Bài tập Tích phân hay

Nguyễn Phú Khánh – ðà Lạt  
lại
( )
( ) ( ) ( )
P x
dx
x a x b x c− − −∫
ðặt : 
( ) ( ) ( ) ( ) ( ) ( )
( )
, ,
.
P x A B c
A B C
x a x b x c x a x b x c
= + + ⇒
− − − − − −
( )( )∫ −−
dx
bxax
xP
3
)(
ðặt : 
( ) ( ) ( ) ( )3 2 3
( )
, , ,
P x A B C D
A B C D
x a x bx a x b x b x b
= + + + ⇒
− −− − − −
( ) ( )
1 1 1 1 1
ln
x
dx dx C
x x xx x
α
α β α β α β βα β
  −
= − = + 
− − − − −− −  
∫ ∫ 
( ) ( )2 2
1 1 1 1 1 1
ln
2 2
x a
dx dx dx C
a x a x a ax a x ax a x a
 − 
 = = − = +   − +− +− +    
∫ ∫ ∫ 
1 3
2 2
2
2 ( 2) ( 2) ( 2)
3
x dx x d x x C− = − − = − +∫ ∫ 
( ) 122 ( 3) ( 3) 3( 3)
dx
x d x x C
x
−−= + + = − + +
+∫ ∫ 
2 2 2
21 1( )
2 2
x x xxe dx e d x e C= = +∫ ∫ 
1
11 22 3 3
3 3 33 3
3 3
1 1 (1 ) 1
(1 ) (1 ) (1 )
3 3 1 21 1
3
x dx x
x d x C x C
x
− +
− +
= + + = + = + +
+ − +
∫ ∫ 
2 1
32 2 3 33 3
3 2
1 2 2
2 2 3 3
3 3
x dx x dx x dx x x C x x C
x
− 
+ = + = + + = + +  
 
∫ ∫ ∫ 
( ) ( ) ( )1 13 cos - 2 sin - 2 3 cos 2 - sin 2x x x dx x dx x dx x dx dx
x x
 
+ = + − + 
 
∫ ∫ ∫ ∫ ∫ 
23 sin 2 cos lnx x x x C= + − + + 
1 1 cos 8 cos2
sin 3 cos5 (sin 8 sin2 ) ( )
2 2 8 2
x x
x xdx x x dx C= − = − + +∫ ∫ 
( )
4
2
2 2
sin 1 1 3 1
cos 2 tan 2 cos2 2 tan sin2
2 4 2 4cos cos
x dx
dx x dx x x xd x x x x C
x x
 
= + − = − + + = − + + 
 
∫ ∫ ∫ ∫ 
( ) ( ) ( )cos 13 sin 2 1 3. sin 2 1 2 1
sin 2
xdx
cotx x dx x d x
x
 − + = − + + ∫ ∫ ∫ 
( ) ( )sin 3 3sin 2 1 2 1 ln | sin | cos(2 1)
sin 2 2
d x
x d x x x C
x
= − + + = + + +∫ ∫ 
Nguyễn Phú Khánh – ðà Lạt  
1
1 1 1 3 3(3 cos 3 ) 3 cos 3 3 sin 3 sin
3 3 ln 3 ln 3
x x
x xx dx xdx dx x C x C
−
−− = − = − + = − +∫ ∫ ∫ 
( ) ( )
1
1 ln 1
1 1 1
x
x
x
x x x
d ed x e
d x d x x e C
e e e
+ 
= − = − = − + + 
+ + + 
∫ ∫ ∫ ∫ 
Tính nguyên hàm của các hàm số sau : 
3
1 1
( )f x
x x
= − 
( ) 2 3x xf x = + 
( ) = + − +( 1)( 1)f x x x x 
( ) = − −37 5
1 1
( )( )f x x x
x x
( ) = + + 22 cos2 2 sin 3f x x x x 
( ) = + + +2 12 sin 2 cosf x x x x
x
Tìm nguyên hàm ( )F x của hàm số 
( ) − −=
3 2
2
1x x
f x
x
 biết ( )− =3 10F ( ) = + + 22 cos2 2 sin 3f x x x x biết π  = − 
 
3
3
F 
Tính nguyên hàm bằng phương pháp ñổi biến số 
 ðịnh lý : Cho hàm số ( )= u u x có ñạo hàm liên tục trên K và hàm số ( )= y f u liên tục 
sao cho ( )  f u x xác ñịnh trên K . Khi ñó nếu F là một nguyên hàm của f , tức là 
( ) ( )= +∫ f u du F u C thì ( ) ( )   = +   ∫ f u x dx F u x C 
2 1
dx
I
x x
=
+
∫ 
ðặt : 2 2 21 1t x t x td t xdx= + ⇒ = + ⇒ = 
( ) 222 2 2
1 1 1
2 1 1111 1
dx xdx tdt dt
I dt
t ttt tx x x x
 
