bài tập tích phân có giải

Chuyên đề: Nguyên hàm – Tích phân
2
1
0
2
xdx
I
x
=
+
∫
3
2
1
1
7
x x dx
I
x
−
=
−
∫
6
3
2
1
2 1 4 1
I dx
x x
=

+ + +
∫
( )
4
4
3
2
1
dx
I
x x
=
+
∫
10
5
5
2 1
dx
I
x x
=
− −
∫
1
8 3
6
0
1I x x= −
∫

1
7
0
3 2
2 1 1
x
I dx
x
+
=
+ +
∫
1
2
8
2
0
4
2 5 2
x x
I dx
x x
+
=
+ +
∫
( )
2
2
9

1
1
2 2
x
I dx
x x
−
=
+ +
∫
2 2
5 2
10
0
1I x x dx= +
∫
11
2
0
sin
3 cos
x x
I dx
x
π
=
+
∫
( )
2

1
12
2
0
1
1
x
I dx
x
+
=
+
∫
( )
3
2
13
3
2
1
1
x x
I dx
x
− +
=
−
∫
2
14

0
sin
sin cos
x
I dx
x x
π
=
+
∫
3
3
15
4
6
sin
cos
x
I dx
x
π
π
=
∫
3
16
3
4
1
cos .sin

I dx
x x
π
π
=
∫
( )
2
3 3
17
0
sin cosI x x dx
π
= +
∫
( )
2
18
2
0
sin 2
2 sin
x
I dx
x
π
=
+
∫
( )

2
19
2
0
sin 2
2 sin
x
I dx
x
π
=
+
∫
( )
4
20
2
0
cos2
sin cos 2
x
I dx
x x
π
=
+ +
∫
( )
2
3 2

21
0
1 sin sinI x xdx
π
= −
∫
3
2
22
0
4sin
1 cos
x
I dx
x
π
=
+
∫
3
2
23
2
0
sin cos
1 cos
x x
I dx
x
π

=
+
∫
ln3
24
0
2
x
dx
I
e
=
+
∫
1
2
25
0
1
x
x
e
I dx
e
=
−
∫
2
26
0

.sin .cosI x x xdx
π
=
∫
( )
27
2
1
ln
1
e
e
x x
I dx
x
+
=
+
∫
10
2
28
1
lgI x xdx=
∫
( ) ( )
2
29
0
2 5 ln 1I x x dx= + +

∫
( )
1
2 2
30
2
0
1
x
x e
I dx
x
=
+
∫
( )
1
2
31
0
ln 1I x x dx= +
∫
( )
2
32
sin 2
x
I x e x dx
−
= +

∫
2
33
2
2
cos
4 sin
x x
I dx
x
π
π
−
+
=
−
∫
4
sin
34
0
(tg cos )
x
I x e x dx
π
= +
∫
2
35
1

3
ln
ln 1
e
x
I dx
x x
=
+
∫
36
1
3 2ln
1 2ln
e
x
I dx
x x
−
=
+
∫
4
37
0
2 1
1 2 1
x
I dx
x

+
=
+ +
∫
2
38
0
sin 2
3 4sin cos2
x
I dx
x x
π
=
+ −
∫
1
3
2
39
2
0
4
x
x
I xe dx
x
 
= −
 ÷

−
 
∫
2
40
2 2
0
sin sin 2
3sin 4
x x
I dx
x cos x
π
+
=
+
∫
( )
0
3
41
1
1
x
I x e x dx
−
−
= + +
∫
( )

ln3
43
3
0
1
x
x
e
I dx
e
=
+
∫
(
)
2
1
3 2
44
0
1
x
I x e x dx= + +
∫
4
45
0
1 cos 2
x
I dx

x
π
=
+
∫
1

Lời Giải: Chuyên đề nguyên hàm – tích phân
1,
( )
( )
2
2 2
3
2
1
0 0
0
2 2 8
2 2 4 2 2 1
3 3
2 2
xdx
I x dx x x
x x
 
 
= = + − = + − + = −
 ÷
 ÷

+ +
 
 
∫ ∫
2,
3 3 3 3 3
2
1 1 1 1 1
( 7) ( 1) ( 1)
1 6 1
1 ( 1) 6
7 7 7 7
x x dx x dx
x x dx x dx
I x d x
x x x x
− − −
− −
= = + = − − −
− − − −
∫ ∫ ∫ ∫ ∫

