Toan 1 - Chuong 8

Nguy¶n h m v t½ch ph¥n b§t �ànhHai ph÷ìng ph¡p t½nh t½ch ph¥n

8.3.Ph²p t½nh nguy¶n h m cõa mët sè h m sè

B i gi£ng: TO�N CAO C�P 1Ch÷ìng VIII: PH�P T�NH NGUY�N H�M

� m Thanh Ph÷ìng, Ngæ M¤nh T÷ðng

Ng y 5 th¡ng 10 n«m 2010

� m Thanh Ph÷ìng, Ngæ M¤nh T÷ðng B i gi£ng: TO�N CAO C�P 1 Ch÷ìng VIII: PH�P T�NH NGUY�N H�M



NËI DUNG CH�NH

Nguy¶n h m v t½ch ph¥n b§t �ành

C¡c ph÷ìng ph¡p t½nh t½ch ph¥n

Ph²p t½nh nguy¶n h m mët sè h m sè




NËI DUNG CH�NH







NËI DUNG CH�NH







Nguy¶n h m cõa h m sè mët bi¸nT½ch ph¥n b§t �ànhB£ng t½ch ph¥n cõa c¡c h m sè thæng döng.

Nguy¶n h m cõa h m sè mët bi¸n

�ành ngh¾a

Cho h m sè y = f (x) x¡c �ành trong kho£ng mð (a; b)H m sè y = F (x) x¡c �ành trong (a; b) �÷ñc gåi l nguy¶n h m cõa

h m sè y = f (x) n¸u y = F (x) câ �¤o h m t¤i måi �iºm thuëc (a; b) v F ′(x) = f (x) hay dF (x) = f (x)dx vîi måi x ∈ (a; b).

Hai nguy¶n h m kh¡c nhau cõa còng mët h m sè ch¿ sai kh¡c nhaumët h¬ng sè.

H m sè y = F (x) l nguy¶n h m cõa h m sè y = f (x) tr¶n kho£ng�âng [a; b] n¸u:

F (x) l nguy¶n h m cõa f (x) tr¶n (a; b) v

F′(a + 0) = f (a);F ′(b − 0) = f (b).

V½ dö. Nguy¶n h m cõa h m sè f (x) = x2 l h m sè F (x) =x3

3. Nguy¶n h m cõa h m sè f (x) = sinx l h m sè

F (x) = −cosx






�ành ngh¾a






F′(a + 0) = f (a);F ′(b − 0) = f (b).



F (x) = −cosx






�ành ngh¾a






F′(a + 0) = f (a);F ′(b − 0) = f (b).



F (x) = −cosx






�ành ngh¾a






F′(a + 0) = f (a);F ′(b − 0) = f (b).



F (x) = −cosx






�ành ngh¾a






F′(a + 0) = f (a);F ′(b − 0) = f (b).



F (x) = −cosx






�ành ngh¾a






F′(a + 0) = f (a);F ′(b − 0) = f (b).



F (x) = −cosx






�ành ngh¾a






F′(a + 0) = f (a);F ′(b − 0) = f (b).



F (x) = −cosx� m Thanh Ph÷ìng, Ngæ M¤nh T÷ðng B i gi£ng: TO�N CAO C�P 1 Ch÷ìng VIII: PH�P T�NH NGUY�N H�M




t½ch ph¥n b§t �ành

�ành ngh¾a

N¸u F (x) l mët nguy¶n h m cõa h m sè f (x) tr¶n (a; b) th¼ tªp hñpc¡c nguy¶n h m cõa h m sè f (x) l F (x) + C ,C l h¬ng sè tòy þ. Hå væsè c¡c nguy¶n h m cõa f (x) �â �÷ñc gåi l t½ch ph¥n b§t �ành cõaf (x), x ∈ (a; b) v k½ hi»u l

∫f (x)dx = F (x) + C

C¡c t½nh ch§t �ìn gi£n

1.(∫

f (x)dx)′

= f (x).

2.d(∫

f (x)dx)

= f (x)dx3.N¸uf (x)l h m kh£ vi th¼

∫f ′(x)dx = f (x) + C

4.N¸uf (x)l h m kh£ vi th¼∫df (x) = f (x) + C

5.∫αf (x)dx = α

∫f (x)dx

6.∫

(f (x) + g(x)) dx =∫f (x)dx+

∫g(x)dx






�ành ngh¾a


∫f (x)dx = F (x) + C


1.(∫

f (x)dx)′

= f (x).

2.d(∫

f (x)dx)


∫f ′(x)dx = f (x) + C


5.∫αf (x)dx = α

∫f (x)dx

6.∫

(f (x) + g(x)) dx =∫f (x)dx+

∫g(x)dx






�ành ngh¾a


∫f (x)dx = F (x) + C


1.(∫

f (x)dx)′

= f (x).

2.d(∫

f (x)dx)

= f (x)dx

3.N¸uf (x)l h m kh£ vi th¼∫f ′(x)dx = f (x) + C


5.∫αf (x)dx = α

∫f (x)dx

6.∫

(f (x) + g(x)) dx =∫f (x)dx+

∫g(x)dx






�ành ngh¾a


∫f (x)dx = F (x) + C


1.(∫

f (x)dx)′

= f (x).

2.d(∫

f (x)dx)


∫f ′(x)dx = f (x) + C


5.∫αf (x)dx = α

∫f (x)dx

6.∫

(f (x) + g(x)) dx =∫f (x)dx+

∫g(x)dx






�ành ngh¾a


∫f (x)dx = F (x) + C


1.(∫

f (x)dx)′

= f (x).

2.d(∫

f (x)dx)


∫f ′(x)dx = f (x) + C


5.∫αf (x)dx = α

∫f (x)dx

6.∫

(f (x) + g(x)) dx =∫f (x)dx+

∫g(x)dx






�ành ngh¾a


∫f (x)dx = F (x) + C


1.(∫

f (x)dx)′

= f (x).

2.d(∫

f (x)dx)


∫f ′(x)dx = f (x) + C


5.∫αf (x)dx = α

∫f (x)dx

6.∫

(f (x) + g(x)) dx =∫f (x)dx+

∫g(x)dx





T½ch ph¥n mët sè h m cì b£n



8.3.Ph²p t½nh nguy¶n h m cõa mët sè h m sèPh÷ìng ph¡p �êi bi¸n sèPh÷ìng ph¡p t½ch ph¥n tøng ph¦n

Ph÷ìng ph¡p �êi bi¸n sè

N¸u tçn t¤i h m hñp f (ϕ(x)) v h m t = ϕ(x) li¶n töc tr¶n �o¤n[a; b] v kh£ vi trong kho£ng (a; b), th¼∫

f (ϕ(x)) · ϕ′(x)dx =∫f (t)dt

∣∣t=ϕ(x)

(1)

N¸u tçn t¤i h m sè ng÷ñc x = ϕ−1(t) cõa h m t = ϕ(x) th¼∫f (t)dt =

∫f (ϕ(x))ϕ′(x)dx

∣∣x=ϕ−1(t)

hay∫f (x)dx =

∫f (ϕ(t))ϕ′(t)dt

∣∣t=ϕ−1(x)