= = = = = − 
− +−−  + +
∫ ∫ ∫ ∫ ∫ 
2
2
1 1 1 1 1
ln ln
2 1 2 1 1
t x
I C C
t x
  − + − = + = +    + + +   
( )32 2
.
1 1
xdx
I
x x
=
+ + +
∫ 
Nguyễn Phú Khánh – ðà Lạt  
ðặt : 2 2 2
2 2
1 1
1 11 . 1 1
xdx tdt dt
t x t x tdt xdx
t t tx x
= + ⇒ = + ⇒ = ⇒ = =
+ ++ + +
22 1 2 1 1 .
1
dt
I t C x C
t
= = + + = + + +
+
∫ 
Tính nguyên hàm của các hàm số sau 
5(5 3)x dx+∫ 
( )202 1x dx+∫ 
( )
1
2 21 x xdx+∫ 
3 3
3
1
4x x
dx
x
+
∫ 
2
1
4 4
dx
x x− +∫ 
2
1
6 9
dx
x x− +∫ 
( )3
1
2 3
dx
x +
∫ 
2
3
9
1-
x
dx
x
∫ 
2
1
(1 )
dx
x x+
∫ 
2
2
3 3(8 27)
x dx
x +
∫ 
2
3 2
6 9
x
dx
x x
+
+ +∫ 
3 2
2 3
2
x
dx
x x x
+
+ −∫ 
23 2
3
x x
dx
x
+ +
+∫ 
2 4 5
2
x x
dx
x
+ +
+∫ 
3
2 2 1
x
dx
x x+ +∫ 
2
2
3 3 3
3 2
x x
dx
x x
+ +
− +∫ 
2
3 2
3 1
2 5 6
x
dx
x x x
+
− − +∫ 
4 21x x dx−∫ 
43 21x x dx−∫ 
2
1
5 6
dx
x x− +∫ 
2
1
1
dx
x x−
∫ 
2 ln 3x
dx
x
+
∫ 
. ln . ln(ln )
dx
x x x∫ 
5ln
dx
x x
∫ 
1
1x
dx
e +∫ 
2
2 1
x
x
e
dx
e +∫ 
21
x
x
e
dx
e−∫ 
21 x
dx
e+
∫ 
6
0
sin .sin 4 .x x dx
π
∫ 
cos sinxe xdx∫ 
s in cosxe xdx∫ 
3 cos sinxe xdx∫ 
Nguyễn Phú Khánh – ðà Lạt  
( )ln
3 ln
ex
dx
x x+∫ 4x x
dx
e e−−∫ 
2x xe e dx−+ +∫ 
3
4
cos
sin
x
dx
x
∫ 
4cos xdx∫ 
( )2 4tan tanx x dx+∫ 
1 sin
1 cos
x
dx
x
+
−∫ 
4sin xdx∫ 
6cos xdx∫ 
2 2sin cosx xdx∫ 
2 5sin cosx xdx∫ 
3 2sin cosx xdx∫ 
5sin os
3 3
x x
c dx∫ 
4sin cosx xdx∫ 
3
4
sin
cos
x
dx
x
∫ 
4
6
sin
cos
x
dx
x
∫ 
4
sin
cos
x
dx
x
∫ 
5
sin
cos
x
dx
x
∫ 
tanxdx∫ 
cotxdx∫ 
3(tan tan )x x dx+∫ 
3(cot + cot )x x dx∫ 
3
2
tan
cos
x
dx
x
∫ 
2
cot
1 sin
x
dx
x+∫ 
4cos
dx
x
∫ 
6sin
dx
x
∫ 
6
1
cos
dx
x
∫ 
1
5 4 cos
dx
x+∫ 
sin cos
dx
x x±∫ 
cos2
1 sin cos
x
dx
x x+∫ 
sin cos
sin2
x x
dx
x x
+
∫ 
1
tan
2 1 2 1
x dx
x x
 
+ 
+ + − 
∫ 
Tính nguyên hàm bằng phương pháp từng phần 
ðịnh lý : Nếu ,u v là hai hàm số liên tục trên K thì −∫ ∫ ( ) '( ) = ( ) ( ) ( ) '( )u x v x dx u x v x v x u x du 
2
2
ln( 1)
.
1
x x x
I dx
x
+ +
=
+
∫ 
2
2
2 2
ln( 1)
ln( 1) .
1 1
x x x x
I dx x x dx
x x
+ +
= = + +
+ +
∫ ∫ 
Nguyễn Phú Khánh – ðà Lạt  
ðặt :
( )2 2
2 2
2 2
1ln 1
1 .
1 1
1 1
x
u x x
dxxdu dx
x
x x xdv dx
x v x