( )
3
3
2
2 2
1
2 4 2
1 6 6

3 3
x I I
′ ′
= − − = −
với
3
2
1
( 1)
7
x dx
I
x
−
′
=
−
∫
Để tính
3
2
1
( 1)
7
x dx
I
x
−
′
=

−
∫
ta đặt
2
1 1x t x t− = ⇒ = +

2
2
2
2
0
2
6
t dt
I
t
′
⇒ = =
−
∫
( )
2
2
2
0
0
6 6
2 1 2 3ln 2 2 3ln(2 3)
6
6

t
dt t
t
t
 
−
 
= + = + = + −
 ÷
 ÷
 ÷
−
+
 
 
∫

2
32 2
48ln(2 3)
3
I⇒ = − −
3,
6
3
2
1
2 1 4 1
I dx
x x

=
+ + +
∫
Đổi biến
2
4 1 4 1 2t x t x tdt dx= + ⇒ = + ⇒ =
5 5 5
3
2 2
3 3 3
( 1) ( 1)
2 1 ( 1) ( 1)
tdt d t d t
I
t t t t
+ +
⇒ = = −
+ + + +
∫ ∫ ∫
( )
5
3
1 3 1
ln 1 ln
1 2 12
t
t
 
 
= + + = −

 ÷
 ÷
+
 
 
4,
( )
( )
( )
( )
4
3 3
4 4 4
3
4
3
3
2 2 2
2
1 1 1
1 1 1 65
ln ln 1 ln 2 ln
3 1 3 3 9
1
x dx d x
dx
I x x
x x
x x
+ − +

 
= = − = − + = −
 
+
+
 
∫ ∫ ∫
5,
10
5
5
2 1
2ln 2 1
dx
I
x x
= =
− −
+
∫
(đổi biến
1t x= −
)
6,
1
8 3
6
0
1I x x= −
∫

Đổi biến
3 2 3 2
1 1 2 3t x t x tdt x dx= − ⇒ = − ⇒ = −
( ) ( )
0 1
2
2 2 6 4 2
6
1 0
2 2
1 2
3 3
I t t dt t t t dt⇒ = − − = − +
∫ ∫
1
6 5 3
0
2 2 16
3 7 5 3 315
t t t
 
= − + =
 ÷
 
2
7,
1
7
0
3 2

2 1 1
x
I dx
x
+
= =
+ +
∫
(đổi biến
2 1 1t x= + +
)
8,
( ) ( )
2
1 1
4 14 1 14 9
2 2 ln 2 ln 3
8
2
3 2 3 2 1 3 2
0 0
2 5 2
x x
I dx dx
x x
x x
+
= = − + = + −
∫ ∫
+ +

+ +
 
 ÷
 
9,
( )
( ) ( )
( )
2
2 2
2
9
3
1 1
2
2 4 2 3
1
2 2
2
x x
x
I dx dx
x x
x
+ − + +
−
= =
+ +
+
∫ ∫

( ) ( ) ( )
2
1 1 3
2 2 2
1
2 2 3 2x x x dx
− −
 
= + − + + +
 
 
∫

( ) ( ) ( )
2
3 1 1
2 2 2
1
2 5
2 2 2 6 2 2 3
3 3
x x x
−
 
= + − + − + = −
 
 
10,
( )
2

3
2 2 3
7 5 3
1
2
5 2 2 2
10
0 1
1
2
1 1
7 5 3
t x
t t t
I x x dx t t dt
= +
 
= + = − = − + =
 ÷
 
∫ ∫
11,
11
2
0
sin
3 cos
x x
I dx
x

π
=
+
∫
Đổi biến
t x dt dx
π
= − ⇒ = −
( ) ( )
( )
( ) ( )
11 11
2 2 2
0 0 0
sin sin cos
3 cos 3 cos 3 cos
t t t t d t
I dt dx I
t t t
π π π
π π π
π
π
− − −
⇒ = = = − −
+ − + +
∫ ∫ ∫
( )
( )
2

1
2
6
3 tan
11
2 2 2
0 1
6
3 1 tan
cos
2 3 cos 2 3 2 3 3tan
6 3
u v
v du
d t
du
I
t u v
π
π
π
π π π π
=
−
−
+
−
⇒ = = = =
+ + +
∫ ∫ ∫