(2)

V½ dö 1. T½nh I =∫ dxsin x

I =∫ dxsin x

=∫ sin xdx

sin2 x= −

∫ dcosx1− cos2x

= −∫ dt1− t2

=12

(∫ dtt − 1

−∫ dtt + 1

)=

12ln(cosx − 1cosx + 1

)+ C=

12ln(tan

x2

)+ C






f (ϕ(x)) · ϕ′(x)dx =∫f (t)dt

∣∣t=ϕ(x)

(1)



∣∣x=ϕ−1(t)

hay∫f (x)dx =


∣∣t=ϕ−1(x)

(2)


I =∫ dxsin x

=∫ sin xdx

sin2 x= −

∫ dcosx1− cos2x

= −∫ dt1− t2

=12

(∫ dtt − 1

−∫ dtt + 1

)=


)+ C=

12ln(tan

x2

)+ C






f (ϕ(x)) · ϕ′(x)dx =∫f (t)dt

∣∣t=ϕ(x)

(1)



∣∣x=ϕ−1(t)

hay∫f (x)dx =


∣∣t=ϕ−1(x)

(2)


I =∫ dxsin x

=∫ sin xdx

sin2 x= −

∫ dcosx1− cos2x

= −∫ dt1− t2

=12

(∫ dtt − 1

−∫ dtt + 1

)=


)+ C=

12ln(tan

x2

)+ C






f (ϕ(x)) · ϕ′(x)dx =∫f (t)dt

∣∣t=ϕ(x)

(1)



∣∣x=ϕ−1(t)

hay∫f (x)dx =


∣∣t=ϕ−1(x)

(2)


I =∫ dxsin x

=∫ sin xdx

sin2 x= −

∫ dcosx1− cos2x

= −∫ dt1− t2

=12

(∫ dtt − 1

−∫ dtt + 1

)=


)+ C=

12ln(tan

x2

)+ C






f (ϕ(x)) · ϕ′(x)dx =∫f (t)dt

∣∣t=ϕ(x)

(1)



∣∣x=ϕ−1(t)

hay∫f (x)dx =


∣∣t=ϕ−1(x)

(2)


I =∫ dxsin x

=∫ sin xdx

sin2 x= −

∫ dcosx1− cos2x

= −∫ dt1− t2

=12

(∫ dtt − 1

−∫ dtt + 1

)=


)+ C=

12ln(tan

x2

)+ C






f (ϕ(x)) · ϕ′(x)dx =∫f (t)dt

∣∣t=ϕ(x)

(1)



∣∣x=ϕ−1(t)

hay∫f (x)dx =


∣∣t=ϕ−1(x)

(2)


I =∫ dxsin x

=∫ sin xdx

sin2 x

= −∫ dcosx1− cos2x

= −∫ dt1− t2

=12

(∫ dtt − 1

−∫ dtt + 1

)=


)+ C=

12ln(tan

x2

)+ C






f (ϕ(x)) · ϕ′(x)dx =∫f (t)dt

∣∣t=ϕ(x)

(1)



∣∣x=ϕ−1(t)

hay∫f (x)dx =


∣∣t=ϕ−1(x)

(2)


I =∫ dxsin x

=∫ sin xdx

sin2 x= −

∫ dcosx1− cos2x

= −∫ dt1− t2

=12

(∫ dtt − 1

−∫ dtt + 1

)=


)+ C=

12ln(tan

x2

)+ C






f (ϕ(x)) · ϕ′(x)dx =∫f (t)dt

∣∣t=ϕ(x)

(1)



∣∣x=ϕ−1(t)

hay∫f (x)dx =


∣∣t=ϕ−1(x)

(2)


I =∫ dxsin x

=∫ sin xdx

sin2 x= −

∫ dcosx1− cos2x

= −∫ dt1− t2

=12

(∫ dtt − 1

−∫ dtt + 1

)

=12ln(cosx − 1cosx + 1

)+ C=

12ln(tan

x2

)+ C






f (ϕ(x)) · ϕ′(x)dx =∫f (t)dt

∣∣t=ϕ(x)

(1)



∣∣x=ϕ−1(t)

hay∫f (x)dx =


∣∣t=ϕ−1(x)

(2)


I =∫ dxsin x

=∫ sin xdx

sin2 x= −

∫ dcosx1− cos2x

= −∫ dt1− t2

=12

(∫ dtt − 1

−∫ dtt + 1

)=


)+ C

=12ln(tan

x2

)+ C






f (ϕ(x)) · ϕ′(x)dx =∫f (t)dt

∣∣t=ϕ(x)

(1)



∣∣x=ϕ−1(t)

hay∫f (x)dx =


∣∣t=ϕ−1(x)

(2)


I =∫ dxsin x

=∫ sin xdx

sin2 x= −

∫ dcosx1− cos2x

= −∫ dt1− t2

=12

(∫ dtt − 1

−∫ dtt + 1

)=


)+ C=

12ln(tan

x2

)+ C






f (ϕ(x)) · ϕ′(x)dx =∫f (t)dt

∣∣t=ϕ(x)

(1)



∣∣x=ϕ−1(t)

hay∫f (x)dx =


∣∣t=ϕ−1(x)

(2)


I =∫ dxsin x

=∫ sin xdx

sin2 x= −

∫ dcosx1− cos2x

= −∫ dt1− t2

=12

(∫ dtt − 1

−∫ dtt + 1

)=


)+ C=

12ln(tan

x2

)+ C




V½ dö 2. T½nh∫ ln(arccos x)dx√

1− x2 arccos x

t = ln(arccos x)⇒ dt =−dx√

1− x2 arccos x

I =∫ ln(arccos x)dx√

1− x2 · arccos x=

12ln2 (arccos x) + C

V½ dö 3. T½nh∫ dxa2 + x2

b¬ng ph÷ìng ph¡p �êi bi¸n.

Sû döng �êi bi¸n d¤ng (2) �°t x = atant, a2 + x2 =a2

cos2t, dx = a.

dtcos2t∫ dx

a2 + x2=∫ a.dtcos2 t

:a2

cos2 t=

1a

∫dt =

1at + C =

1aarctan

xa

+ C

Chó þ. Thæng th÷íng khi g°p biºu thùc:.√a2 − x2, ta �°t x = asint

.1

a2 + x2, ta �°t x = atant

v sû döng ph²p �êi bi¸n sè d¤ng (2)





1− x2 arccos x


1− x2 arccos x







cos2t, dx = a.

dtcos2t∫ dx


:a2

cos2 t=

1a

∫dt =

1at + C =

1aarctan

xa

+ C


.1







1− x2 arccos x


1− x2 arccos x







cos2t, dx = a.

dtcos2t∫ dx


:a2

cos2 t=

1a

∫dt =

1at + C =

1aarctan

xa

+ C


.1







1− x2 arccos x


1− x2 arccos x







cos2t, dx = a.

dtcos2t∫ dx


:a2

cos2 t=

1a

∫dt =

1at + C =

1aarctan

xa

+ C


.1






Ph²p ph¥n �o¤n

Gi£ sû hai h m u = u(x), v = v(x) li¶n töc tr¶n [a; b] v kh£ vi trong(a; b).

