Chuyên đề: Nguyên hàm-Tích phân (Luyện Thi Đại Học và Cao Đẳng)

Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng 
 Nguyªn hµm - tÝch ph©n vµ c¸c øng dông 
a.tÝnh tÝch ph©n b»ng ®Þnh nghÜa 
Ph−¬ng ph¸p: 
1. §Ó x¸c ®Þnh nguyªn hµm cña hµm sè f(x), Chóng ta cÇn chØ ra ®−îc hµm sè F(x) 
sao cho: F’(x) = f(x). 
• ¸p dông b¶ng c¸c nguyªn hµm c¬ b¶n, c¸c hµm sè s¬ cÊp . 
• Neáu gaëp daïng caên thöùc ñöa veà daïng soá muõ phaân theo coâng thöùc: , ( 0)
n
m n mx x m= ≠ 
• Neáu gaëp daïng ( )nP xx thöïc hieän pheùp chia theo coâng thöùc: 
1, ( ); , ( )
m m
m n
n n n m
x xx m n m n
x x x
−
−= > = < . 
• Coâng thöùc ñoåi bieán soá (loaïi 2): 
Tích phaân daïng: ( )( ) . '( )f g x g x dx∫ Ñaët g(x) = u => g’(x)dx = du 
( ( )) '( ) ( )f g x g x dx f u du=∫ ∫ . 
2. Mét sè d¹ng c¬ b¶n: 
1. Sö dông c«ng thøc c¬ b¶n:
1. Daïng : ñaët u = ax + b ⇒du = adx dx=( ) ( 1, 0)ax b dx aα α+ ≠ ≠∫ ⇒ 1 dua 
( )
( ) 1!1( )
1 ( 1)
ax buax b dx u du C C
a a a
αα
α α
α α
++ ++ = = + = ++ +∫ ∫ 
2. Daïng : ñaët ( ) 1 , ( 0, 1)n nax b x dx aα α−+ ≠∫ ≠
1 1
1 1
1
1. .
1 (( )
( 1) ( 1)
nu=ax n n
n
n n
b du a n x dx x dx du
an
u ax bax b x dx u du C C
an na na
α α
α α
α α
− −
+ +
−
+ ⇒ = ⇒ =
++ = = + =+ +∫ ∫ ) +
3. Daïng: ). cos sin ( 1) a xdxα α ≠ −∫ ( Ñaët 
11cos sin ) cos sin cos
( 1)
u x du xdx x xdx u du x Cα α αα
+−= ⇒ = − ⇒ = − = ++∫ ∫ 
 ). cos ( 1) sin xb xdxα α ≠ −∫ (Ñaët 
11sin cos sin
1
 du=cos xdx sin xu x xdx u du xα α αα
+= ⇒ ⇒ = = ++∫ ∫ C 
4. Daïng: 1 ln ( 0)dx ax b C a
ax b a
= + + ≠+∫ 
Neáu gaëp : ( )P x
ax b+ vôùi baäc : laøm baøi toaùn chia. ( ) 1P x ≥
GV: NguyÔn Thanh S¬n 1
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng 
5. Daïng: 2cos ( )
dx
x a btgx+∫ Ñaët 
2 2
1 1 1; l
cos cos ( )2
dx 
co s
bdx dx duu a btgx du du a btgx C
x x b x a btgx b u b
= + ⇒ = ⇒ = = = + ++∫ ∫ n
2. Coâng thöùc: 
( ) '( )
ln
u
u x u aa u x dx a du C
a
= = +∫ ∫ 
3. Coâng thöùc ñoåi bieán soá (loaïi 1): 
Tích phaân daïng: ( )( ) . '( )f g x g x dx∫ Ñaët g(x) = u => g’(x)dx = du 
( ( )) '( ) ( )f g x g x dx f u du=∫ ∫ 
4. Coâng thöùc : 
2
2
2
1). ln .( 0)