12,
( )
( )
2
1 1
1
2
12
2 2
0
0 0
1
2
1 ln 1 ln 2 1
1 1
x
x
I dx dx x x
x x
+
 
= = + = + + = +
 ÷
+ +
 
∫ ∫
13,
( )
( ) ( )
( ) ( )

3
2
3 3
2
13
3 3 2
2 2
2
1 1 1
1 1 1 5
ln 1 ln 2
1 4
1 1 2 1
x x
x x
I dx dx x
x
x x x
 
− + − +
− +
= = = − − − = +
 
−
− − −
 
 
∫ ∫
14,
2

14
0
sin
sin cos
x
I dx
x x
π
=
+
∫
( )
2
0
sin cos '
1
1
2 sin cos 4
x x
dx
x x
π
π
+ 
= − =
 ÷
+
 
∫
15,

3 2
3 3
3
15
4 4 3
6
6 6
sin 1 cos 1 1 14 26 3
cos
cos cos 3cos cos 3 27
x x
I dx d x
x x x x
π π
π
π
π π
−
 
= = − = + = −
 ÷
 
∫ ∫
3
16,
( )
( )
3 3 3
16
3 2 3

2 3
4 4 4
sin
1 cos
cos .sin cos .sin
1 sin sin
d x
x
I dx dx
x x x x
x x
π π π
π π π
= = =
−
∫ ∫ ∫

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

t t
 
= = + +
 ÷
−
−
 
∫ ∫
3
2
2
2
2
2
1 1 1 1
ln ln 1 ln3
2 2 2 3
t t
t
 
= − − − = +
 ÷
 
17,
( )
2 2 2
3 3 3 3
17
0 0 0
sin cos sin cosI x x dx xdx xdx

π π π
= + = +
∫ ∫ ∫

( ) ( )
2 2
3 3
2 2
2 2
0 0
0 0
cos sin 4
sin cos cos sin cos sin
3 3 3
x x
xd x xd x x x
π π
π π
   
= − + = − − + − =
 ÷  ÷
   
∫ ∫
18,
( ) ( )
( )
2 2
18
2 2
0 0

sin 2 sin
2 sin
2 sin 2 sin
x x
I dx d x
x x
π π
= =
+ +
∫ ∫

( )
( )
( )
( )
2
1 1
2 2
0 0
0
2 2
2
2 2 2 ln 2 2ln 2 1
2
2 2
t
t
dt dt t
t
t t

+ −
 
= = = + + = −
 ÷
+
 
+ +
∫ ∫
20,
( )
( ) ( )
( )
4 4
20
2 2
0 0
sin cos cos sin
cos2
sin cos 2 sin cos 2
x x x x
x
I dx dx
x x x x
π π
+ −
= =
+ + + +
∫ ∫

Đặt

sin cos 2 cos sint x x dt x x= + + ⇒ = −
, khi
0 3; 2 2
4
x t x t
π
= → = = → = +
Do đó:
2 2
2 2 2 2
20
2 2
3 3
3
2 1 2 2 2 2 5
ln ln 2
3 3
t
I dt dt t
t t t t
+
+ +
− +
   
= = − = + = + −
 ÷  ÷
   
∫ ∫
21,
( )

2 2 2
3 2 2 5
21
0 0 0
1 sin sin sin sinI x xdx xdx xdx
π π π
= − = −
∫ ∫ ∫

( ) ( )
2 2 2
2
2
2 2 4
0 0 0
0
2
3 5
0
1 cos 2 sin 2
1 cos cos 1 2cos cos cos
2 2 4
2cos cos 8
cos
4 3 5 4 15
x x x
dx x d x x x d x
x x
x
π π π

π
π
π π
−
 
= + − = − + − +
 ÷
 
 
= + − + = −
 ÷
 
∫ ∫ ∫
4
22,
( )
3 2
2 2
22
0 0
4sin 4sin
cos
1 cos 1 cos
x x
I dx d x
x x
π π
= =
+ +
∫ ∫