N¸u tçn t¤i∫v .u′dx th¼ tçn t¤i

∫u.v ′dx . Ngo i ra∫

u · v ′dx = u · v −∫v · u′dx

hay∫u · dv = u · v −

∫v · du




Ph²p ph¥n �o¤n

Gi£ sû hai h m u = u(x), v = v(x) li¶n töc tr¶n [a; b] v kh£ vi trong(a; b).

N¸u tçn t¤i∫v .u′dx th¼ tçn t¤i

∫u.v ′dx . Ngo i ra∫

u · v ′dx = u · v −∫v · u′dx

hay∫u · dv = u · v −

∫v · du




Chó þ

Chó þ. Thæng th÷íng khi g°p biºu thùc t½ch ph¥n d¤ng:∫Pn(x) ln xdx∫Pn(x) arcsinxdx∫Pn(x) arccos xdx

Ta �°t dv = Pn(x)dx , ph¦n cán l¤i l u

Khi g°p ∫Pn(x)exdx∫Pn(x)sinxdx∫Pn(x)cosxdx

�°t u = Pn(x), dv l ph¦n cán l¤i.




Chó þ

Chó þ. Thæng th÷íng khi g°p biºu thùc t½ch ph¥n d¤ng:∫Pn(x) ln xdx∫Pn(x) arcsinxdx∫Pn(x) arccos xdx

Ta �°t dv = Pn(x)dx , ph¦n cán l¤i l uKhi g°p ∫

Pn(x)exdx∫Pn(x)sinxdx∫Pn(x)cosxdx

�°t u = Pn(x), dv l ph¦n cán l¤i.




V½ dö

V½ dö 1. T½nh I =∫arccos2 xdx

�°t u = arccos2x ⇒ du =−2 arccos xdx√

1− x2, dv = dx ⇒ v = x

⇒ I = x arccos2 x −∫ −2x arccos x√

1− x2dx = x arccos2 x + I1

u = arccos x ⇒ du =−dx√1− x2

dv =xdx√1− x2

⇒ v =∫ xdx√

1− x2= −√1− x2 + C

I1 = −√1− x2 arccos x −

∫dx = −

√1− x2 arccos x − x + C2




V½ dö

V½ dö 1. T½nh I =∫arccos2 xdx

�°t u = arccos2x ⇒ du =−2 arccos xdx√

1− x2, dv = dx ⇒ v = x

⇒ I = x arccos2 x −∫ −2x arccos x√

1− x2dx = x arccos2 x + I1

u = arccos x ⇒ du =−dx√1− x2

dv =xdx√1− x2

⇒ v =∫ xdx√

1− x2= −√1− x2 + C

I1 = −√1− x2 arccos x −

∫dx = −

√1− x2 arccos x − x + C2




8.3.1. Nguy¶n h m cõa h m sè húu t�Nguy¶n h m cõa mët sè h m sè væ t� �ìn gi£nNguy¶n h m cõa h m sè l÷ñng gi¡c.

Nguy¶n h m cõa h m sè húu t�

∫ Pn(x)

Qm(x)dx , Pn,Qm l �a thùc bªc n,m câ h» sè thüc

1. Chia tû cho mü v �÷a v· t½ch ph¥n c¡c ph¥n thùc thüc sü,

ch¯ng h¤n∫ Rk(x)

Qm(x)dx , vîi 0 ≤ k < m.

2. Ph¥n t½ch mü ra thøa sè bªc nh§t v bªc hai:Qm(x) = (x − a1)

s1 ... (x − ak)sk ·(x2 + p1x + q1

)t1 · · · (x2 + pvx + qv)tv

3.Ph¥n t½chPk(x)

Qm(x)=

Pn(x)

(x − a1)s1 (x2 + p1x + q1)

t1

=A1

(x − a1)+

A2

(x − a1)2

+ · · ·+ As1

(x − a1)s1

+ · · ·+ B1x + C1

(x2 + p1x + q1)+

B2x + C2

(x2 + p1x + q1)2

+ · · ·+ Bt1x + Ct1

(x2 + p1x + q1)t1






∫ Pn(x)





2. Ph¥n t½ch mü ra thøa sè bªc nh§t v bªc hai:

Qm(x) = (x − a1)s1 ... (x − ak)

sk ·(x2 + p1x + q1

)t1 · · · (x2 + pvx + qv)tv

3.Ph¥n t½chPk(x)

Qm(x)=

Pn(x)

(x − a1)s1 (x2 + p1x + q1)

t1

=A1

(x − a1)+

A2

(x − a1)2

+ · · ·+ As1

(x − a1)s1

+ · · ·+ B1x + C1

(x2 + p1x + q1)+

B2x + C2

(x2 + p1x + q1)2

+ · · ·+ Bt1x + Ct1

(x2 + p1x + q1)t1






∫ Pn(x)






s1 ... (x − ak)sk ·(x2 + p1x + q1

)t1 · · · (x2 + pvx + qv)tv

3.Ph¥n t½chPk(x)

Qm(x)=

Pn(x)

(x − a1)s1 (x2 + p1x + q1)

t1

=A1

(x − a1)+

A2

(x − a1)2

+ · · ·+ As1

(x − a1)s1

+ · · ·+ B1x + C1

(x2 + p1x + q1)+

B2x + C2

(x2 + p1x + q1)2

+ · · ·+ Bt1x + Ct1

(x2 + p1x + q1)t1






∫ Pn(x)






s1 ... (x − ak)sk ·(x2 + p1x + q1

)t1 · · · (x2 + pvx + qv)tv

3.Ph¥n t½chPk(x)

Qm(x)=

Pn(x)

(x − a1)s1 (x2 + p1x + q1)

t1

=A1

(x − a1)+

A2

(x − a1)2

+ · · ·+ As1

(x − a1)s1

+ · · ·+ B1x + C1

(x2 + p1x + q1)+

B2x + C2

(x2 + p1x + q1)2

+ · · ·+ Bt1x + Ct1

(x2 + p1x + q1)t1






∫ Pn(x)






s1 ... (x − ak)sk ·(x2 + p1x + q1

)t1 · · · (x2 + pvx + qv)tv

3.Ph¥n t½chPk(x)

Qm(x)=

Pn(x)

(x − a1)s1 (x2 + p1x + q1)

t1

=A1

(x − a1)+

A2

(x − a1)2

+ · · ·+ As1

(x − a1)s1

+ · · ·+ B1x + C1

(x2 + p1x + q1)+

B2x + C2

(x2 + p1x + q1)2

+ · · ·+ Bt1x + Ct1

(x2 + p1x + q1)t1





5. �÷a t½ch ph¥n c¦n t½nh v· c¡c t½ch ph¥n cì b£n sau:

