 = 2 cos t .
2
3

2

(1 + 2 cos t ) 2sin t
4 (2 cos t )2

dt =

( 3 + 4 cos t + 2 cos2t ) dt =

2

+

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

Bi tp Nguyờn hm - Tớch phõn cú li gii

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Dng 3: Tớch phõn tng phn
Cõu 56. I =

3

x 2 1dx

2

x
dx
u = x 2 1 du =
t

2
x

1
dv = dx
v = x

I = x x2 1
=5 2

3

2

I=

3
2

3

x.

2

x 2 1dx

x
x2 1
3

dx = 5 2

dx

x2 1

2

3

2
x 1 +

2

dx
2
x 1
1

= 5 2 I ln x + x 2 1

5 2
1
ln ( 2 + 1) + ln 2
2
4

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
cos 4 x
1
Ta cú: I =
dx =
dx = 2
dx =
+C
sin 4 x
sin 4 x
2sin 4 x
sin 2 4 x

cos2 x +
8

Cõu 59. I =
dx
sin 2 x + cos 2 x + 2

1 + cos 2 x +
1

4 dx
Ta cú: I =

2 2 1 + sin 2 x +

4

cos 2 x +

1
dx

4

dx +
=

2
2 2 1 + sin 2 x +

sin x + + cos x +

4

8
8

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Bi tp Nguyờn hm - Tớch phõn cú li gii

cos 2 x +

1
dx

4 dx + 1

=

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

Cõu 60. I =

dx

2+

3 sin x cos x

3

I=

1
2

1
dx
1
dx
= I=
=
.

4

4
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

=

cos

6

6

0

1
sin x sin

3

dx =

6

dx =

1
2

6

dx

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

dx

x
x

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

x
x

cos

sin

2+ 6
2 6
16
16

dx = ln sin x
=
dx +

2 6
20
x
20
x

sin
cos +
2 6
2 6

6
0

x
ln cos +
2 6

6
0

Cõu 62. I =

2

(sin

4

x + cos4 x )(sin 6 x + cos6 x )dx .

0

Ta cú: (sin 4 x + cos4 x )(sin6 x + cos6 x ) =

33 7
3
33
+ cos 4 x + cos8 x I =
.
64 16
64
128

Cõu 63. I =

2

cos2 x(sin

4

x + cos4 x )dx

0

2

0

1

1 2

0

1

I = cos2 x 1 sin2 2 x dx = 1 sin2 2 x d (sin 2 x ) = 0
2
2
2

Cõu 64. I =

2

3

(cos

x 1) cos2 x.dx

0

Bieõn soaùn: Thay Tran Sú Tuứng - Trang 11

= .....

Bi tp Nguyờn hm - Tớch phõn cú li gii

5
cos xdx =

2
(1 sin x )

2

A =

0

2

d (sin x ) =

0

8
15

2

2
cos x.dx =

B=

2

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0

12

(1 + cos 2 x ).dx =

20
4

8

.
15 4

Vy I =

2

cos

Cõu 65. I =

2

x cos 2 xdx

0

2

I = cos2 x cos2 xdx =
0

2

1
12
(1
+
cos
2
x
)
cos2
xdx
=
(1 + 2 cos2 x + cos 4 x )dx
2 0
4 0

2
1
1

= ( x + sin 2 x + sin 4 x ) =
4
4
8
0

2
0

Cõu 66. I =

4sin3 x
dx
1 + cos x

4 sin3 x 4sin3 x (1 cos x )
=
= 4sin x 4sin x cos x = 4sin x 2sin 2 x
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

2
2
2
2
2 4
0
0

2

0

3

2
2
x
x
= 2 sin + dx sin + dx = 4 2
2 4
2 4
0
3

2

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

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Bi tp Nguyờn hm - Tớch phõn cú li gii

.