2
). ln
du u aa C a
u a a u a
dub u u k C
u k
α
−= + ≠− +
= + + ++
∫
∫
5. Coâng thöùc : 
2
2 2ln
2 2
x x k kx kdx x x k C++ = + + + +∫ 
3. Mét sè d¹ng th−êng gÆp: 
1. Tích phaân daïng: 2 2 2 21).
(mx+n)dx dx (mx+n)dx 2). 3). 4). dx
ax bx c ax bx c ax bx c ax bx c+ + + + + + +∫ ∫ ∫ ∫ + 
Tuyø vaøo moãi daïng aùp duïng caùc coâng thöùc tính tích phaân chæ trong baûng sau: 
 Töû soá baäc nhaát Töû soá haèng soá 
Maãu soá khoâng caên lndu u C
u
= +∫ 2 2 1 ln2 −= +− +∫ du u a Cu a a u a 
Maãu soá coù caên 2du u C
u
= +∫ 22 ln= + + ++∫
du u u k C
u k
Söû duïng haèng ñaúng thöùc: 
2 2 2
2 2
2
( ) ( )
2 2
2 2
a ax ax x
b bax bx a x
a a
+ = + −
⎡ ⎤⎛ ⎞ ⎛ ⎞+ = + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
GV: NguyÔn Thanh S¬n 2
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng 
4. TÝch ph©n cña c¸c ph©n thøc h÷u tØ:
3 2
ax b A B C
cx dx ex x x m x n
+ = + ++ + − − 
Giaûi daïng naøy ta coù hai caùch: 
− Caùch 1: Ñoàng nhaát hai veá: Cho taát caû caùc heä soá chöùa x cuøng baäc baèng nhau. 
− Caùch 2: Gaùn cho x nhöõng giaù trò baát kyø. Thöôøng thì ta choïn giaù trò ñoù laø 
nghieäm cuûa maãu soá 
5. TÝch ph©n cña c¸c hµm sè l−îng gi¸c: 
1. Daïng: 
cos , , 1). sin , cosn n1 1 sin cosaxdx= sinaxdx=- , 2). co s
a a
nxdx xdx ax C ax C xdx+ +∫ ∫ ∫ ∫ ∫
Phöông phaùp: 
™ n = chaün : haï baëc 
2
2
1 cos 2cos
2
1 cos2sin
2
1sin cos sin 2
2
xx
xx
x x x
+⎧ =⎪⎪ −⎪ =⎨⎪⎪ =⎪⎩
™ n leõ: 
Vieát: 2 1 2 2cos cos cos (1 sin ) cosp p pxdx x xdx x dx+ = = − 
Ñaët sin cosu x du x= ⇒ = dx
2. Daïng: sin cosm nu ud∫ u
u
a. m,n cung chaün: haï baäc. 
b. m,n leû (moät trong hai soá leû hay caû hai cuøng leû). 
™ Neáu m leû: Ta vieát: thay 1sin sin sinm mu u−=
1
2 2 2 2sin 1 cos (1 cos ) sinm va sin
m
u u u u
−
= − = − u 
™ Neáu m, n leû: laøm nhö treân cho soá muõ naøo beù 
3. Daïng: hay ntg xdx∫ cot ng xdx∫ 
Chuù yù: 2 22( ) (1 ) (1 )cos 2
dx 
co s
dxd tgx tg x dx tg x dx tgx C
x x
= = + ⇒ = + = +∫ ∫ 
™ Töông töï: 
2 2
2(cot ) (1 ) (1 )sin 2
dx 
sin
dxd gx cotg x dx cotg x dx cotgx C
x x
= − = − + ⇒ = + = − +∫ ∫
™ Ngoaïi tröø: sin ln cos
cos
(u=cosx)xdxtgxdx x C
x
= = +∫ ∫ 
Ñeå tính: ntg xdx∫
Phöông phaùp: 
Laøm löôïng 2( 1)tg x + xuaát hieän baèng caùch vieát: 
GV: NguyÔn Thanh S¬n 3
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng 
2 2 2 2 2 4 2 1 2* ( 1) ( 1) ... ... ( 1) ( 1) ( 1) 1 n n n ntg x tg x tg x tg tg x tg x− − −= + − + + + + − + + n− 
2 1 2 3 2 2 5 2 2 2 1* ( 1) ( 1) ... ... ( 1) ( 1) ( 1) n n n n ntg x tg x tg x tg tg x tgx tg x tgx− − − − −= + − + + + + − + + −
4. Daïng: hay 2( 1)tg x dx+∫ 2cos ndx x∫ 
Ta vieát: 2 2 1 2( 1) ( 1) ( 1)ntg x dx tg x tg x dx−+ = + +∫ ∫ 
Ñaët u = tgx 2 2( 1) ( 1)2 n (tg x+1) dx ndu tg x dx u du−= + ⇒ = + 1∫ ∫ 
Chuù yù: 2 22
1 1 (1
cos 2n
dx, 
co s
ntg x tg x dx)
x x
= + = +∫ ∫ 
5. Daïng: 
cos
m
n
cotg x, or 
sin x
m
n
tg x dx dx
x∫ ∫ 
Phöông phaùp: 
™ Neáu n chaün : Thay 
22 2 2 21 (1 ) ; (1 ) (1 ) ( 1)
cos cos
mtg 
n n n
m m
n n
xdxtg x tg x tgx dx tg x tgx tgx dx
x x
−= + ⇒ = + = + +∫ ∫ ∫
Ñaët: 
22 2(1 )
m
2
n
tg x du=(1+tg x)dx 
cos x
n
mu tgx dx u u du
−= ⇒ ⇒ = +∫ ∫ 
™ Neáu m leû vaø n leû : 
1
1 .cos cos cos
m
n
tgx tg x tgx
x x x
−
−= Ñaët 1cos
tgx du=
cosx
u dx
x
= ⇒ 
Thay: 
1 1
2 12 2
2 2 1
1 1 11 ( 1) . . ( 1)
cos cos cos cosn
tgmx 
cos x
m m
n
n
tgxtgx dx dx u u du
x x x x
− −
−
−= − ⇒ = − = −∫ ∫ ∫
6. Daïng: sin cos ; sinmxsinnxdx ; cosmxcosnxdxmx nxdx∫ ∫ ∫
Aùp duïng caùc coâng thöùc bieán ñoåi: 
[ ]
[ ]
[ ]
sin( ) sin( )
cos( ) s( )
cos( ) cos( )