 += + + +  = =⇒ 
+ + += 
 +  = +
( ) ( )2 2 2 21 ln 1 1 ln 1 .I x x x xdx x x x x C= + + + − = + + + − +∫ 
Tính nguyên hàm của các hàm số sau 
s inx xdx∫ (2 1)sinx xdx+∫ sin
3
x
x dx∫ 
-xxe dx∫ ( 1) xx e dx+∫ 3
3
cos 1
1
x
dx
x
+
+
∫ 
e s inx xdx∫ sin2xe xdx∫ 
lnx xdx∫ 2 lnx xdx∫ 
2 1xe dx+∫ 
2tanx xdx∫ 
2tanx xdx∫ 
2ln( )x x dx+∫ 1ln
1
x
x dx
x
−
+∫ ( )
2
3
21
x dx
x−
∫ 
Tích Phân 
Ví dụ : 
Nguyễn Phú Khánh – ðà Lạt  
( )
3
3 3
4 4
2 2 2
1
2 3
2
dx d x x
x
= = = −∫ ∫ 
7 7 7
33 3
2 16
3 3 ( 3) ( 3) 3
3 3
x dx x d x x x − = − − = − − =  ∫ ∫ 
4 4 4
0
0 0
(25 3 ) 2 2
25 3 ( 13 5)
3 325 3 3 25 3
d xdx
x
x x
− − −
= = − = −
− − −
∫ ∫ 
( )21 1 1
2
2 2 0
0 0
12 1
ln 1 ln 3
1 1
d x xx
dx x x
x x x x
+ ++
= = + + =
+ + + +∫ ∫ 
( )
( )
( )
( )
( )
1 1
1 1 2 2 22 2
2 2
3 3 2
2 2 2
0 0
0 0
1 15 5 5 5 1 35
. .
2 2 2 4 361 1 1
d x xxdx
x x x
−
− −
= − = − = =
−− − −
∫ ∫ 
( )
5 5 5 5 5 5 55
2 2
2 2 2 2 2 2
1 2 1 1
( 1 ) 2 2 1 2 ln 1 3 4 ln 2
1 1 1 1
x
dx dx dx dx dx d x x x
x x x x
+
= − + = − + = − − − = − − − = − −
− − − −∫ ∫ ∫ ∫ ∫ ∫
( ) ( )
( ) ( )
4
4 4 4
2
3 3 3
3
21 2 1 1 4
ln ln
1 2 1 33 2 1 2
xx x dxdx
dx
x x xx x x x
  −− − − −
 = = + = =
 − − −− + − −  
∫ ∫ ∫ 
0
0 0
2
1 1 1
1 1 1 1 3 1 3
ln ln
2 3 1 2 1 2 24 3
dx x
dx
x x xx x− − −
  −
= − = = 
− − −− +  
∫ ∫ 
11 1 1
20 0 0 0
2
ln 2 ln 2 ln 3
2 24
xdx dx dx
x x
x xx
= + = + + − =
+ −−∫ ∫ ∫ 
( ) ( )
11 1 1
20 0 0 0
3 2 1 20 1 1
ln 1 20 ln 6 ln2 20 ln 5 10 ln 6
7 1 6 7 75 6
x dx dx
dx x x
x xx x
 +
= + = + + − = + − 
+ −− −  
∫ ∫ ∫ 
4 4 4
4
4
4 4 4
sin (cos )
tan ln 0
x d x
xdx dx cosx
cos x cosx
π π π
π
π
π π π
−
− − −
−
= = = − =∫ ∫ ∫ 
( )
2 2
2 2
2
4
4 4
1
1 1
4sin
cot xdx dx cotx x
x
π π
π
π
π π
π 
= − = − − = − 
 
∫ ∫ 
( )
23 3
3
2 2 2
4
4 4
3 3 2 11 3
3 tan 2 5
3cos cos sin
cot x
dx dx x cotx
x x x
π π
π
π
π π
 −
= − = + = − 
 
∫ ∫ 
( )
3 3
3
2 2 2 2
6
6 6
1 1 4 3
tan
3sin cos sin cos
dx
dx x cotx
x x x x
π π
π
π
π π
 
= + = − = 
 
∫ ∫ 
2 2 2
3 3 3
2 9
0
0 0
1 1 1
. ( ) (1 )
2 2 2
x x xx e dx e d x e e− − − −
−
= − = = −
−∫ ∫ 
Nguyễn Phú Khánh – ðà Lạt  
32 2
2
1
1 1
( 2) 2
ln(2 ) ln
2 12 2
xx
x
x x
d e e ee dx
e
ee e −− −
+ +
= = + =
++ +∫ ∫ 
2 2
2(ln )
ln ln ln2
ln ln
e e
e
e
e e
dx d x
x
x x x
= = =∫ ∫ 
4 5
4
1 1 1
ln (ln ) 1
(ln ) (ln )
5 5
e
e e
x x
dx x d x
x
= = =∫ ∫ . Ngoài ra , ta có thể ñặt lnt x= 
( ) ( )
1 2
2 1 2 2 2
2 2 1 2 1
1 1 1 5
2 2
x x
x dx x dx x dx x x
−
−
− − − − −
   