1.∫ dx

(x − a)n=

1

(n − 1) (x − a)n−1+ C , n 6= 1

2.∫ (Mx + n) dx

x2 + px + q=

M2

∫2x+p

x2+px+qdx +

(N − Mp

2

)∫ dxx2 + px + q

3.In =∫ dx

(x2 + a2)nu =

1(x2 + a2)n

⇒ du =−2nxdx

(x2 + a2)n+1

dv = dx ⇒ v = x

In =x

(x2 + a2)n+ 2n

∫ x2dx

(x2 + a2)n+1

In =x

(x2 + a2)n+ 2n

∫ (x2 + a2 − a2)dx

(x2 + a2)n+1

In =x

(x2 + a2)n+ 2n

∫ dx(x2 + a2)n

− 2na2∫ dx

(x2 + a2)n+1

In =x

(x2 + a2)n+ 2nIn − 2na2In+1 H» thùc truy hçi

In+1 =1

2na2

(x

(x2 + a2)n+ (2n − 1) In

)I1 =

∫ dxx2 + a2

=1aarctan

xa

+ C






1.∫ dx

(x − a)n=

1

(n − 1) (x − a)n−1+ C , n 6= 1

2.∫ (Mx + n) dx

x2 + px + q=

M2

∫2x+p

x2+px+qdx +

(N − Mp

2

)∫ dxx2 + px + q

3.In =∫ dx

(x2 + a2)nu =

1(x2 + a2)n

⇒ du =−2nxdx

(x2 + a2)n+1

dv = dx ⇒ v = x

In =x

(x2 + a2)n+ 2n

∫ x2dx

(x2 + a2)n+1

In =x

(x2 + a2)n+ 2n

∫ (x2 + a2 − a2)dx

(x2 + a2)n+1

In =x

(x2 + a2)n+ 2n

∫ dx(x2 + a2)n

− 2na2∫ dx

(x2 + a2)n+1

In =x


In+1 =1

2na2

(x

(x2 + a2)n+ (2n − 1) In

)I1 =

∫ dxx2 + a2

=1aarctan

xa

+ C� m Thanh Ph÷ìng, Ngæ M¤nh T÷ðng B i gi£ng: TO�N CAO C�P 1 Ch÷ìng VIII: PH�P T�NH NGUY�N H�M





1.∫ dx

(x − a)n=

1

(n − 1) (x − a)n−1+ C , n 6= 1

2.∫ (Mx + n) dx

x2 + px + q=

M2

∫2x+p

x2+px+qdx +

(N − Mp

2

)∫ dxx2 + px + q

3.In =∫ dx

(x2 + a2)nu =

1(x2 + a2)n

⇒ du =−2nxdx

(x2 + a2)n+1

dv = dx ⇒ v = x

In =x

(x2 + a2)n+ 2n

∫ x2dx

(x2 + a2)n+1

In =x

(x2 + a2)n+ 2n

∫ (x2 + a2 − a2)dx

(x2 + a2)n+1

In =x

(x2 + a2)n+ 2n

∫ dx(x2 + a2)n

− 2na2∫ dx

(x2 + a2)n+1

In =x


In+1 =1

2na2

(x

(x2 + a2)n+ (2n − 1) In

)I1 =

∫ dxx2 + a2

=1aarctan

xa

+ C� m Thanh Ph÷ìng, Ngæ M¤nh T÷ðng B i gi£ng: TO�N CAO C�P 1 Ch÷ìng VIII: PH�P T�NH NGUY�N H�M




V½ dö 1. T½nh I =∫ dx

(x − 2)3

I =∫ d(x − 2)

(x − 2)3=∫

(x − 2)−3d(x − 2)

= −12

(x − 2)−3+1 + C =−1

2(x − 2)2+ C

V½ dö 2. T½nh I =∫ dxx2 + 2x + 5

I =∫ dx

(x + 1)2 + 22=∫ d (x + 1)

(x + 1)2 + 22=

12arctan

x + 12

+ C

V½ dö 3. T½nh I =∫ (x + 4)dx

(x − 2)(x + 1)x + 4

(x − 2)(x + 1)=

Ax − 2

+B

x + 1Qui �çng, �çng nh§t hai v¸, t¼m �÷ñc A = 2,B = −1.

I =∫ 2dxx − 2

−∫ dxx + 1

= 2 ln(x − 2)− ln(x + 1) + C = ln∣∣∣∣ (x − 2)2

x + 1

∣∣∣∣+ C






(x − 2)3

I =∫ d(x − 2)

(x − 2)3=∫

(x − 2)−3d(x − 2)

= −12

(x − 2)−3+1 + C =−1

2(x − 2)2+ C

V½ dö 2. T½nh I =∫ dxx2 + 2x + 5

I =∫ dx

(x + 1)2 + 22=∫ d (x + 1)

(x + 1)2 + 22=

12arctan

x + 12

+ C

V½ dö 3. T½nh I =∫ (x + 4)dx

(x − 2)(x + 1)x + 4

(x − 2)(x + 1)=

Ax − 2

+B


I =∫ 2dxx − 2

−∫ dxx + 1

= 2 ln(x − 2)− ln(x + 1) + C = ln∣∣∣∣ (x − 2)2

x + 1

∣∣∣∣+ C






(x − 2)3

I =∫ d(x − 2)

(x − 2)3=∫

(x − 2)−3d(x − 2)

= −12

(x − 2)−3+1 + C =−1

2(x − 2)2+ C

V½ dö 2. T½nh I =∫ dxx2 + 2x + 5

I =∫ dx

(x + 1)2 + 22=∫ d (x + 1)

(x + 1)2 + 22=

12arctan

x + 12

+ C

V½ dö 3. T½nh I =∫ (x + 4)dx

(x − 2)(x + 1)x + 4

(x − 2)(x + 1)=

Ax − 2

+B


I =∫ 2dxx − 2

−∫ dxx + 1

= 2 ln(x − 2)− ln(x + 1) + C = ln∣∣∣∣ (x − 2)2

x + 1

∣∣∣∣+ C






(x − 2)3

I =∫ d(x − 2)

(x − 2)3=∫

(x − 2)−3d(x − 2)

= −12

(x − 2)−3+1 + C =−1

2(x − 2)2+ C

V½ dö 2. T½nh I =∫ dxx2 + 2x + 5

I =∫ dx

(x + 1)2 + 22=∫ d (x + 1)

(x + 1)2 + 22=

12arctan

x + 12

+ C

V½ dö 3. T½nh I =∫ (x + 4)dx

(x − 2)(x + 1)x + 4

(x − 2)(x + 1)=

Ax − 2

+B


I =∫ 2dxx − 2

−∫ dxx + 1

= 2 ln(x − 2)− ln(x + 1) + C = ln∣∣∣∣ (x − 2)2

x + 1

∣∣∣∣+ C






(x − 2)3

I =∫ d(x − 2)

(x − 2)3=∫

(x − 2)−3d(x − 2)

= −12

(x − 2)−3+1 + C =−1

2(x − 2)2+ C

V½ dö 2. T½nh I =∫ dxx2 + 2x + 5

I =∫ dx

(x + 1)2 + 22=∫ d (x + 1)

(x + 1)2 + 22=

12arctan

x + 12

+ C

V½ dö 3. T½nh I =∫ (x + 4)dx

(x − 2)(x + 1)

x + 4(x − 2)(x + 1)

=A

x − 2+

Bx + 1

Qui �çng, �çng nh§t hai v¸, t¼m �÷ñc A = 2,B = −1.