1 sinmxcosnx=
2
1 sinmxsinnx=
2
1 cosmxcosnx=
2
m n x m n x
m n x co m n x
m n x m n x
• + +
• − −
• − +
−
+
+
GV: NguyÔn Thanh S¬n 4
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng 
I. TÝnh c¸c tÝch ph©n bÊt ®Þnh. 
Bµi 1: Dïng c¸c c«ng thøc c¬ b¶n tÝnh c¸c tÝch ph©n sau: 
1/ 2
1(3x 2x )dx
x
+ −∫ 2/ 2x 3 dxx−∫ 
3/ 4
32( x )dx
x
−∫ 4/ 3 4 1(3 x 4 x )dxx− +∫ 
5/ 
x
x
3 2
ee (2 )dx
3 x
−
−∫ 6/ x 2x 3x2 .3 4 dx∫
7/ cos (1 t )x gx dx+∫ 8/ 22(4sin x )dxcos x−∫ 
9/ 2
x2cos dx
2∫ 10/ 2 2dxcos x sin x∫ 
Bµi 2: TÝnh c¸c tÝch ph©n sau ®©y: 
1/ 2/ 10x(x 1) dx−∫ 21 2( )x 1 (x 1)−+ +∫ dx 
3/ 2x x 9dx+∫ 4/ 2 24 8x dx(x 1)+∫ 
5/ 
3. xe dx
x∫ 6/ ∫ xx dx2ln 
7/ 8/ sin 7x.cos3x.dx∫ 4cos xdx∫
9/ 3
sin x dx
cos x∫ 10/ 2 2cos2x dxsin x.cos x∫ 
II: TÝnh c¸c tÝch ph©n x¸c ®Þnh sau: 
Ph−¬ng ph¸p: 
( ) ( ) ( ) ( )
b
a
b
a
f x dx F x F b F a= = −∫ . 
1. C¸c ph−¬ng ph¸p tÝnh tÝch ph©n. 
• ¸p dông b¶ng c¸c nguyªn hµm c¬ b¶n, c¸c hµm sè s¬ cÊp . 
• TÝnh tÝch ph©n b»ng ph−¬ng ph¸p ph©n tÝch. 
• TÝnh tÝch ph©n b»ng ph−¬ng ph¸p ®æi biÕn d¹ng I. 
• TÝnh tÝch ph©n b»ng ph−¬ng ph¸p ®æi biÕn d¹ng II. 
• TÝnh tÝch ph©n b»ng ph−¬ng ph¸p ®æi biÕn d¹ng III. 
• TÝnh tÝch ph©n b»ng ph−¬ng ph¸p tÝch ph©n tõng phÇn. 
• TÝnh tÝch ph©n b»ng ph−¬ng ph¸p sö dông nguyªn hµm phô. 
• Mét sè thñ thuËt ®æi biÕn kh¸c, tÝch ph©n chøa biÓu thøc gi¸ trÞ tuyÖt ®èi... 
2. Chøng minh bÊt ®¼ng thøc tÝch ph©n 
GV: NguyÔn Thanh S¬n 5
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng 
§Ó chøng minh bÊt ®¼ng thøc tÝch ph©n , ta th−êng sö dông chñ yÕu 4 tÝnh chÊt 
sau: víi c¸c hµm sè f(x), g(x) liªn tôc trªn [a;b] ta cã: 
1. NÕu [ ]( ) 0, ;f x x a≥ ∀ ∈ b th× ( ) 0b
a
f x dx ≥∫
2. NÕu [ ]( ) ( ), ;f x g x x a b≥ ∀ ∈ th× ( ) ( )b b
a a
f x dx g x dx≥∫ ∫ 
DÊu ®¼ng thøc chi x¶y ra khi f(x) = g(x), [ ];x a b∀ ∈ 
3. NÕu [ ]( ) , ;m f x M x a b≤ ≤ ∀ ∈ th× 
( ) ( ) ( )
b
a
m b a f x dx M b a− ≤ ≤∫ − 
 4. ( ) ( ) .
