TỔNG HỢP TÍCH PHÂN

T×m hä nguyªn hµm cña hµm sè gx.b.

Loại 1: Tích phân hàm số đa thức Bài 1: I =

1

3 4 5 0

∫ I =

1

2 3 0

∫ I =

2 1 0

I =1 3

0

xdx(2x 1)+

∫ I =

3 2

2 1

1dx(1 x) x+

2 0

xdx

∫ I =

3 3 2 1

xdx

∫ I =1 3

0

4xdx(x +1)

∫ I =1 3

0

3dx

∫ I=2 2 2

0

1dx(4 x )+

−+

Loại 3: Tích phân hàm số vô tỉ Bài 1: I =

2 1

0

x

dx(x 1) x 1+ +

2 1

x 1dxx

∫ I =

7 3 3 0

x 1

dx3x 1

++

∫

I =2

3 0

x 1

dx3x 2

++

∫ I =

1

3 3

4 3

3

dxx

−

∫ I = 2 2

2 2

1dx

x 1 x+

∫

I =2 3

2 5

1dx

dx

++

∫ I = 7 3

0

xdx

1 x+

∫

I =

2 3 2

2 0

xdx

1 x−

∫ I =

2

2 2

3

1dx

∫ I =

3 2

2 1

2

1dx

3

1dx

∫ I =2

3 1

1dx

x 1 x+

∫ I =

2 1

3 0

I =

1

2 0

1dx

4 x−

∫ I =

3

2 1

1

dx4x x−

2 0

xdx

1 x−

∫ I =

2 1

2 2 2

1 x

dxx

−

∫ I =

2 1

2 0

4 x dx+

∫ I =

3 2 2

1dx

∫ I =

0 2 1

dxx

+

∫

I =

2 1

2 0

2 0

x x

dxx

−

∫ I =

1 2

2 1

2

1

dx(3 2x) 5 12x 4x

1 3 1 2

xdx

π x dx I =∫4

0 6

cos1

π

∫

I =4

3 0

1dxcos x

π

π∫ I = ∫3

4

2 2

2

cos2sin1

π

I =3 2

6 0

sin x

dxcos x

dxsin x cot gx

π π

3 4

∫ I =

0

2 2

sin 2x

dx(2 sin x)

2 0

sin x

dx(sin x 3)

π π

π

π

−

−+

∫ I =2

0

cos x

dxsin x cos x 1

∫ I =2

0

sin 2x sin x

dxcos3x 1

π

++

∫

I =4 3

2 0

sin x

dxcos x

++

π

++

0

sin x.cos x

dxcos x 1

π

+

∫

I =2

6

1 sin 2x cos 2x

dxcos x sin x

2 sin x

π+

0

sin 2x

dxsin x 2cos x

tan x

dxcos x cos x 1

π

3 3 2

3

sin x sin x

cot gx dxsin x

π π

2 tgx

π+

sin xdx

π

∫ I =

1 0

B i 1: à I =1 x

0

1dx

∫ I =2 x

1

1dx

1 e− −

∫ I =2 x2x

0

edx

x 0

1dx

∫ I =1 2x

0

1dx

e

dx(e +1)

e −1

∫ I =

1 4x 2x 2

2x 0

dx

1 e

++

1dx

∫ I =1 2x x

0

1dx

∫

I =

x 2 1

2x 0

(1 e )

dx

1 e

++

∫ I =

1

1 x 3 a

edxx

∫

I = 1 2x

1

1dx

x e

dx(x 2)+

ln x

dxx(ln x 1)+

∫ I =

e 1

sin(ln x)

dxx

∫ I =

e 1

ln x 2 ln x

dxx

+

∫

I =e

1

1 3ln x ln x

dxx

+

2

e e

ln xdxx

∫ I =1 2

2 0

1

.ln xdxx

e

ln x

dx(x 1)+

∫ I =

1 2 0

+

∫ I=

e

1 e

xdx

1 2

−∫ +

B i 4: à I =

2 1

x.sin coscos A-02 I = x(e x x )dx.

x B-03 =∫ −

1

2 x

x

e

dxeI

02 T59 = 4

1 cos 20

x dx x

e

I x B-04 I = ∫3 +

1 3

x x

dx A-04 dx

x

xx

I=∫2 −++

4

41

2 1

π

= +

ln 2 3 1

dx x x

x I

I cos D – 07: I = dx

x

x x

0

)1(

0 3 4 sin cos 2

2 sin

π

dx x x

x I

A1- 08 : = ∫ +

3

2 /

x dx x

x x dx x

ln x

dx x

3 ln ( 1)

x dx x

+ +

+ + +

1

ln (2 ln )

e

x dx

1

3 (2 ) ln

x x dx

x

π

∫+

(C§SP Kon Tum 2004) T 17 = 1

10

dx x e

∫+ +

− (C§ KTKT CÇn Th¬ A2005) T31 =

ln21

e x dx x

x

+

∫+

xdx x

∫+

(C§ SP Sãc Tr¨ng 2005)T 35 = 3 .sin2

2sin 2 cos0

x dx x

(C§ SP Qu¶ng B×nh 2005) T 46 =

21

23

0 ( 1)

dx x

+

∫+ (C§ SP Hµ TÜnh AB2002) T69 =

∫ (C§ KTKT Th¸i B×nh 2002) T75 =3

2 2 10

∫+ +

(C§ SP KT Vinh 2002) T76 = 2 4cos 3sin 1

x

+

∫+ (C§ GTVT 2003) T79 =

ln(1 )1

x dx x

++

∫

(C§ SP T©y Ninh 2003)

a TÝnh tÝch ph©n: T89=

1cos(ln )

t

x x dx

∫

sin cos

π

=∫

IV/ Các đề thi ĐH trước năm 2001

1/ (§H Quèc Gia Hµ Néi & HV Ng©n Hµng A2001- 2002)

2/ (§H Quèc Gia Hµ Néi & HV Ng©n Hµng D2001 - 2002)

T×m hä nguyªn hµm: 93 tan( )cot( )

1 sinln

tr-21/ (§H Y Hµ Néi 2001- 2002) a

3 2 117

∫

0

sinsin 3 cos

xdx I

xdx J

3 2

cos 2cos 3 sin

x

=+

∫

31/ (§H Quèc Gia Hµ Néi (khèiD) HV Ng©n Hµng D2000- 2001)

135

cos cos

4

dx T

x x

J =∫x −x dx, n = 0, 1,2,

b TÝnh I n+1 theo I vµ t×m n lim n 1

I I

f = ∫π f x dx= 51/ (§H LuËt, X©y Dùng Hµ Néi 00- 01)

a TÝnh: 1 3

0

31

dx x

∫ b (Ph©n ban) TÝnh:sin 3

π π

55/ (§H SP Vinh D, G, M00- 01)

2 0

a T×m hä nguyªn hµm cña hµm sè g(x).

b TÝnh tÝch ph©n: 2

2

( )1

x

g x dx e

xdx I

xdx J

−∫ +

67/ (§H §µ L¹t A, B99- 00) a

1

2 ln2

dx x

1 4 6 0

11

x

dx x

++

dx

x + x+

∫ b (CB) TÝnh:1

+

∫ b (CB) TÝnh:1

11

x dx x

++

1

x

dx x

++

x

π π

41

x

dx x

3 2