( ) ( )
2
2
2
0
0
cos
4 1 cos cos 4 cos 2
2
x
x d x x
π
π
 
= − = − = −
 ÷
 
∫
23,
( )
3 2
2 2
2
23
2 2
0 0
sin cos 1 cos
1 cos
1 cos 2 1 cos

x x x
I dx d x
x x
π π
= = − +
+ +
∫ ∫

( )
2
2
1
1
1 1 1 1 ln 2
ln
2 2 2
t
dt t t
t
− −
= = − =
∫
24,
( )
( )
( )
ln3 ln3 ln3
24
0 0 0
1 1 1

2 2 2
2
x
x
x x x
x x
d e
dx
I d e
e e e
e e
 
= = = −
 ÷
+ +
+
 
∫ ∫ ∫

ln3
0
1 1 3 1 1 6
ln ln ln ln
2 2 2 5 2 2 5
x
x
e
e
 
= = − =

 ÷
+
 
25,
( ) ( )
1 1 1
2
25
0 0 0
1
1 1
1 1 1
x x
x x x
x x x
e e
I dx d e e d e
e e e
 
= = = − + −
 ÷
− − −
 
∫ ∫ ∫
( ) ( )
( )
1
3 1
2 2
0

2 2
1 2 1 1 2
3 3
x x
e e e e
 
= − + − = − +
 ÷
 
26,
( )
2 3 3 3
26
0
0 0 0
1 1
.sin .cos cos cos cos
3 3
I x x xdx xd x x x xdx
π π π
π
 
= = − = − −
 ÷
 
∫ ∫ ∫

( )
( )
3

2
0
0
1 1 sin
1 sin sin sin
3 3 3 3 3 3
x
x d x x
π
π
π π π
 
= + − = + − =
 ÷
 
∫
27,
( )
( )
27
2
1 1
ln 1
ln
1
1
e e
e e
x x
I dx x x d

x
x
+
 
= = − +
 ÷
+
 
+
∫ ∫
( ) ( )
1
1
1 1
ln ln
1 1
e
e
e
e
x x d x x
x x
 
= − + + +
 ÷
+ +
 
∫

1 1

1 1 1 2 1 2
1 1
1 1 1 1
e e
e e
e e
dx dx
e x x e x e
− −
 
= − + + + = + =
 ÷
+ + + +
 
∫ ∫
28,
( ) ( )
10 10 10
10
2 2 2 2 2 2 2
28
1
1 1 1
1 1
lg lg lg lg
2 2
I x xdx xd x x x x d x
 
= = = −
 ÷

 
∫ ∫ ∫

( )
( )
10 10 10
10
2 2 2
1
1 1 1
10
2 2
1
1 2 1 1
100 lg 50 lg 50 lg lg
2 ln10 2ln10 2ln10
50 1 50 99
50 50
ln10 2ln 10 ln10 4ln 10
x xdx xd x x x x d x
xdx
   
= − = − = − −
 ÷  ÷
   
= − + = − −
∫ ∫ ∫
∫
29,
( ) ( ) ( )

( )
2 2
2
29
0 0
2 5 ln 1 ln 1 5I x x dx x d x x= + + = + +
∫ ∫
5
( )
( )
( )
2 2
2
2
2
0
0 0
2
2
0
5 4
5 ln 1 14ln 3 4
1 1
14ln 3 4 4ln 1 18ln3 10
2
x x
x x x dx x dx
x x
x
x x

+
 
= + + − = − + −
 ÷
+ +
 
 
= − + − + = −
 ÷
 
∫ ∫
30,
( )
1 1
2 2
2 2
30
2
0 0
1
1
1
x
x
x e
I dx x e d
x
x
 
= = −

 ÷
+
 
+
∫ ∫

( ) ( )
1
1 1 1 1
2 2 2 2 2
1
2 2 2 2 2 2
0
0 0 0 0
0
1
2 2 2 2
0
1
2
1 1 2 2 2
1 1
2 2 2 2 2 2
x
x x x x x
x
x e e e e
d x e xe dx xd e xe e dx
x x
e e e e

= − + = − + = − + = − + −
+ +
 
= − = − − =
 ÷
 
∫ ∫ ∫ ∫
31,
( ) ( )
( )
1 1
2 3
31
0 0
1
ln 1 ln 1
3
I x x dx x d x= + = +
∫ ∫

( )
( )
1 1
3
1
3 2
0
0 0
1
3 2

0
1 1 ln 2 1 1
ln 1 1
3 3 1 3 3 1
ln 2 1 2ln 2 5
ln 1
3 3 3 2 3 18
x
x x dx x x dx
x x
x x
x x
 