b b
a a
f x dx f x dx≤∫ ∫ 
 Bµi 1: TÝnh c¸c tÝch ph©n x¸c ®Þnh sau: 
1/ 2/ 
2
2 3 4
0
(3x 2x 4x )dx− +∫
1
3 2
1
( x 3x) dx
−
− +∫
3/ 
4 x
4
0
(3x e )dx−∫ 4/ 
2 2
3
1
x 2x dx
x
−∫ 
5/ 
0 2
1
x x 5 dx
x 3−
− −
−∫ 6/ 
5
2
dx
x 1 x 2− + −∫ 
7/ 
1 2x
x
0
e 4 dx
e 2
−
+∫ 8/ 
32
0
4sin x dx
1 cosx
π
+∫ 
9/ 
3
0
sin x.cos3xdx
π
∫ 10/ 24 2
6
2tg x 5 dx
sin x
π
π
+∫ 
11/ 
2
0
cos2x dx
sin x cosx
π
−∫ 12/ 
4
2
0
sin ( x)dx
4
π
π −∫ 
Bµi 2: TÝnh c¸c tÝch ph©n cã chøa trÞ tuyÖt ®èi sau: 
1/ 
2
2
x 1 dx
−
−∫ 2/ 4 2
1
x 6x 9d− +∫ x 
3/ 
4
2
1
x 3x 2 d
−
− +∫ x 4/ 
1
x
1
e 1 d
−
−∫ x 
GV: NguyÔn Thanh S¬n 6
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng 
5/ 
3
3
(3 x )dx
−
+∫ 6/ 
0
2
2
x x 1 dx
−
+∫ 
7/ 
0
cosx dx
π∫ 8/ 
3
4
4
cos2x 1dx
π
π
+∫ 
9/ 
0
cosx sinxdx
π∫ 10/ 3 x
0
2 4 d−∫ x 
Bµi 3: Chøng minh c¸c B§T sau: 
1/ 
3
0
3 x 1dx≤ + ≤∫ 6 2/ 
1 2
0
4 51
2 2
x dx+≤ ≤∫ 
3/ 
2
2
0
dx1 2
x 1
≤ ≤+∫ 4/ 
2
2
4
53 sin xdx
2 4
π
π
π π≤ + ≤∫ 
5/ 
3
4
2
4
dx
4 3 2sin x
π
π
π π≤ ≤−∫ 2 6/ 
2
2
0
3 tg x 3dx
4 2
π
π π≤ + ≤∫ 
7/ 
2
2
sin x 2
0
e dx e
2
π
ππ ≤ ≤∫ 8/ 2 2x 1 2x
1 1
e dx e dx+ ≤∫ ∫ 
9/ 
2 2
3 2
0 0
sin xdx sin xdx
π π
≤∫ ∫ 10/ 2 2
0 0
sin 2xdx 2 sin xdx
π π
≤∫ ∫ 
B: Ph−¬ng ph¸p ®æi biÕn: 
 Ph−¬ng ph¸p: 
1. Daïng: 
1 1
( , )n mR x x dx∫ Ñaët 1 mn mn-1 x=t dx=mnt dtmnt x= ⇒ ⇒ 
2. Daïng: 
1 1
( ) , ( )n mR ax b ax b dx
⎡ ⎤+ +⎢ ⎥⎣ ⎦∫ 
Ñaët 
1
mn mn-1mn mnt=(ax+b) ax+b=t dx= t dt 
a
⇒ ⇒ 
3. Daïng : dxR(lnx)
x∫ ñaët ln dx du = xu x= ⇒ ( )dx R(lnx) x R u du⇒ =∫ ∫ 
GV: NguyÔn Thanh S¬n 7
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng 
4. Daïng: ñaët xR(e )dx∫
( ) duR u
u
⇒ ⇒ ⇒ =∫ ∫x x xduu=e du=e dx dx= R(e )dxu 
5. Daïng : 2( , )R x ax bx c dx+ +∫ 
Ñöa tam thöùc 2ax bx c+ + veà daïng: hay. 2 2 2u +m ,u -m2 2 2m -u
Ñoåi tích phaân thaønh 1 trong caùc daïng sau: 
.2 2
2 2
2 2
1). R(u, m -u )du
2). R(u, m +u )du.
3). R(u, m -u )du.