+ = − − + + = − − + + =   
   
∫ ∫ ∫ 
2 25 5 5 5
1
5
1 1 1 1
1 2 (1 ) 1 1
4
1 1 1 1
x x x x x
dx dx dx dx x
x x x x
− − − −
−
−
− − − −
− + − − −
= = = = =
− − − −∫ ∫ ∫ ∫ 
1 2
2 1 2 3 3
2 2 2
0 0 1 10
1 (1 ) ( 1) ( ) 2
3 3
x x
x dx x dx x dx x x
 
− = − + − = − + − = 
 
∫ ∫ ∫ 
3 4
4 3 4 3 2 3 2
2 2 2
0 0 3 0 3
49
 6 ( 6) ( 6) 6 6
3 2 3 2 3
x x x x
x x dx x x dx x x dx x x
   
− − = − + + + − − = − + + + − − =   
   
∫ ∫ ∫ 
( )
2
2
0
20 0
2
cos cos cos sin sin 2x dx xdx x dx x x
π
π π π
π
π
π
= + − = − =∫ ∫ ∫ 
2 2 2
0 0 0
1 cos2 2 sin 2 sin sin 4 2xdx x dx xdx xdx
π π π π
π
 
− = = − =  
 
∫ ∫ ∫ ∫ 
2
0 0 0
2
1 cos2
cos cos cos 2
2
x
dx x dx xdx xdx
π
π π π
π
+
= = − =∫ ∫ ∫ ∫ 
( )
0
2
2
sin2
.
2 sin
x
I dx
xπ−
=
+
∫ 
ðặt : sin cost x dt xdx= ⇒ = 
ðổi cận 
0 0
1
2
x t
x t
π
 = ⇒ =


= − ⇒ = −
Nguyễn Phú Khánh – ðà Lạt  
( )
( )
( ) ( )
( )
0 0 0
2 2 2
1 1 1
2 2 1 2
2 2 2 2
22 2 2
ttdt
I dt d t
tt t t− − −
 + −  = = = − +
 ++ + +  
∫ ∫ ∫ 
( )
0
1
2
2 ln 2 2 ln2 2.
2
I t
t
−
 
= + + = − 
+ 
8
2
3 1
dx
x x +
∫ 
ðặt : 
2
2
1
1
x xdx tdt
t x dt dx dx
t xx
= + ⇒ = = ⇒ =
+
ðổi cận 
3 2
8 3
x t
x t
 = ⇒ =

= ⇒ =
( ) 222 2 2
1 1 1
2 1 1111 1
dx tdt tdt dt
dt
t ttt tx x x x
 
= = = = − 
− +−−  + +
( )
338 3
2
2 23 2
1 1 1 1 1 1 1 3
ln 1 ln 1 ln ln .
2 1 1 2 2 1 2 21
dx t
dt t t
t t tx x
   −
− = − − + = =   
− + +   +
∫ ∫ 
1
2
0 1
dx
x x+ +∫ 
2
2 1 31
2 4
x x x
 
+ + = + + 
 
ðặt ( )22
1 3 3 1 3 1 3
tan ; tan 1 tan
2 4 2 2 4 2 2 2cos
x t t dx d t dt t dt
t
π π    
 + = ∈ − ⇒ = − = = +         
ðổi cận 
1
0 tan
63
1 tan 3
3
x t t
x t t
π
π

= ⇒ = ⇒ =

 = ⇒ = ⇒ =

Nguyễn Phú Khánh – ðà Lạt  
Vậy 
( )
( )
2
1 3 3
2
20
6 6
3
1 tan
2 3 32
3 3 91
1 tan
4
t dt
dx
dt
x x
t
π π
π π
π
+
= = =
+ + +
∫ ∫ ∫ 
1
4 2
0
.
4 3
dx
x x+ +∫ 
( ) ( )4 2 2 22 2
1 1 1 1 1
24 3 1 31 3x x x xx x
 
= = − 
+ + + ++ +  
1 1 1
4 2 2 2
0 0 0
1
24 3 1 3
dx dx dx
A B
x x x x
 
= − = +  + + + + 
∫ ∫ ∫ 
1
2
0 1
dx
A
x
=
+∫ 
ðặt : ( ) ( )
2
2
2 2
1 tan
tan , 1 tan
2 2 1 1 tan
t dtdx
x t t dx t dt dt
x t
π π +−
= < < ⇒ = + ⇒ = =
+ +
ðổi cận 
0 0
1
4
x t
x t
π
 = ⇒ =


= ⇒ =
4
4
0
0
4
A dt t
π
π π
= = =∫ 
1
2
0 3
dx
B
x
=
+∫ 
ðặt : 
( ) ( )
2
2
2 2
1 tan1 1
3 tan , 3 1 tan
2 2 1 1 tan3 3
t dtdx
x t t dx t dt dt
x t
π π +−
= < < ⇒ = + ⇒ = =
+ +
ðổi cận 
0 0
1
6
x t
x t
π
 = ⇒ =