sin 2 xdx
3 + 4sin x cos2 x
2sin x cos x
1
Ta cú: I =
+C
dx . t t = sin x I = ln sin x + 1 +
2
sin x + 1
2sin x + 4 sin x + 2
dx
Cõu 70. I =
sin3 x.cos5 x
dx
dx
I = 3
= 8 3
3
2
sin x. cos x. cos x
sin 2 x. cos 2 x

3
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 + t2
Cõu 69. I =

I = 2

dt
2t

=

t2 + 1
1
t2
tan2 x
dt = (t + )dt = + ln t + C =
+ ln tan x + C
t
t
2
2

1 + t2
Cõu 72. I =

2011

sin 2011 x sin 2009 x
sin 5 x

cot xdx

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

2011 2011 2011 2011
=
t
+
t
+C
4024
8046

8046

2011
2011
=
cot 2011 x +
cot 2011 x + C
4024
8046

Cõu 73. I =

2

sin 2 x.cos x
dx
1
cos
x
+
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

x tan xdx

0

3

Ta cú: I = sin2 x.
0

sin x
dx =
cos x

(1 cos2 x )sin x
dx . t t = cos x

x
cos
0
3

Bieõn soaùn: Thay Tran Sú Tuứng - Trang 13

Bi tp Nguyờn hm - Tớch phõn cú li gii

www.mathvn.com

1
2

1 u2
3
du = ln 2
u
8
1

I =

sin

Cõu 75. I =

2

x (2 1 + cos2 x )dx

2

Ta cú: I = 2sin2 xdx sin2 x 1 + cos2 xdx = H + K
2

2

2

2

+ H = 2sin2 xdx = (1 cos 2 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

2

Cõu 76. I =

3

dx

sin2 x.cos4 x

4

3

I = 4.

dx
sin 2 2 x.cos2 x

. t t = tan x dt =

dx
cos2 x

.

4

I=

3

(1 + t 2 )2 dt
t2

1

3

=

1

3

1
1
t3
8 34
2
+
2
+
t
dt
=

+
2
t
+

=
2

3 1
3
t
t

Cõu 77. I =

2

sin 2 x

( 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

dx = 2

sin x cos x

2
0 (2 + sin x )

dx . t t = 2 + sin x .
3

3

1 2

2
3 2
dt = 2 dt = 2 ln t + = 2 ln
2
t t
t 2
2 3

2

Cõu 78. I =

6

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

2
3

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Bi tp Nguyờn hm - Tớch phõn cú li gii

i cn: x = 0 t = 1; x =

Ta c I =

3
2

1
2

2t 1

1

t=

6

dt =

1
2 2

ln

3
2
2t 2
2t + 2

1
3
2

=

1
2 2

ln

32 2
52 6

Cõu 79. I =

2

2

t t = sin2 x I =

sin x
3
e .sin x.cos x. dx
0

11 t
1
e (1 t )dt = e 1 .

20
2

2

Cõu 80. I = sin x sin2 x +

1
dx
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=

sin 4 x

3
1 sin 2 2 x
4

0

3
dx . t t = 1 sin 2 2 x I =
4

1
4

4
2 1
3 t dt = 3 t

1

1
1
4

=

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

;
6

3
1

sin x = sin x + =
sin x + cos x
6 6
2
6 2
6

sin x dx
2
6
3
3
1 2
dx

I=
=
+

16 0

6
16 0

cos3 x
cos2 x
6
6

Cõu 83. I =

sin x 1 cos2 x

4

cos2 x

dx

3

4

I=

=

0

sin x
2

cos x

1 cos2 x .dx =

sin x
2

cos x

3
2

4

4

dx +

0

sin 2 x
2

cos x

dx =

sin x
2

cos x

sin x dx =

3

0

sin x
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

Bi tp Nguyờn hm - Tớch phõn cú li gii

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6

Cõu 84. I =

sin x +

1
3 cos x

0

dx

sin x +
1
1
1
1
3 dx .