∫
∫
∫
Neáu döôùi daáu tích phaân coù chöùa 
2 2 m -u• ñaët 2 2u=msint m -u =mcost⇒ 
2 2m +u• ñaët 2 2 mu=mtgt m +u =
cost
⇒ 
2 2 u -m• ñaët 2 2mu= u -m =mtgt
cost
⇒ 
6. Daïng :
2( )
dx
mx n ax bx c+ +∫ + Gaëp tích phaân naøy ñaët:
1t=
mx+n
Bµi 1: TÝnh c¸c tÝch ph©n sau b»ng ph−¬ng ph¸p ®æi biÕn lo¹i I 
1/ 
1
2
0
2x dx
1 x+∫ 2/ 
4
2
0
x x 9dx+∫ 
3/ 
10
2
dx
5x 1−∫ 4/ 
1
0
x 1 xdx−∫ 
5/ 
5
0
x. x 4dx+∫ 6/ 7 3
0
x dx
x 1+∫ 
7/ 
5
3 2
0
x . x 4dx+∫ 8/ 
2 2
3 3
0
3x dx
1 x+∫ 
9/ 
2
x
1
dx
1 e−−∫ 10/ 
4
x
1
dx
x.e∫ 
11/ 
tgx 24
2
0
e dx
cos x
π
+∫ 12/ e
1
1 3ln x dx
x
+∫ 
GV: NguyÔn Thanh S¬n 8
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng 
13/ 
e 2
1
1 ln x dx
x
+∫ 14/ 6
0
1 4sin x.cos xdx
π
+∫ 
15/ 
4
2
6
1cot gx(1 )dx
sin x
π
π
+∫ 16/ 2 2
0
cos x.sin 2xdx
π
∫ 
17/ 
/ 6
2 2
0
sin 2x dx
2sin x cos x
π
+∫ 18/ 
/ 2 3
2
0
cos x.sin x dx
1 sin x
π
+∫ 
19/ 
8
2
3
1 dx
x x 1+∫ 20/ 
/ 3
3
0
cos x.sin x.dx
π∫ 
Bµi 2 : TÝnh tÝch ph©n b»ng ph−¬ng ph¸p ®æi biÕn lo¹i II: 
1/ 
0
2
1
1 x dx
−
−∫ 2/ 
3
2
2 3
0
1 dx
(1 x )−∫ 
3/ 
2
2 2
1
x 4 x dx−∫ 4/ 1 2
5
dx
x 4x− 7+ +∫ 
5/ 
2
2
0 4
dx
x +∫ 6/ 
4 / 3 2
3
2
x 4 dx
x
−∫ 
7/ 
1
2
2
dx
x x 1
−
− −∫ 8/ 
6
2
2 3
dx
x x 9−∫ 
9/ 
6
2
1
dx
x x 1− + +∫ 10/ 
3 2
2
1
9 3x dx
x
+∫ 
11/ 
1/ 2
1
1 xdx
1 x−
+
−∫ 12/ 
2
2
x 2dx
x 1
+
−∫ 
13/ 
1
2 2
0
dx
(x 1)(x 2)+ +∫ 14/ 
3
2
0
dx
x 3+∫ 
 Bµi 3 : TÝnh tÝch ph©n c¸c hµm sè höu tØ: 
1/ 
2
1
dx
x(2x 1)+∫ 2/ 
2
2
1
dx
x 6x 9− +∫ 
3/ 
2
1
6x 7 dx
x
+∫ 4/ 1 4 2
0
x dx
x x 1+ +∫ 
GV: NguyÔn Thanh S¬n 9
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng 
5/ 
4
2
3
x 1 dx
x 3x 2
+
− +∫ 6/ 
1
2
0
xdx
(x 1)+∫ 
7/ 
6
2 2
0
sin 2xdx
2sin x cos x
π
+∫ 8/ 
3
2
6
cosx dx
sin x 5sin x 6
π
π − +∫ 
9/ 
2
0
dx
(x 1)(x 2)+ +∫ 10/ 
3 2
2
1
9 3x dx
x
+∫ 
11/ 
1/ 2
2
0
dx
4x 4x 3− −∫ 12/ 
4 3 2
4
2
(x x x 1)dx
x 1
+ − +
−∫ 
13/ 
2
0
dx
(x 1)(x 2)+ +∫ 14/ 
2001
2 2001
x dx
(x 1)+∫ 
15/ 
1/ 2
4 2
0
dx
x 2x− +∫ 1 16/ 
1
3
0
3dx
1 x+∫ 
c: Ph−¬ng ph¸p tÝch ph©n tõng phÇn: 
 Coâng thöùc: . . .
b b
b
a
a a
u dv u v v du= −∫ ∫ 
• Coâng thöùc cho pheùp thay moät tích phaân udv∫ phöùc taïp baèng 1 
tích phaân ñôn giaûn hôn. vdu∫
• Coâng thöùc duøng khi haøm soá döôùi daáu tích phaân coù daïng: 
− Daïng tích soá: 
− Haøm soá logaric. 