= ⇒ =
6 6
0 0
1 1
3 3 6 3
B dt t
π π
π
= = =∫ 
1
4 2
0
1
.
2 44 3 6 3
dx
x x
π π 
= − 
+ +  
∫ 
Nguyễn Phú Khánh – ðà Lạt  
1
0
, 0I x x a dx a= − >∫ 
1a• ≥ khi ñó 
1
1 3 2
0 0
1
( ) .
3 2 2 3
x ax a
I x x a dx
−
= − − = + = −∫ 
0 1a• < < khi ñó 
1
1 3 2 3 2
0 0
( ) ( )
3 2 3 2
a
a
a a
x ax x ax
I x x a dx x x a dx
   −
= − − + − = + + −   
   
∫ ∫ 
3 3 3 3  ... 
3
2
cos2
cos 3 sin
x
dx
x x
π
π −
∫ 
Nguyễn Phú Khánh – ðà Lạt  
3
6
1 sin2 cos2
sin cos
x x
dx
x x
π
π
+ +
+∫ 
2
4 4
0
cos2 .(cos sin ).x x x dx
π
+∫ 
( )
2
10 10 4 4
0
cos sin cos sinx x x x dx
π
+ −∫ 
2
0
sin 7 cos 6
4 sin 3 cos 5
x x
dx
x x
π
+ +
+ +∫ 
3
6
1
sin sin
6
dx
x x
π
π π + 
 
∫ 
Tính tích phân bằng phương pháp từng phần 
Công thức tích phân từng phần : = −∫ ∫( ) '( ) ( ) ( ) ( ) '( )
b b
b
a
a a
u x v x dx u x v x v x u x dx 
 Dạng 1 
β
α
 
 
 
 
 
∫
sin
( )
ax
ax
f x cosax dx
e
 ðặt 
 = =
 
    
   ⇒ = =    
    
    
∫
( ) '( )
sin sin
cos
ax ax
u f x du f x dx
ax ax
dv ax dx v cosax dx
e e
Dạng 2: 
β
α
∫ ( )ln( )f x ax dx ðặt 
 = = 
⇒ =  =  ∫
ln( )
( )
( )
dx
u ax du
x
dv f x dx
v f x dx
Ví dụ : 
2
1
ln
e
A x xdx= ∫ 
1
2
0
xB x e dx= ∫ 
4
4
1
lnI x xdx= ∫ 
2
0
sinJ x xdx
π
= ∫ 
2
1
ln
e
A x xdx= ∫ 
ðặt: 
2 3
ln
3
dx
duu x
x
dv x dx x
v

= = 
⇒ =  =

3 3 3 3 3 3
2 2
1 11 1
1 1 2 1
ln = ln
3 3 3 9 3 9 9
e e
e e
x e x e e e
A x xdx x x dx
− +
= − = − = − =∫ ∫ 
1
2
0
xB x e dx= ∫ 
Nguyễn Phú Khánh – ðà Lạt  
ðặt 
2 2
xx
du xdxu x
v edv e dx
  == 
⇒  ==  
1 1 1
1
2 2
0
0 0 0
2 2 ,x x x xB x e dx x e xe dx e C C xe dx= = − = − =∫ ∫ ∫ 
ðặt 
x x
u x du dx
dv e dx v e
 = = 
⇒ = =  
1
1 1
0 0
0
2 2 2x x xB e xe e dx e e e
 
= − − = − + = −  
 
∫ 
4
4
1
lnI x xdx= ∫ 
ðặt:
4 5
1
ln
5
du dxu x
x
dv x dx x
v

= = 
⇒ =  =

5 5
5 55 5 5
4
1 1 11
1 1 1 3124
. ln . 625 ln 5 625 ln 5 625 ln 5
5 5 5 5 5 25
x x x
I x dx x dx
x
 
= − = − = − = − 
 
∫ ∫ 
2
0
sinJ x xdx
π
= ∫ 
ðặt: 
sin cos
u x du dx
dv xdx v x
 = = 
⇒ = = −  
( )
2 2
2 2
0 0
0 0
cos cos cos sin 1J x x x dx xdx x
π π
π π
= − − − = = =∫ ∫ 
ln2
0
xI xe dx−= ∫ ( )
1
2
0
ln 1J x x dx= +∫ 
ln2
0
xI xe dx−= ∫ 
ðặt: 
x x
u x du dx
dv e dx v e− −
 = = 
⇒ = = −  
( ) ( ) ( )
ln 2 ln2 ln2ln2
ln2
0 0
0 0
1 1 1 1
ln2. ln2 ln2
2 2 2 2
x x x xI xe e dx e e d x e− − − − −= − − − = − − − = − − = −∫ ∫ 
( )
1
2
0
ln 1J x x dx= +∫ 
Nguyễn Phú Khánh – ðà Lạt  
ðặt: ( )
2
2
2
2
ln 1 1
1
2
xdx
duu x
x
dv xdx
v x