I =∫ 2dxx − 2

−∫ dxx + 1

= 2 ln(x − 2)− ln(x + 1) + C = ln∣∣∣∣ (x − 2)2

x + 1

∣∣∣∣+ C






(x − 2)3

I =∫ d(x − 2)

(x − 2)3=∫

(x − 2)−3d(x − 2)

= −12

(x − 2)−3+1 + C =−1

2(x − 2)2+ C

V½ dö 2. T½nh I =∫ dxx2 + 2x + 5

I =∫ dx

(x + 1)2 + 22=∫ d (x + 1)

(x + 1)2 + 22=

12arctan

x + 12

+ C

V½ dö 3. T½nh I =∫ (x + 4)dx

(x − 2)(x + 1)x + 4

(x − 2)(x + 1)=

Ax − 2

+B


I =∫ 2dxx − 2

−∫ dxx + 1

= 2 ln(x − 2)− ln(x + 1) + C = ln∣∣∣∣ (x − 2)2

x + 1

∣∣∣∣+ C






(x − 2)3

I =∫ d(x − 2)

(x − 2)3=∫

(x − 2)−3d(x − 2)

= −12

(x − 2)−3+1 + C =−1

2(x − 2)2+ C

V½ dö 2. T½nh I =∫ dxx2 + 2x + 5

I =∫ dx

(x + 1)2 + 22=∫ d (x + 1)

(x + 1)2 + 22=

12arctan

x + 12

+ C

V½ dö 3. T½nh I =∫ (x + 4)dx

(x − 2)(x + 1)x + 4

(x − 2)(x + 1)=

Ax − 2

+B


I =∫ 2dxx − 2

−∫ dxx + 1

= 2 ln(x − 2)− ln(x + 1) + C = ln∣∣∣∣ (x − 2)2

x + 1

∣∣∣∣+ C





Chó þ

C¡ch t¼m h» sè A,B trong (*) nhanh:�º t¼m A, nh¥n hai v¸ (*) vîi (x − 2) rçi thay x = 2 v o.�º t¼m B , nh¥n hai v¸ (*) vîi (x + 1) rçi thay x = −1 v o.

V½ dö 4. T½nh I =∫ 2x3 + x2 + 5x + 1

(x2 + 3)(x2 − x + 1)dx

2x3 + x2 + 5x + 1(x2 + 3)(x2 − x + 1)

=Ax + Bx2 + 3

+Cx + D

x2 − x + 1Qui �çng, �çng nh§t hai v¸ ta �÷ñc A = 0;B = 1;C = 2;D = 0.

I =∫ dxx2 + 3

+∫ 2xdxx2 − x + 1





Chó þ


V½ dö 4. T½nh I =∫ 2x3 + x2 + 5x + 1

(x2 + 3)(x2 − x + 1)dx

2x3 + x2 + 5x + 1(x2 + 3)(x2 − x + 1)

=Ax + Bx2 + 3

+Cx + D


I =∫ dxx2 + 3

+∫ 2xdxx2 − x + 1





Chó þ


V½ dö 4. T½nh I =∫ 2x3 + x2 + 5x + 1

(x2 + 3)(x2 − x + 1)dx

2x3 + x2 + 5x + 1(x2 + 3)(x2 − x + 1)

=Ax + Bx2 + 3

+Cx + D

x2 − x + 1

Qui �çng, �çng nh§t hai v¸ ta �÷ñc A = 0;B = 1;C = 2;D = 0.

I =∫ dxx2 + 3

+∫ 2xdxx2 − x + 1





Chó þ


V½ dö 4. T½nh I =∫ 2x3 + x2 + 5x + 1

(x2 + 3)(x2 − x + 1)dx

2x3 + x2 + 5x + 1(x2 + 3)(x2 − x + 1)

=Ax + Bx2 + 3

+Cx + D


I =∫ dxx2 + 3

+∫ 2xdxx2 − x + 1





Chó þ


V½ dö 4. T½nh I =∫ 2x3 + x2 + 5x + 1

(x2 + 3)(x2 − x + 1)dx

2x3 + x2 + 5x + 1(x2 + 3)(x2 − x + 1)

=Ax + Bx2 + 3

+Cx + D


I =∫ dxx2 + 3

+∫ 2xdxx2 − x + 1





Chó þ


V½ dö 4. T½nh I =∫ 2x3 + x2 + 5x + 1

(x2 + 3)(x2 − x + 1)dx

2x3 + x2 + 5x + 1(x2 + 3)(x2 − x + 1)

=Ax + Bx2 + 3

+Cx + D


I =∫ dxx2 + 3

+∫ 2xdxx2 − x + 1





V½ dö 5. T½nh I =∫ dxx2 + 2x + 5

I =∫ dx

(x + 1)2 + 22=∫ d (x + 1)

(x + 1)2 + 22=

12arctan

x + 12

+ C =∫ dxx2 + 3

+∫ (2x − 1) + 1

x2 − x + 1dx

= 1√3arctan

x√3

+ ln(x2 − x + 1) +2√3arctan

2x − 1√3

+ CV½ dö 6. T½nh

I =∫ 4x2 − 8x

(x − 1)2(x2 + 1)2dx

P(x)

(x2 + 1)2(x − 1)2=

Ax − 1

+B

(x − 1)2+ Cx+D

x2+1+

Ex + F

(x2 + 1)2(*)

T¼m �÷ñc: A = 2,B = −1,C = −2,D = −1,E = −2,F = 4.∫ (−2x + 4)dx

(x2 + 1)2=∫ −2xdx

(x2 + 1)2+∫ 4dx

(x2 + 1)2

Dòng h» thùc truy hçi, t½nh I2 =∫ 4dx

(x2 + 1)2





V½ dö 5. T½nh I =∫ dxx2 + 2x + 5

I =∫ dx

(x + 1)2 + 22=∫ d (x + 1)

(x + 1)2 + 22=

12arctan

x + 12

+ C =∫ dxx2 + 3

+∫ (2x − 1) + 1

x2 − x + 1dx

= 1√3arctan

x√3

+ ln(x2 − x + 1) +2√3arctan

2x − 1√3

+ C

V½ dö 6. T½nh

I =∫ 4x2 − 8x

(x − 1)2(x2 + 1)2dx

P(x)

(x2 + 1)2(x − 1)2=

Ax − 1

+B

(x − 1)2+ Cx+D

x2+1+

Ex + F

(x2 + 1)2(*)