= + − = − − + −
 ÷
+ +
 
 
= − − + − + = −
 ÷
 
∫ ∫
32,
( )
2 2 2
32
sin 2 sin 2
x x
I x e x dx x e dx x xdx
− −

= + = +
∫ ∫ ∫

( )
( )
( )
( )
( )
2 2 2 2
2 2 2 2
2 2
1 1
cos 2 2 cos 2 cos 2
2 2
1 1 1 1 1
cos 2 2 sin 2 cos 2 2 sin 2 sin 2
2 2 2 2 2
1 1 1
2 1 cos2 sin 2 cos 2
2 2 4
x x x
x x x x x
x
x d e x d x x e xe dx x x x xdx
x e x x xd e xd x x e x x xe e dx x x xdx
x x e x x x x x
− − −
− − − − −
−
= − − = − + − +

= − − − + = − − − + + −
= − + + − +
∫ ∫ ∫ ∫
∫ ∫ ∫ ∫
33,
2 2 2
33
2 2 2
2 2 2
cos cos
4 sin 4 sin 4 sin
x x x x
I dx dx dx
x x x
π π π
π π π
− − −
+
= = +
− − −
∫ ∫ ∫
Ta có:
( )
2 2 2
2 2 2
2 2 2
0
4 sin 4 sin 4 sin
t x
x t t

A dx dt dt A A
x t t
π π π
π π π
−
=−
− −
= = = − = − ⇒ =
− − − −
∫ ∫ ∫
6
( )
2 2
2
2
2
2 2
cos 1 1 1 1 2 sin ln 3
sin ln
4 sin 4 2 sin 2 sin 4 2 sin 2
x x
B dx d x
x x x x
π π
π
π
π π
−
− −
+

 
= = − + = − = −
 ÷
− − + −
 
∫ ∫
Vậy
33
ln3
2
I A B= + = −
34,
4 4 4
sin sin
34
0 0 0
sin
(tan cos ) cos
cos
x x
x
I x e x dx dx e xdx
x
π π π
= + = +
∫ ∫ ∫

( )
4
2

sin sin
4
4 2
0
0
0
2 ln 2
ln cos ln 1
2 2
x x
x d e e e
π
π
π
= − + = − + = + −
∫
35,
2
35
1
3
ln
ln 1
e
x
I dx
x x
=
+
∫

Đặt
2
1
ln 1 ln 1 2t x t x tdt dx
x
= + ⇒ = + ⇒ =
( )
( ) ( ) ( )
2
2
2 3
35
1
1 1
2 2
1
2 2
2 2 1 1 1
3 3
t
I tdt t d t t
t
−
⇒ = = − − = − =
∫ ∫
36,
36
1
3 2ln
1 2ln

e
x
I dx
x x
−
=
+
∫
Đặt
2
1
2ln 1 2ln 1t x t x tdt dx
x
= + ⇒ = + ⇒ =
( )
2
2 2
2 3
2
36
1 1
1
4 10 2 11
4 4
3 3 3
t t
I t dt t dt t
t
 
−

⇒ = = − = − = −
 ÷
 
∫ ∫
37,
4
37
0
2 1
1 2 1
x
I dx
x
+
=
+ +
∫
Đặt
( ) ( )
2
1 2 1 1 2 1 1t x t x dx t dt= + + ⇒ − = + ⇒ = −
( )
4
4 4
2
37
2 2
2
1 1
1 2 2 ln ln 2 2

2
t t
I t dt t dt t t
t t
 
−
 
⇒ = − = − + = − + = −
 ÷
 ÷
 
 
∫ ∫
38,
( )
2 2
38
2
0 0
sin sin
sin 2
3 4sin cos 2 2 4sin 2sin
xd x
x
I dx dx
x x x x
π π
= =
+ − + +
∫ ∫