= = +  +⇒ 
=  =

( ) ( )
1 21 1 1 131
2 2 2
2 2 2
0
0 0 0 00
11 1 1 1 1
. ln 1 ln 2 ln 2
2 2 2 2 21 1 1
d xx x
J x x dx xdx dx x
x x x
+
 = + − = − + = − +  + + +∫ ∫ ∫ ∫
( )
1
2
0
1 1 1 1 1 1 1
ln2 ln 1 ln2 ln2 ln2
2 2 2 2 2 2 2
J x= − + + = − + = − . 
2
0
(2 1)cosI x xdx
π
= −∫ 1
ln
e
e
J x dx= ∫ 
2
0
(2 1)cosI x xdx
π
= −∫ 
ðặt 
2 1 2
cos sin
u x du dx
dv xdx v x
 = − = 
⇒ = =  
2 2
2 2
0 0
0 0
(2 1)cos =(2 1)sin2 2 sin 1 2 cos 3I x xdx x x xdx x
π π
π π
π π= − − − = − + = −∫ ∫ 
1
ln
e
e
J x dx= ∫ 
Vì 
ln 1;
ln 1
ln ;1
x khi x e
x
x khi x
e
  ∈  
=  
− ∈  
 
( )
1 1
1 1 1 1
2
ln ln ln ln ... 2
e e
e e
J x dx xdx xdx xdx
e
= − + = − = = −∫ ∫ ∫ ∫ 
( )
0
32
1
1xx e x dx
−
+ +∫ 
5 cos2 .x x dx
π
π−
∫ 
Nguyễn Phú Khánh – ðà Lạt  
( )
4
0
1 cosx xdx
π
−∫ 
( )
2
0
1 sin2x xdx
π
+∫ 
2
0
( os )s inx c x xdx
π
+∫ 
4
0
( sin )
1 cos
x x dx
x
π
+
+∫ 
3
0
.sin .x x dx
π
∫ 
4
2
0 cos
x
dx
x
π
∫ 
3
2
4
tanx xdx
π
π
∫ 
3
2
0
sin
cos
x x
dx
x
π
+
∫ 
2
3
4
cos
sin
x xdx
x
π
π
∫ 
3 4
0
tan .
cos2
x dx
x∫ 
4
0
1 cos2
x
dx
x
π
+∫ 
4
0
cos xdx∫ 
2
0
sinx xdx
π
∫ 
2
4
0
cosx xdx
π
∫ 
( )
2
2
0
sin sinx x dx
π
+∫ 
3
6
( cos sin ).x x dx
π
π
−∫ 
4
1
xe dx∫ 
2
cos
0
sin2xe xdx
π
∫ 
2
3
0
sin 5xe xdx
π
∫ 
2
2
0
cosx xdx
π
∫ 
( )
2
2
0
2 1 cosx xdx
π
−∫ 
( )
4
2
0
sinx x dx
π
∫ 
( )
1
2
0
sinxe x dxπ∫ 
( )2
1
2
1
sinx xe x e x dx
−
+∫ 
1
9
3
2 5
0
1
5
sin (2 1) 4 1
x x dx
x x
 
+ +  + − 
∫ 
2
2
0
xxe dx∫ 
( )
1
0
2 xx e dx−∫ 
( )
1
2
0
1 xx e dx+∫ 
2
1
3
0
xx e dx∫ 
( )
1
2
2
0
1 xx e dx+∫ 
( )
1
2 2
0
4 2 1 xx x e dx− −∫ 
0
2
1
2 2.x x dx
−
+ +∫ 
1
2
0
4 x dx−∫ 
2
2
sin 3
0
sin cosxe x xdx
π
∫ 
Nguyễn Phú Khánh – ðà Lạt  
2
1
3
1
.xx e dx
−
∫ 
2
ln 2
5
0
xx e dx∫ 
1
1
( )ln
e
x xdx
x
+∫ 
( )2
1
ln
e
x x dx∫ 
2
1
( ln )
e
x x xdx∫ 
( )
3
2
2
1
ln
1
x x
dx
x +
∫ 
1
cos(ln )
e
x dx
π
∫ 
( )
2
2
1
sin log x dx
π
∫ 
( )
2
0
cos ln 1 cosx x dx
π
+∫ 
( )
3
2
2
ln x x dx−∫ 
1
ln
e
x xdx∫ 
( )
2
1
2 lnx xdx−∫ 
2
1
ln
e
x xdx∫ 
3 2
1
ln
e
x xdx∫ 
( )
1
2
0
ln 1x x dx+∫ 
( )
3
2
0
ln 5x x dx+∫ 
10
2
1
lgx xdx∫ 
2
1
ln
e
x
dx
x
∫ 
( )21
ln
1
e
e
x
dx
x +
∫ 
( )2
2
1
ln 1 x
dx
x
+
∫ 
3
3
1
ln
e
x
dx
x
∫ 
1
2
0
ln( 1)x x dx+∫ 
2
2
1
ln( )x x dx+∫ 
1
2
2
1
2
1
. ln .
1
x
x dx
x
−
 −
 