− Haøm soá löôïng giaùc. 
* Daïng vôùi f(x) laø haøm nx f(x) , ln ,sin ,cos .xe x x x
• Khi tính choïn: 
− Haøm soá phöùc taïp ñaët baèng u. 
− Haøm soá cos tích phaân ñöôïc cho trong baûng tích phaân thöôøng 
duøng laøm dv
Bµi 1: Dïng ph−¬ng ph¸p tÝch ph©n tõng phÇn h·y tÝnh: 
GV: NguyÔn Thanh S¬n 10
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng 
1/ 2/ 
0
xsin xdx
π∫ 1 2 2x
0
(x 1) e dx+∫
 3/ 
4
2
6
x sin 2xdx
π
π
∫ 4/ e 2
1
(x ln x) dx∫
5/ 
4
2
0
x(2cos x 1)dx
π
−∫ 6/ 3 2
4
xdx
sin x
π
π
∫ 
7/ 
e
2
1/ e
ln x dx
(x 1)+∫ 8/ 
4
x
1
e dx∫ 
9/ 
2
4
0
x cos xdx
π
∫ 10/ 3 2
0
ln(x x 1)dx+ +∫ 
11/ 12/ 
1
2 2x
0
(x 1) .e dx+∫ 2 2
0
(x 1).sin x.dx
π
+∫ 
13/ 
2
2
1
ln(1 x) dx
x
+∫ 14/ 4
0
x.sin x.cos x.dx
π
∫ 
Bµi 2: TÝnh c¸c tÝch ph©n sau: 
1/ 
e
2
1
ln x dx
x∫ 2/ 
2e
1
x ln xdx∫ 
 3/ 
2e
1
ln x dx
x
⎛ ⎞⎜ ⎟⎝ ⎠∫ 4/ 
e
2
1
ln xdx∫
 5/ 6/ 
e
2
1
(x ln x) dx∫ 2 x
0
e (x sin x)dx
π
+∫ 
7/ 8/ x 2
0
e sin ( x)dx
π
π∫ x
0
xe sin dx
2
π∫ 
 9/ 
x(1 sin x)e dx
1 cos x
+
+∫ 10/ 
2 2 2
2
3
1 x dx
x
+∫ 
 D: øng dông h×nh häc cña tÝch ph©n 
GV: NguyÔn Thanh S¬n 11
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng 
Bµi 1: TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi c¸c ®−êng (P): y = x2 - 2x + 2 ;tiÕp tuyÕn (d) 
cña nã t¹i ®iÓm M(3;5) vµ Oy. 
Bµi 2: TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi c¸c ®−êng (P): y = x2 + 2x vµ ®−êng th¼ng (d): 
y = x + 2. 
 Bµi 3: Cho hµm sè y = 
23x 5x 5
x 1
− +
− (C) . TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi (C) ; tiÖm 
cËn cña nã vµ x = 2 ; x= 3. 
Bµi 4: Cho hµm sè y = ( )( )2x 1 x 2+ − (C) . TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi (C) vµ 
®−êng th¼ng : x - y + 1 = 0. 
Bµi 5: Cho hµm sè y = 
4
2x 3x
2 2
− − (C) . TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi (C) vµ trôc 
hoµnh. 
Bµi 6: TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi c¸c ®−êng (P): y2 = 4x vµ ®−êng th¼ng d : 4x 
- 3y - 4 = 0 . 
Bµi 7: TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi c¸c ®−êng (P): y2 + x - 5 = 0 vµ ®−êng th¼ng d 
: x + y - 3 = 0 . 
 Bµi 8: TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi c¸c ®−êng y = 0 ; y = tgx ; y = cotgx 
. (0 x )≤ ≤ π
Bµi 9: TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi c¸c ®−êng (C): x2 + y2 = 8 vµ ®−êng (P): y2 = 
2x . 
Bµi 10: TÝnh thÓ tÝch h×nh trßn xoay do h×nh ph¼ng giíi h¹n bëi c¸c ®−êng : y = 
4
x
 vµ y = -x + 5 quay quanh Ox. 
Bµi 11: Cho hµm sè y = 
2x 3x
x 2
3+ +
+ (C) . Gäi (H) lµ phÇn h×nh ph¼ng giíi h¹n bëi (C) 
trôc Ox vµ hai ®−êng th¼ng x = -1 , x = 0. TÝnh thÓ tÝch khèi trßn xoay t¹o thµnh khi (H) 
quay mét vßng xung quanh Ox. 