I=
dx =
dx =
20

20

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

6

6

6

1
2

1
1
1
t t = cos x + dt = sin x + dx I =
dt = ln 3
2
3
3
2 0 1 t
4

2

Cõu 85. I =

1 3 sin 2 x + 2 cos2 xdx

0

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

Cõu 86. I =

2

sin xdx

(sin x + cos x )3
0

t x =

2

t dx = dt I =

2

cos tdt

=

cos xdx

(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 x + cos x )
sin ( x + )
4
dx

Cõu 87. I =

2

7sin x 5cos x

(sin x + cos x )3 dx

0

Xột: I1 =
t x =

2

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

2

Tớnh I1 + I2 =

0

I1 = I 2 =

dx

( sin x + cos x )

2

=

2

dx

0

2 cos2 ( x )
4

=

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

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Bi tp Nguyờn hm - Tớch phõn cú li gii

t x =

2

t dx = dt I =

2

3cos t 2sin t

(cos t + sin t )3
0

2I = I + I =

2

Cõu 89. I =

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

dx =

0

2

1

(sin x + cos x )2 dx = 1

I=

0

x sin x

1 + cos2 x dx

0

t x = t dx = dt I =

0

2I =

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
0 1 + cos t
d (cos t )

dt =

cos4 x sin x

2

Cõu 90. I =

sin t

( t )sin t

cos3 x + sin3 x dx
0

t x =

2

0

t dx = dt I =

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

4

cos x sin x + sin x cos x
sin3 x + cos3 x

0

dx =

2

0

3

3

sin x cos x (sin x + cos x )
sin3 x + cos3 x

dx =

12
1
sin 2 xdx =

20
2

1
4

I= .

2

0

1

cos2 (sin x ) tan

Cõu 91. I =

t x =

2

2

(cos x ) dx

t dx = dt

2

2

1
tan 2 (sin t ) dt =
tan 2 (sin x ) dx
2
2

cos (cos t )
cos (cos x )
0
0

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

.

Cõu 92. I =

4

cos x sin x

0

3 sin 2 x

dx

Bieõn soaùn: Thay Tran Sú Tuứng - Trang 17

1
.
2

Bi tp Nguyờn hm - Tớch phõn cú li gii

t u = sin x + cos x I =

2

1

www.mathvn.com

du
4u

2

. t u = 2sin t I =

4

2 cos tdt

2

4 4sin t

6

4

= dt =

12

6

Cõu 93. I =

3

sin x

2

cos x 3 + sin x

0

4 cos2 x . Ta cú: cos2 x = 4 t 2 v dt =

t t = 3 + sin2 x =

I=

3

3

0

=

dx

sin x

.dx =

cos x 3 + sin2 x
15
2

1 t+2
ln
4 t2

3

2
3

2

sin x
2
3

+ Tớnh I1 =

3

2
3

+ Tớnh I 2 =

3

Vy: I =

1
15 + 4
ln
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
3

=

3

15
2

dx =

3

dt
4t

2

=

1
4

sin x cos x
3 + sin2 x

dx .

15
2

3

1
1

dt
t+2 t2

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

(

)

dx

dx
.
1 + sin x

dx +

3

u = x

du = dx
dx
I1 =
dx . t
2
dv
=
v = cot x
3
sin x

sin 2 x

x

2

dx
= 3
1 + sin x
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

I=

Cõu 96. I =

6

0
6

0

dx
2

2sin x cos x

0

I=

cos2 x + 4sin2 x

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
6

4 dx = tan x + 1 dx . t t = tan x dt = 1 dx = (tan 2 x + 1)dx

2
cos 2 x
cos2 x
0 (tan x + 1)

Bieõn soaùn: Thay Tran Sú Tuứng - Trang 18

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Bi tp Nguyờn hm - Tớch phõn cú li gii