T¼m �÷ñc: A = 2,B = −1,C = −2,D = −1,E = −2,F = 4.∫ (−2x + 4)dx

(x2 + 1)2=∫ −2xdx

(x2 + 1)2+∫ 4dx

(x2 + 1)2


(x2 + 1)2





V½ dö 5. T½nh I =∫ dxx2 + 2x + 5

I =∫ dx

(x + 1)2 + 22=∫ d (x + 1)

(x + 1)2 + 22=

12arctan

x + 12

+ C =∫ dxx2 + 3

+∫ (2x − 1) + 1

x2 − x + 1dx

= 1√3arctan

x√3

+ ln(x2 − x + 1) +2√3arctan

2x − 1√3

+ CV½ dö 6. T½nh

I =∫ 4x2 − 8x

(x − 1)2(x2 + 1)2dx

P(x)

(x2 + 1)2(x − 1)2=

Ax − 1

+B

(x − 1)2+ Cx+D

x2+1+

Ex + F

(x2 + 1)2(*)

T¼m �÷ñc: A = 2,B = −1,C = −2,D = −1,E = −2,F = 4.∫ (−2x + 4)dx

(x2 + 1)2=∫ −2xdx

(x2 + 1)2+∫ 4dx

(x2 + 1)2


(x2 + 1)2





V½ dö 5. T½nh I =∫ dxx2 + 2x + 5

I =∫ dx

(x + 1)2 + 22=∫ d (x + 1)

(x + 1)2 + 22=

12arctan

x + 12

+ C =∫ dxx2 + 3

+∫ (2x − 1) + 1

x2 − x + 1dx

= 1√3arctan

x√3

+ ln(x2 − x + 1) +2√3arctan

2x − 1√3

+ CV½ dö 6. T½nh

I =∫ 4x2 − 8x

(x − 1)2(x2 + 1)2dx

P(x)

(x2 + 1)2(x − 1)2=

Ax − 1

+B

(x − 1)2+ Cx+D

x2+1+

Ex + F

(x2 + 1)2(*)

T¼m �÷ñc: A = 2,B = −1,C = −2,D = −1,E = −2,F = 4.∫ (−2x + 4)dx

(x2 + 1)2=∫ −2xdx

(x2 + 1)2+∫ 4dx

(x2 + 1)2


(x2 + 1)2





V½ dö 5. T½nh I =∫ dxx2 + 2x + 5

I =∫ dx

(x + 1)2 + 22=∫ d (x + 1)

(x + 1)2 + 22=

12arctan

x + 12

+ C =∫ dxx2 + 3

+∫ (2x − 1) + 1

x2 − x + 1dx

= 1√3arctan

x√3

+ ln(x2 − x + 1) +2√3arctan

2x − 1√3

+ CV½ dö 6. T½nh

I =∫ 4x2 − 8x

(x − 1)2(x2 + 1)2dx

P(x)

(x2 + 1)2(x − 1)2=

Ax − 1

+B

(x − 1)2+ Cx+D

x2+1+

Ex + F

(x2 + 1)2(*)

T¼m �÷ñc: A = 2,B = −1,C = −2,D = −1,E = −2,F = 4.∫ (−2x + 4)dx

(x2 + 1)2=∫ −2xdx

(x2 + 1)2+∫ 4dx

(x2 + 1)2


(x2 + 1)2





�º t¼m c¡c h» sè A,B ,C , ... nhanh, câ thº sû döng khai triºnHeaviside:Tø (*) ta câ 4x2 − 8x = A(x − 1)(x2 + 1)2 + B(x2 + 1)2++(Cx + D)(x − 1)2(x2 + 1) + (Ex + F )(x − 1)2

Thay x = 1, t¼m �÷ñc B = −1.Thay x = −1, c¥n b¬ng ph¦n thüc, £o: E = −2,F = 4.�¤o h m 2 v¸, ch¿ quan t¥m sè h¤ng kh¡c 0 khi x = iThay x = i , t¼m �÷ñc C = −2,D = −1.





�º t¼m c¡c h» sè A,B ,C , ... nhanh, câ thº sû döng khai triºnHeaviside:Tø (*) ta câ 4x2 − 8x = A(x − 1)(x2 + 1)2 + B(x2 + 1)2++(Cx + D)(x − 1)2(x2 + 1) + (Ex + F )(x − 1)2

Thay x = 1, t¼m �÷ñc B = −1.Thay x = −1, c¥n b¬ng ph¦n thüc, £o: E = −2,F = 4.�¤o h m 2 v¸, ch¿ quan t¥m sè h¤ng kh¡c 0 khi x = iThay x = i , t¼m �÷ñc C = −2,D = −1.





T½ch ph¥n câ chùa...

∫R

x ,(ax + bcx + d

)p1

q1 ,

(ax + bcx + d

)p2

q2 , · · ·

dx

C¡ch gi£i: �êi bi¸n tn =ax + bcx + d

, n l bëi chung nhä nh§t cõa q1, q2

T½ch ph¥n câ chùa√ax2 + bx + c

Bi¸n �êi biºu thùc d÷îi d§u c«n v· d¤ng αt2 + β






∫R

x ,(ax + bcx + d

)p1

q1 ,

(ax + bcx + d

)p2

q2 , · · ·

dxC¡ch gi£i: �êi bi¸n tn =

ax + bcx + d









∫R

x ,(ax + bcx + d

)p1

q1 ,

(ax + bcx + d

)p2

q2 , · · ·


ax + bcx + d









∫R

x ,(ax + bcx + d

)p1

q1 ,

(ax + bcx + d

)p2

q2 , · · ·


ax + bcx + d








V½ dö 7. T½nh I =∫ dx√

2x − 1− 4√2x − 1

�êi bi¸n 2x − 1 = t4 ⇒ 2dx = 4t3dt

I =∫ 2t2dt

t − 1= 2

∫ (t + 1 +

1t − 1

)dt = t2 + 2t + ln |t − 1|+ C

V½ dö 8. T½nh I =∫ x + 1 + 3

√(x + 1)2 + 6

√x + 1

(x + 1)(1 + 3√x + 1)

dx

�êi bi¸n x + 1 = t6 ⇒ dx = 6t5dt

I = 6∫ (t6 + t4 + t)t5dt

t6(1 + t2)= 6

∫t3dt + 6

∫ dtt2 + 1

=

32

3√x2 + 6 arctan 6

√x + C






2x − 1− 4√2x − 1

�êi bi¸n 2x − 1 = t4 ⇒ 2dx = 4t3dt

I =∫ 2t2dt

t − 1= 2

∫ (t + 1 +

1t − 1

)dt = t2 + 2t + ln |t − 1|+ C

V½ dö 8. T½nh I =∫ x + 1 + 3

√(x + 1)2 + 6

√x + 1

(x + 1)(1 + 3√x + 1)