 Bµi 12: Cho hµm sè y = 
2x x
x 1
1+ +
+ (C) . Gäi (H) lµ phÇn h×nh ph¼ng giíi h¹n bëi (C) trôc 
Ox vµ hai ®−êng th¼ng x = 0, x = 1. TÝnh thÓ tÝch khèi trßn xoay t¹o thµnh khi (H) quay 
mét vßng xung quanh Ox. 
 Bµi 13: TÝnh thÓ tÝch vËt thÓ trßn xoay ®−îc t¹o thµnh do h×nh ph¼ng (D) giíi h¹n bëi : 
y = x , y = 2 - x vµ y = 0 khi ta quay quanh (D) quanh Oy. 
Bµi 14: TÝnh thÓ tÝch vËt thÓ trßn xoay ®−îc t¹o thµnh do h×nh ph¼ng (D) giíi h¹n bëi : 
GV: NguyÔn Thanh S¬n 12
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng 
y = , x = 1 vµ y = 0 ( xxe 0 x 1≤ ≤ ) khi ta quay quanh (D) quanh Ox. 
 Bµi 15: TÝnh thÓ tÝch vËt thÓ trßn xoay ®−îc t¹o thµnh do h×nh ph¼ng (D) giíi h¹n bëi : y = 
sinx , y = cosx , x = 
2
π
 vµ (0 x )
2
π≤ ≤ khi ta quay quanh (D) quanh Ox. 
Bµi 16: TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi c¸c ®−êng sau: 
1/ vµ x = -1; x = 2. 2y 0; y x 2x= = −
 2/ 2y x 4x 3= − + vµ y x 3= + 
 3/ 
2xy 4
4
= − vµ 
2xy
4 2
= 
 4/ 
ln xy ;y 0;x 1
2 x
= = = x e vµ = . 
 5/ 2y x x 1;Ox= + vµ x 1= . 
E. D¹ng th−êng gÆp trong c¸c k× thi §H-C§ 
 Bµi 1: TÝnh c¸c tÝch ph©n sau: 
 1/ 
1 3
2
0 1
x dx
x +∫ 2/ 
ln3
3
0 ( 1)
x
x
e dx
e +∫ 
 3/ 
0
2 3
1
( x 1)x e x
−
+ +∫ dx 4/ 2 6 3 5
0
1 cos .sin .cos .x x x d
π
−∫ x 
 5/ 
2 3
2
5 4
dx
x x +∫ 6/ 
1
3 2
0
1x x dx−∫ 
 7/ 
24
0
1 2sin
1 2sin 2
x dx
x
π
−
+∫ 8/ 
ln5 2
ln 2 1
x
x
e dx
e −∫ 
 9/ 
ln5
ln 2
( 1).
1
x x
x
e e dx
e
+
−∫ 10/ − + −∫
2
2 2
0
(3x 1) x 3x 4 dx 
 Bµi 2: Cho hµm sè: f(x) = 3 .( 1)
xa bx e
x
++ 
 T×m a, b biÕt f’(0)=-22 vµ 
1
0
( ) 5f x dx =∫ 
 Bµi 3: TÝnh c¸c tÝch ph©n sau: 
GV: NguyÔn Thanh S¬n 13
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng 
 1/ 
2
2
0
x x dx−∫ 2/ 21 3
0
. xx e dx∫ 
 3/ 
2
1
1 ln .