1

1

3

I =

1 3 1 3
.
=
=
2
t +1 0
2
(t + 1)
dt

0

Cõu 97. I =

3

cot x
dx
sin x.sin x +

4
6

3

I = 2

cot x
sin 2 x (1 + cot x )

dx . t 1 + cot x = t

1
sin 2 x

dx = dt

6

3 +1

I= 2

3 +1

t 1
dt = 2 ( t ln t )
t

2

= 2
ln 3
3

3 +1
3 +1
3

3

Cõu 98. I =

3

dx

sin2 x.cos4 x

4

3

Ta cú: I = 4.

dx
2

2

sin 2 x.cos x

. t t = tan x dx =

dt
1 + t2

4

3 (1 + t 2 )2 dt
3 1
1
t3
2
(
2
)
(
2
)
I
=
=
+
+
t
dt
=

+
t
+

2
2
t
3
t
1
1 t
1

3
=

8 34
3

Cõu 99. I =

4

sin x

5sin x.cos2 x + 2 cos x dx
0

4

tan x

1

5tan x + 2(1 + tan2 x ). cos2 x dx . t t = tan x ,

Ta cú: I =

0

1

t

I =

0 2t

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

sin 2 xdx

4

Cõu 100. I =

dt =

1

1 t

dt
2

2t + 5

1+ t

t 2 dt

1

dt
2

. t

I=

1 t

t 1
2

2

2t + 5

= tan u I1 =

1
2

= 2 + ln
0

du =

2
3

1

3

1 t

dt
2

2t + 5

2 3
. Vy I = 2 + ln
.
8
3 8

4

Bieõn soaùn: Thay Tran Sú Tuứng - Trang 19

Bi tp Nguyờn hm - Tớch phõn cú li gii

sin 2 x
dx .
sin 3 x

2

Cõu 101. I =

www.mathvn.com

6

I=

2

2

sin x

3sin x 4sin3 x

dx =

2

sin x

4 cos2 x 1 dx

6

6

t t = cos x dt = sin xdx I =

0

3
2

Cõu 102. I = 2

sin x cos x
1 + sin 2 x

4

dt
2

4t 1

=

1
4

3
2

0

dt
t2

1
4

=

1
ln(2 3)
4

dx

Ta cú: 1 + sin 2 x = sin x + cos x = sin x + cos x (vỡ x ; )
4 2

sin x cos x
dx . t t = sin x + cos x dt = (cos x sin x )dx
sin x + cos x

I = 2
4

21

I =

1

t
2

2

dt = ln t 1 =

1
ln 2
2

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

cos x 1 + cos2 x

Ta cú: I =
3

4

tan xdx

0

cos2 x tan2 x + 2

tdt
I=
=
t
2

. t t = 2 + tan 2 x t 2 = 2 + tan 2 x tdt =

3

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
32
2 t

t t = cos x sin x + 3 I =

Bieõn soaùn: Thay Tran Sú Tuứng - Trang 20

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Cõu 106. I =

4
0

Bi tp Nguyờn hm - Tớch phõn cú li gii

sin 4 x
2

dx

4

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 107. I =

4

sin 4 x

1 + cos2 x dx
0

Ta cú: I =

4

2

2sin 2 x (2 cos x 1)
2

1 + cos x

0

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

tan x + 1
dt
1 3
=
.
dx . t t = tan x I =
2
2
(tan
x
+
1)
(
t
+
1)
2
0
0
2

6

Ta cú: I =

Cõu 109. I =

tan 3 x
0 cos 2 x dx
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

Cõu 110. I =

2

cos x

7 + cos 2 x

0

I=

dx

1
2

2

0

cos x dx
22 sin2 x

=

3

Cõu 111.

4

4

dx
sin3 x.cos5 x

Ta cú:

3

1

4

4

3

sin x
3

cos x

t t = tan x I =

dx =
.cos8 x
3 3
t 4 dt

3

4

4

1

.