+ 
∫ 
2
0
2 cos 4x xdx
π
∫ 
( )3
2
6
ln sin
cos
x
dx
x
π
π
∫ 
Nguyễn Phú Khánh – ðà Lạt  
1
2
2
0
1 1
ln
11
x
dx
xx
+
−−∫ 
( )23
2
0
ln 1
1
x x x
dx
x
+ +
+
∫ 
( )
4
3
0
sin cos ln cosx x x dx
π
∫ 
Một số tích phân tổng hợp 
3
2
3
sin
cos
x x
dx
x
π
π−
∫ 
2
2
0
.cos .
1 sin
x x dx
x
π
+∫ 
1 2
2
0 ( 1)
xx e
dx
x +∫ ñặt 
2
2( 1)
xu x e
dx
dv
x
 =


= +
( )
1
0 1 1
nn n
dx
x x+ +
∫ 
3 8
4 3
2 ( 1)
x dx
x −∫ ñặt 
5
3
4 3( 1)
u x
x dx
dv
x
 =


= −
 ( )
2
1
3
0
1 2 x xx e dx−−∫ 
2
0
1 sin
1 cos
xxe dx
x
π
+
+∫ 
( )1 cos4
0
ln 1 sin
1 cos
x
x
dx
x
π +
+
+∫ 
( )1 cos2
0
1 sin
ln
1 cos
x
x
dx
x
π +
+
+∫ 
1 1 1 12 2 2
1 22 2 2 2 2 2 2
0 0 0 0
1
(1 ) (1 ) 1 (1 )
dx x x dx x dx
dx I I
x x x x
+ −
= = − = −
+ + + +∫ ∫ ∫ ∫ 
Tính 
1
1 2
0 1
dx
I
x
=
+∫ bằng phương pháp ñổi biến số 
 Tính
1 2
2 2 2
0 (1 )
x dx
I
x
=
+∫ bằng phương pháp từng phần : ñặt 2 2(1 )
u x
x
dv dx
x
 =

 = +
Nguyễn Phú Khánh – ðà Lạt  
Tìm 0x > sao cho 
1 2
2
0
1
( 1)
xt e
dx
t
=
+∫ 
Dạng ñặc biệt 
( )
1
2
1
ln 2007x x dx
−
+ +∫ 
( )
4
0
ln 1 tanx dx
π
+∫ 
4 42 2
4 4 4 4
0 0
cos . sin .
cos sin cos sin
x dx x dx
x x x x
π π
=
+ +∫ ∫ 
20092
2009 2009
0
sin
sin cos
x
dx
x x
π
+∫ 
20102
2010 2010
0
sin
sin cos
x
dx
x x
π
+
∫ 
2
3 4
2
sin
4 5
x
dx
x− +
∫ 4 3
0
.cos .sinx x xdx
π
∫ 2
0
.sin .cosx x xdx
π
∫ 
2
0
.sin
.
1 cos
x x
dx
x
π
+∫ 20
sin
9 4 cos
x x
dx
x
π
+∫ 0 1 sin
x
dx
x
π
+∫ 
0
sin
2 cos
x x
dx
x
π
+∫ 
2
0
. sin
.
3 cos
x x
dx
x
π
+∫ 
1 2
1
1
.
1 2x
x
dx
−
−
+∫ 
1
2
1 ( 1)( 1)
x
dx
e x− + +
∫ 
7 5 34
2
4
3 7 1
cos
x x x x
dx
x
π
π−
− + − +
∫ 
2
2
2
cos
.
4 sin
x x
dx
x
π
π
−
+
−∫ 
22
2
sin
2 1x
x x
dx
π
π− +
∫ 
2
2
sin .sin2 .cos5
1x
x x x
dx
e
π
π− +
∫ 
6 64
4
sin cos
6 1x
x x
dx
π
π−
+
+∫ 
2sin
.
3 1x
x
dx
π
π− +
∫ 
1 4
1 2 1
x
x
dx
− +
∫ 6 7
8
8
sinx xdx
π
π−
∫ 
1 4
1
.
1 2x
x
dx
− +
∫ 
2
2
2
cos
4 sin
x x
dx
x
π
π−
+
−∫ 
1
4 2
1 1
x dx
x x− − +
∫ 
3 2
3
1
1 2x
x
dx
−
+
+∫ 
2
2
sin sin 3 cos5
1 x
x x x
dx
e
π
π
−
+∫ 
Nguyễn Phú Khánh – ðà Lạt  
2
3
0
7 cos 6 sin
(sin cos )
x x
dx
x x
π
−
+∫ 
2
2
2
cos . ln( 1)x x x dx
π
π−
+ +∫ 
1
2
1
ln( 1 )x x dx
−
+ +∫ 
2
0
( sin cos )x x dx
π
−∫ 
Tìm t ñể 4
0
3
4 sin 0
2
t
x dx
 