e x x dx
x
+∫ 4/ 3 1(cos )1x dxx x+ + −∫ 
 5/ 
1 2
0 ( 1) 1
x dx
x x+ +∫ 6/ 
2
0
sin .sin 2 .sin 3 .x x x d
π
∫ x 
 7/ 
2
4 4
0
cos 2 (sin cos )x x x
π
+∫ dx 8/ 2 5
0
cos .x dx
π
∫ 
 9/ 
+
+∫
3 5 3
2
0
x 2x
dx
x 1
 10/ 
1
2 3
0
(1 x ) dx−∫ 
 Bµi 3: TÝnh c¸c tÝch ph©n sau: 
 1/ 
2
3 3
0
( cos sin )x x dx
π
−∫ 2/ 
3 7
8 4
2 1 2
x dx
x x+ −∫ 
 3/ 2 2
1
ln
e
x xdx∫ 4/ 3
1
lne xdx
x∫ 
 5/ 
2
0
4cos 3sin 1
4sin 3cos 5
x x dx
x x
π
− +
+ +∫ 6/ 
9
3
1
1x xdx−∫ 
 7/ 
2
3
0
1
3 2
x dx
x
+
+∫ 8/ 
1
2
0
( 2 ) xx x e dx−+∫ 
 9/ 
π
+∫ 46
0
1 tg x
dx
cos2x
 10/ 
−
−
+ + +∫
3
1
x 3
dx
3 x 1 x 3
 Bµi 4: TÝnh c¸c tÝch ph©n sau: 
 1/ 
2
0 2 2
xdx
x x+ + −∫ 2/ 
2
1 2 1
dx
x x +∫ 
 3/ 
1
2
0
ln(1 )
1
x dx
x
+
+∫ 4/ 
2
0
sin
sin cos
x dx
x x
π
+∫ 
 5 
0
.sinx xdx
π∫ 6/ 2 2 3
0
sin .cos .x x dx
π
∫ 
GV: NguyÔn Thanh S¬n 14
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng 
 7/ 
1
1 3ln .lne x x dx
x
+∫ 8/ 
3
3 2
0
1x x dx+∫ 
 9/ 
− +
+∫
2 4
2
0
x x 1
dx
x 4
 10/ + −∫
3 7
8 4
2
x
dx
1 x 2x
 Bµi 5: TÝnh c¸c tÝch ph©n sau: 
 1/ 
3 5 3
2
0
2
1
x x dx
x
+
+∫ 2/ 
3 3
0
1 ln .x x dx
x
+∫ 
 3/ 
1
2
0
( 1) xx e dx+∫ 4/ 3 2
4
cos 1 cos
tgx dx
x x
π
π +∫ 
 5/ 
22
1
1
2
x dx
x−
−⎛ ⎞⎜ ⎟+⎝ ⎠∫ 6 20
sin
1 cos
x x dx
x
π
+∫ 
 7/ 
1
0 1
x
dx
e+∫ 8/ 
4
2
0
.x tg xdx
π
∫ 
 9/ 
π
+∫2 4 4
0
cos2x(sin x cos x)dx 10/ 
π
⎛ ⎞+⎜ ⎟⎝ ⎠∫
4
0
x
1 tgxtg sin xdx
2
 Bµi 6: TÝnh c¸c tÝch ph©n sau: 
 1/ 
5
3
( 2 2 )x x d
−
+ − −∫ x 2/ 2 2 2
0
.
( 2)
xx e dx
x +∫ 
 3/ 
4
1
2
5 4
dx
x− + +∫ 4/ 
1
2 2
0
(4 2 1). xx x e d− −∫ x 
 5/ 
2
2 2
0
4x x dx−∫ 6/ 1 2
0 2 5
dx
x x 2+ +∫ 
 7/ 
2
0
sin 2
cos 1
x dx
x
π
+∫ 8/ 
1
2
0 ( 1)
x dx
x +∫ 
 9/ 
π
+∫4 sin x
0
(tgx e cosx)dx 10/ 
π
+∫
2
2 20
sin x
dx
x
sin x 2cosx.cos
2
 Bµi 7: TÝnh c¸c tÝch ph©n sau: 
GV: NguyÔn Thanh S¬n 15
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng 
 1/ 
20042
2004 2004
0
sin
sin cos
x dx
x x
π
+∫ 2/ 
32
0
4sin
1 cos
x dx
x
π
+∫ 
 3/ 
2
0
sin 2 .cos
1 cos
x xdx
x
π
+∫ 4/ 
2
0
sin 2 sin
1 3cos
x x dx
x
π
+
+∫ 
 5/ 
2
sin
0
( cos ) cos .xe x x
π
+ dx∫ 6/ 3 2
6
cos
sin 5sin 6
x dx
x x
π
π − +∫ 
 7/ 
2
2
1
xdx
x x+ −∫ 8/ 
2
0
co x dxs
7 cos 2x
π
+∫ 
 9/ (
−
+ + )0 2x 3
1
e x 1 dx∫ x 10/ 
π
∫ 23 2
0
xsin x
dx
sin2xcos x
 Bµi 8: TÝnh c¸c tÝch ph©n sau. 
 1/ 
1
2004
1
sin .x x dx
−
∫ 2/ 2
0
.sin .cos .x x x∫ dxπ
 3/ 
2
3
0
.cos .x x dx
π∫ 4/ 42 4 4
0
cos x
cos sinx x
π
+∫ 
 5/ 
3
2
0
sin
cos
x xdx
x
π
+∫ 6/ 1 2
0
.x tg xdx∫ 
 7/ CM: 
02
0
2
sin sinx xdx dx
x x
π
π
>∫ 8/ CM: ∫ 4 4
0
2
sin cos
dx
x x
π
π π+∫< < 
 9/ 
π
∫ e 10/ 2 3x
0
sin5xdx
π
∫ x c x 
2
4
0
os dx
 Chóc c¸c em lµm bµi tèt ! 
GV: NguyÔn Thanh S¬n 16