( )
( )
1
1
2
0
0
1 1 1 1 2ln 2 1
ln 1 ln 2
2 1 2 2 4
2 1
tdt
dx t
t
t
−
   
= = + + = − =
 ÷  ÷
+
   
+
∫
39,
( ) ( )
1 1 1
3 2
2 2 2
39

2 2
0 0 0
1 1
4
2 2
4 4
x x
x x
I xe dx xd e d x
x x
 
= − = + −
 ÷
− −
 
∫ ∫ ∫
7
4
1
4
3
2 2 2 2
2
2
3
0
3
1 1 4 1 1 1 2 1 1 32 61
8 6 3 3 3
2 2 2 2 2 2 2 3 4 2 3 4 12

x
x
e t e e e
xe dt t t
t
 
   
− +
 
= − − = + − − = − − = + −
 ÷
 ÷  ÷
 ÷
 
   
 
∫
40,
2 2 2
40
2 2 2 2 2 2
0 0 0
sin sin 2 sin sin 2
3sin 4 3sin 4 3sin 4
x x x x
I dx dx dx
x cos x x cos x x cos x
π π π
+
= = +

+ + +
∫ ∫ ∫
Có:
( )
1
2 2
2 2 2 2
0 0 0
cos
sin
3sin 4 3 3
d x
x dt
A dx
x cos x cos x t
π π
= = − =
+ + +
∫ ∫ ∫
Đặt
( )
2
3 tan 3 1 tant u dt u du= ⇒ = +

( )
( )
2
6 6 6
6
2

2
0 0 0
0
3 1 tan
sin
1 1 sin 1
ln ln3
cos 1 sin 2 1 sin 2
3 3tan
u du
d u
du u
A
u u u
u
π π π
π
+
+
⇒ = = = = =
− −
+
∫ ∫ ∫
( )
( )
2
2 2
2
2
2 2 2

0
0 0
4 sin
sin 2
2 4 sin 2 2 3
3sin 4 4 sin
d x
x
B dx x
x cos x x
π π
π
−
= = − = − − = −
+ −
∫ ∫
Vậy
( )
40
ln3
2 2 3
2
I A B= + = + −
41,
( )
0 0 0
3 3
41
1 1 1
1 1

x x
I x e x dx xe dx x x dx A B
− −
− − −
= + + = + + = +
∫ ∫ ∫
( ) ( )
3
1
0 0 0 0 1
7 4
1
0
3 3
3
1
1 1 1 1 0
0
9
2 1, 1 3 1 3
7 4 28
t x
x x x x
t t
A xe dx xd e xe e dx e B x x dx t t dt
= +
− − − −
−
− − − −
 

= = − = − + = − = + = − = − = −
 ÷
 
∫ ∫ ∫ ∫ ∫
Vậy
41
37
2
28
I A B e= + = −

2
2 2
43
3 2
2
2 2
2 2 2
2 1
tdt dt
I
t t t
⇒ = = = − = −
∫ ∫
43,
( )
ln3
43
3
0

1
x
x
e
I dx
e
=
+
∫
.
Đặt
2
1 1 2
x x x
t e t e tdt e dx= + ⇒ = + ⇒ =
44,
(
)
2 2
1 1 1
3 2 3 3 2
44
0 0 0
1 1
x x
I x e x dx x e dx x x dx A B= + + = + + = +
∫ ∫ ∫
Ta có:
( )
( )

2 2 2 2
1 1 1
1
3 2 2 2
0
0 0 0
1 1
2 2
x x x x
A x e dx x d e x e e d x
 
= = = −
 ÷
 
∫ ∫ ∫

2
1
0
1 1 1
2 2 2 2 2 2
x
e e e
e
   
= − = − − =
 ÷  ÷
   
( )
2

2
1 2
5 3
1
3 2 2 2
0 1
1
2 2 2
1 1
5 3 15
t x
t t
B x x dx t t
= +
 
+
= + = − = − =
 ÷
 
∫ ∫
Vậy
44
1 2 2 2 17 4 2
2 15 30
I A B
+ +
= + = + =
45,
( )
4 4 4 4

4
45
2
0
0 0 0 0
1 1
tan tan tan
1 cos 2 2cos 2 2
x x
I dx dx xd x x x xdx
x x
π π π π
π
 
 ÷
= = = = −
 ÷
+
 
∫ ∫ ∫ ∫
8

4
0
1 1 2 1
ln cos ln ln 2
8 2 8 2 2 8 4
x
π
π π π

= + = + = −
9