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
1 + cos 2 x

cos x (1 + cos2 x ) + sin x
x.sin x
dx = x.cos x.dx +
dx = J + K

2
2

1
+
cos
x
1
+
cos
x
0
0

Ta cú: I = x
0

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 )

0

2K =

1 + cos2 x

t t = cos x K =

Vy I =

4

2

1 + cos t

dx =
dt

2 1

=

4

2

4

( x ).sin x

0

sin x.dx
2

dt =

x

1 + cos2 x

K=

dx

sin x.dx

2 0 1 + cos2 x

t t = tan u dt = (1 + tan2 u)du

1 + t2 ,

1 + tan 2 u

2

1

(1 + tan u)du

2

( t ).sin t

0 1 + cos

2

4

dt =

0

( x + x ).sin x

0

t x = t dx = dt

dx .

( t ).sin( t )

K=

K=

x

4

du =

2

. u 4 =

4

2
4

2

Cõu 113. I =

2

cos x

sin 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

3

dt
4t

2

=

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

Dng 3: i bin s dng 2

2

Cõu 114. I = sin x sin2 x +

1
.dx
2

6

Bieõn soaùn: Thay Tran Sú Tuứng - Trang 22

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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

3sin
0

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

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

sin
x
x
x
x
x
0
0
0
0
0
2

2

I =

2

2

1

2

3sin x
3dt
+ Tớnh I1 =
dx . t t = cos x dt = sin xdx I1 =
2
3 + cos x
3 + t2
0
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

+ ln 3

Cõu 116. I =

4

6

tan x
2

cos x 1 + cos x

dx

Ta cú: I =

4

6

tan x
cos2 x

t u = tan x du =

dx =

1
2

cos x
1

4

+1

6

tan x
2

cos x tan x + 2
1

dx I =

2

cos x

2

1

u
u2 + 2

dx

dx . t t = u2 + 2 dt =

u
u2 + 2

3

I =

3

3

dt = t

7
3

7
3

Cõu 117. I =

2

7

= 3

3

=

3 7
3

.

sin x +
4

dx
2sin x cos x 3

4

Ta cú: I =

1

2

2

4

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
2
2 tan u + 2
2

Bieõn soaùn: Thay Tran Sú Tuứng - Trang 23

1

dt

2 t2 + 2
0

1
2

1

du .

Bi tp Nguyờn hm - Tớch phõn cú li gii

www.mathvn.com

Dng 4: Tớch phõn tng phn

3

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

3

3

dx
4
=
J , vi J =
cos x
3

3

3

tớnh J ta t t = sin x. Khi ú J =

Vy I =

3

3
2

dx
=
cos x

3

dx
cos x

3

1 t 1
1 t 2 = 2 ln t + 1
3
dt

2

4
2 3
ln
.
3
2+ 3

Cõu 119. I =

2

1 + sin x

0

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

e dx

2

x
I=
+ e tan dx = e 2
x 0
2
0 2 cos2
2
x

Bieõn soaùn: Thay Tran Sú Tuứng - Trang 24

3
2

3
2

= ln

2 3
2+ 3

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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
t

1
dv =
dx
=

v
2

1 + sin 2 x
(1 + sin 2 x )

4

4

1

1
1
1
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

1 2
2
= + .
tan x 4 = + .
0 + 1) =

(
16 2 2
4
16 2 2
4 16

0

TP4: TCH PHN HM S M - LOGARIT
Dng 1: i bin s
Cõu 121. I =

e2 x
1 + ex

dx

t t = e x e x = t 2 e x dx = 2tdt .
t3
2
2
I = 2
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

t t = e2 x + 9 I =

dt
t2 9

=

1 t 3
1
ln
+ C = ln
6 t+3
6

e2 x + 9 3
e

2x

+9 +3

+C

ln(1 + x 2 ) x + 2011x
dx
2
x 2 +1

ln (ex + e)

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 )

dx

e

d (e x + ln x )

1

e x + ln x

J=

= ln e x + ln x

Bieõn soaùn: Thay Tran Sú Tuứng - Trang 25

e
1

= ln

ee + 1
e

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