dx

�êi bi¸n x + 1 = t6 ⇒ dx = 6t5dt

I = 6∫ (t6 + t4 + t)t5dt

t6(1 + t2)= 6

∫t3dt + 6

∫ dtt2 + 1

=

32

3√x2 + 6 arctan 6

√x + C






2x − 1− 4√2x − 1

�êi bi¸n 2x − 1 = t4 ⇒ 2dx = 4t3dt

I =∫ 2t2dt

t − 1= 2

∫ (t + 1 +

1t − 1

)dt = t2 + 2t + ln |t − 1|+ C

V½ dö 8. T½nh I =∫ x + 1 + 3

√(x + 1)2 + 6

√x + 1

(x + 1)(1 + 3√x + 1)

dx

�êi bi¸n x + 1 = t6 ⇒ dx = 6t5dt

I = 6∫ (t6 + t4 + t)t5dt

t6(1 + t2)= 6

∫t3dt + 6

∫ dtt2 + 1

=

32

3√x2 + 6 arctan 6

√x + C






2x − 1− 4√2x − 1

�êi bi¸n 2x − 1 = t4 ⇒ 2dx = 4t3dt

I =∫ 2t2dt

t − 1= 2

∫ (t + 1 +

1t − 1

)dt = t2 + 2t + ln |t − 1|+ C

V½ dö 8. T½nh I =∫ x + 1 + 3

√(x + 1)2 + 6

√x + 1

(x + 1)(1 + 3√x + 1)

dx

�êi bi¸n x + 1 = t6 ⇒ dx = 6t5dt

I = 6∫ (t6 + t4 + t)t5dt

t6(1 + t2)= 6

∫t3dt + 6

∫ dtt2 + 1

=

32

3√x2 + 6 arctan 6

√x + C






2x − 1− 4√2x − 1

�êi bi¸n 2x − 1 = t4 ⇒ 2dx = 4t3dt

I =∫ 2t2dt

t − 1= 2

∫ (t + 1 +

1t − 1

)dt = t2 + 2t + ln |t − 1|+ C

V½ dö 8. T½nh I =∫ x + 1 + 3

√(x + 1)2 + 6

√x + 1

(x + 1)(1 + 3√x + 1)

dx

�êi bi¸n x + 1 = t6 ⇒ dx = 6t5dt

I = 6∫ (t6 + t4 + t)t5dt

t6(1 + t2)= 6

∫t3dt + 6

∫ dtt2 + 1

=

32

3√x2 + 6 arctan 6

√x + C





Nguy¶n h m cõa h m l÷ñng gi¡c

1.∫R (sin x , cos x)dx Trong �â R(u, v) l h m húu t� theo bi¸n u, v .

C¡ch gi£i chung: �°t t = tan(x2

), x ∈ (−π, π)

⇒ x = 2 arctan t ⇒ dx = 2dt

1 + t2

sin x =2t

1 + t2, cos x =

1− t2

1 + t2∫R (sin x , cos x)dx = 2

∫R(

2t1 + t2

,1− t2

1 + t2

)dt

1+t2








), x ∈ (−π, π)


1 + t2

sin x =2t

1 + t2, cos x =

1− t2


∫R(

2t1 + t2

,1− t2

1 + t2

)dt

1+t2








), x ∈ (−π, π)


1 + t2

sin x =2t

1 + t2, cos x =

1− t2


∫R(

2t1 + t2

,1− t2

1 + t2

)dt

1+t2





V½ dö. T½nh I =∫ dx3 sin x + 4 cos x + 5

�êi bi¸n

t = tan(x2

), x ∈ (−π, π)

⇒ dx = 2dt

1 + t2sin x =

2t1 + t2

, cos x =1− t2

1 + t2

I = 2∫ dt6t + 4(1− t2) + 5(1 + t2)

= 2∫ dtt2 + 6t + 9

= 2∫

(t + 3)−2d(t + 3) =−2t + 3

+ C =−2

tan(x/2) + 3+ C






�êi bi¸n

t = tan(x2

), x ∈ (−π, π)

⇒ dx = 2dt

1 + t2sin x =

2t1 + t2

, cos x =1− t2

1 + t2

I = 2∫ dt6t + 4(1− t2) + 5(1 + t2)

= 2∫ dtt2 + 6t + 9

= 2∫

(t + 3)−2d(t + 3) =−2t + 3

+ C =−2

tan(x/2) + 3+ C






�êi bi¸n t = tan(x2

), x ∈ (−π, π)

⇒ dx = 2dt

1 + t2; sin x =

2t1 + t2

, cos x =1− t2

1 + t2

I = 2∫ dt6t + 4(1− t2) + 5(1 + t2)

= 2∫ dtt2 + 6t + 9

= 2∫

(t + 3)−2d(t + 3) =−2t + 3

+ C =−2

tan(x/2) + 3+ C

Trong nhi·u tr÷íng hñp, c¡ch gi£i tr¶n kh¡ cçng k·nh







), x ∈ (−π, π)

⇒ dx = 2dt

1 + t2; sin x =

2t1 + t2

, cos x =1− t2

1 + t2

I = 2∫ dt6t + 4(1− t2) + 5(1 + t2)

= 2∫ dtt2 + 6t + 9

= 2∫

(t + 3)−2d(t + 3) =−2t + 3

+ C =−2

tan(x/2) + 3+ C








), x ∈ (−π, π)

⇒ dx = 2dt

1 + t2; sin x =

2t1 + t2

, cos x =1− t2

1 + t2

I = 2∫ dt6t + 4(1− t2) + 5(1 + t2)

= 2∫ dtt2 + 6t + 9

= 2∫

(t + 3)−2d(t + 3) =−2t + 3

+ C =−2

tan(x/2) + 3+ C






∫R (sin x , cos x)dx

1 1) R (− sin x , cos x) = −R (sin x , cos x) �°t t = cos x , x ∈(−π2,π

2

)2 2) R (sin x ,− cos x) = −R (sin x , cos x) �°t t = sin x , x ∈ (0, π)

3 3) R (− sin x ,− cos x) = R (sin x , cos x) �°t

t = tan x , x ∈(−π2,π

2

)4 4)

∫sinp x · cosqx · dx �°t t = sin x ho°c t = cos x

Ho n to n t÷ìng tü cho c¡c h m Hyperbolic: sinh x , cosh x





∫R (sin x , cos x)dx

1 1) R (− sin x , cos x) = −R (sin x , cos x) �°t t = cos x , x ∈(−π2,π

2

)2 2) R (sin x ,− cos x) = −R (sin x , cos x) �°t t = sin x , x ∈ (0, π)

3 3) R (− sin x ,− cos x) = R (sin x , cos x) �°t

t = tan x , x ∈(−π2,π

2

)4 4)