− = 
 
∫ 
ỨNG DỤNG TÍCH PHÂN 
Tính diện tích hình phẳng giới hạn bởi các ñường : 
3
0
2; 1
y x
y
x x
 =

=
 = − =
2
3
0
y x
x
y

= + +

 =
2 2
1 1
;
sin cos
;
6 3
y y
x x
x x
π π

= =

 = =

2sin
0
y x
y x x
x π
 =

= +
 ≤ ≤
sin 2 cos
3
0;
y x x
y
x x π
 = −

=
 = =
2 3sin .cos
 y 0
0,
2
y x x
x x
π

 =

=

 = =

2( 1)
sin
0
x y
y x
x
 = +

=
 =
1
x
x
y e
y e
x
−
 =

=
 =
2 2
4
y x
y x
 = +
 = −
2 2
2
y x x
y x
 = +
 = +
2 2
3
y x x
y x
 = − +
 = −
2
2
y x
x y
 =

= −
2
2 2
6
16
y x
x y
 =

+ =
2
2
2 2
3 6
0; 4
y x x
y x x
x x
 = −

= + −
 = =
Nguyễn Phú Khánh – ðà Lạt  
2
2 6
0; 1
x
y
x x
x x

=
 −
 = =
2
2
4
4
4 2
x
y
x
y

 = −

 =

( )
2
2
0
ax y
a
ay x
 =
>
=
2( 1)
sin
y x
x yπ
 = +
 =
2 4 5
2 4
4 11
y x x
y x
y x
 = − +

= − +
 = −
2
2
2
2 1
2
y x
y x x
y
 =

= − −
 =
2
2
6 5
4 3
3 15
y x x
y x x
y x
 = − + −

= − + −
 = −
2
2
2 2
4 5
1
y x x
y x x
y
 = − +

= + +
 =
2
2
2
4 4
8
y x
y x x
y
 =

= − −
 =
2 2
2 2 1 0
0
y x
x y
y
 =

+ + =
 =
4
0;
1
2
; 0
1
x
x
x
y y
x

 =

=


= =
−
2
2
8
8
y x
x
y
y
x

=

=

 =

2
2
27
27
y x
x
y
y
x

=

=

 =

5( 1)
1
x
y x x
y e
x
 = +

=
 =
25
0
0; 3
xy
y
x y x
− =

=
 = = −
Nguyễn Phú Khánh – ðà Lạt  
2 1
2
y x
x
 = −
 =
2 1
2
x y
x
 = −

=
23 2
0
y x x
y
 = − − +
 =
2 5 6
6
y x x
y
 = − +

=
2 4 3
3
y x x
y
 = − +

=
2 1
5
y x
y x
 = −

= +
2 3 2
2
y x x
y
 = − +

=
2 5 6
1
y x x
y x
 = − +

= +
2 4 3
3
y x x
y x
 = − +

= +
2 4 3
3
y x x
y x
 = − +

= −
2
2
1
7
y x
y x
 = −

= − +
2
2
3 2y x x
y x
 = − +

= −
siny x
y x π
 =

= −
lg
0
1
, 10
10
y x
y
x x

 =

=

 = =

2 31 2 sin
2
12
1
0,
2
x
y
x
y
x x
π
π

= −


= +

 = =


1
0
y x
y
x
y
x e
 =

 =

 =
 =
 2 3
2
(4 )
4
y x
y x
 = −

=
2
2 2
2
27 8( 1)
y x
y x
 =

= −
Nguyễn Phú Khánh – ðà Lạt  
Thể tích khối tròn xoay ñược tạo bởi khi quay hình phẳng giới hạn bởi các ñường sau quay 
quanh trục Ox 
2y x
y x
 =

=
2
10 3
y x
y x
 =
 = −
2
2
1
1
2
y
x
x
y

= +
 =

2
2 3
4
(4 )
y x
y x
 =

= −
4 4cos sin
0
;
2
y x x
y
x x
π
π

 = +

=

 = =

tan
0
0;
3
y x
y
x x
π

 =

=

 = =

Thể tích khối tròn xoay ñược tạo bởi khi quay hình phẳng giới hạn bởi các ñường sau quay 
quanh trục 
1. Ox 
2. Oy 
2( 2)
4
y x
y
 = −
 =
22
0
y x x
y
 = −
 =
2 sin (0 )
0
y x x
y
π = ≤ ≤

=
2
1
1
0, 0, 1
y
x
y x x

=
 +
 = = =
4
4
0; 2
y
x
x x

=
−
 = =
2 2, 4
4
y x y x
y
 = =
 =
1
2
0; 0
y x
y
x y
 = −

=
 = =
. ln
0
1;
y x x
y
x x e
 =

=
 = =
2( 0)
3 10
1
y x x
y x
y
 = >

= − +
 =