∫sinp x · cosqx · dx �°t t = sin x ho°c t = cos x

Ho n to n t÷ìng tü cho c¡c h m Hyperbolic: sinh x , cosh x





C¡c v½ dö

V½ dö 1. T½nh I =∫ (2 sin x + 3 cos x)dx

sin2 x cos x + 9 cos3 x

R (− sin x ,− cos x) = R (sin x , cos x) �êi bi¸n

t = tan(x), x ∈ (−π/2, π/2)⇒ dt =dx

cos2 xChia tû v mü cho cos3 x

I =∫ (2 tan x + 3)d(tan x)

tan2 x + 9=∫ 2t + 3t2 + 9

dt =∫ 2tt2 + 9

dt +∫ 3t2 + 32

dt

= ln(t2 + 9) + arctant3

+ C = ln(tan2 x + 9) + arctantan x3

+ C

V½ dö 2. T½nh I =∫cos3x · sin8 xdx

�êi bi¸n t = sin x ⇒ dt = cos xdxI =

∫cos2x · sin8 x (cos xdx) =

∫ (1− sin2 x

)sin8 x (cos xdx)

=∫

(1− t2)t8dt =t9

9− t11

11+ C =

sin9 x9− sin11 x

11+ C





C¡c v½ dö


sin2 x cos x + 9 cos3 xR (− sin x ,− cos x) = R (sin x , cos x) �êi bi¸n

t = tan(x), x ∈ (−π/2, π/2)⇒ dt =dx

cos2 x

Chia tû v mü cho cos3 x

I =∫ (2 tan x + 3)d(tan x)

tan2 x + 9=∫ 2t + 3t2 + 9

dt =∫ 2tt2 + 9

dt +∫ 3t2 + 32

dt



+ C




∫ (1− sin2 x

)sin8 x (cos xdx)

=∫

(1− t2)t8dt =t9

9− t11

11+ C =

sin9 x9− sin11 x

11+ C





C¡c v½ dö



t = tan(x), x ∈ (−π/2, π/2)⇒ dt =dx


I =∫ (2 tan x + 3)d(tan x)

tan2 x + 9=∫ 2t + 3t2 + 9

dt =∫ 2tt2 + 9

dt +∫ 3t2 + 32

dt



+ C




∫ (1− sin2 x

)sin8 x (cos xdx)

=∫

(1− t2)t8dt =t9

9− t11

11+ C =

sin9 x9− sin11 x

11+ C





C¡c v½ dö



t = tan(x), x ∈ (−π/2, π/2)⇒ dt =dx


I =∫ (2 tan x + 3)d(tan x)

tan2 x + 9=∫ 2t + 3t2 + 9

dt =∫ 2tt2 + 9

dt +∫ 3t2 + 32

dt



+ C




∫ (1− sin2 x

)sin8 x (cos xdx)

=∫

(1− t2)t8dt =t9

9− t11

11+ C =

sin9 x9− sin11 x

11+ C





C¡c v½ dö



t = tan(x), x ∈ (−π/2, π/2)⇒ dt =dx


I =∫ (2 tan x + 3)d(tan x)

tan2 x + 9=∫ 2t + 3t2 + 9

dt =∫ 2tt2 + 9

dt +∫ 3t2 + 32

dt



+ C




∫ (1− sin2 x

)sin8 x (cos xdx)

=∫

(1− t2)t8dt =t9

9− t11

11+ C =

sin9 x9− sin11 x

11+ C





C¡c v½ dö



t = tan(x), x ∈ (−π/2, π/2)⇒ dt =dx


I =∫ (2 tan x + 3)d(tan x)

tan2 x + 9=∫ 2t + 3t2 + 9

dt =∫ 2tt2 + 9

dt +∫ 3t2 + 32

dt



+ C




∫ (1− sin2 x

)sin8 x (cos xdx)

=∫

(1− t2)t8dt =t9

9− t11

11+ C =

sin9 x9− sin11 x

11+ C





V½ dö 3. T½nh I =∫

(sinh2 x · cosh3 x)dx

R (− sinh x ,− cosh x) = −R (sinh x , cosh x)�êi bi¸n t = sinh(x)⇒ dt = cosh(x))dxI =

∫(sinh2x cosh2 x)(cosh x)dx

=∫sinh2 x(sinh2 x + 1)(cosh x)dx

=∫t2(t2 + 1)dt =

t6

6+

t3

3+ C =

sinh6 x6

+sinh3 x

3+ C





V½ dö 3. T½nh I =∫

(sinh2 x · cosh3 x)dxR (− sinh x ,− cosh x) = −R (sinh x , cosh x)

�êi bi¸n t = sinh(x)⇒ dt = cosh(x))dxI =



=∫t2(t2 + 1)dt =

t6

6+

t3

3+ C =

sinh6 x6

+sinh3 x

3+ C





V½ dö 3. T½nh I =∫

(sinh2 x · cosh3 x)dxR (− sinh x ,− cosh x) = −R (sinh x , cosh x)�êi bi¸n t = sinh(x)⇒ dt = cosh(x))dx

I =∫

(sinh2x cosh2 x)(cosh x)dx=∫sinh2 x(sinh2 x + 1)(cosh x)dx

=∫t2(t2 + 1)dt =

t6

6+

t3

3+ C =

sinh6 x6

+sinh3 x

3+ C





V½ dö 3. T½nh I =∫

(sinh2 x · cosh3 x)dxR (− sinh x ,− cosh x) = −R (sinh x , cosh x)�êi bi¸n t = sinh(x)⇒ dt = cosh(x))dxI =



=∫t2(t2 + 1)dt =

t6

6+

t3

3+ C =

sinh6 x6

+sinh3 x

3+ C





V½ dö 3. T½nh I =∫

(sinh2 x · cosh3 x)dxR (− sinh x ,− cosh x) = −R (sinh x , cosh x)�êi bi¸n t = sinh(x)⇒ dt = cosh(x))dxI =



=∫t2(t2 + 1)dt =

t6

6+

t3

3+ C =

sinh6 x6

+sinh3 x

3+ C





Nguy¶n h m d¤ng I =∫ a1 sin x + b1 cos x

a sin x + b cos xdx

Ph¥n t½ch a1 sin x + b1 cos x = A (a sin x + b cos x)′ + B (a sin x + b cos x)

�çng nh§t hai v¸{

Ab − aB = a1Aa + Bb = b1

gi£i t¼m A,B .

I =∫ A(a sin x + b cos x)′dx

a sin x + b cos x+∫Bdx

= A ln(a sin x + b cos x) + Bx + C






a sin x + b cos xdx




gi£i t¼m A,B .









a sin x + b cos xdx




gi£i t¼m A,B .









a sin x + b cos xdx




gi£i t¼m A,B .









a sin x + b cos xdx




gi£i t¼m A,B .








V½ dö. T½nh I =∫ (2 sin x + 3 cos x)dx

sin x + 4 cos x

Ph¥n t½ch 2 sin x + 3 cos x = A(sin x + 4 cos x) + B(sin x + 4 cos x)′

2 sin x + 3 cos x = (A− 4B) sin x + (4A + B) cos x{A− 4B = 24A + B = 3

⇔{

A = 1B = −1/4

I =∫ A(sin x + 4 cos x)

sin x + 4 cos xdx +

∫ B(sin x + 4 cos x)′

sin x + 4 cos xdx

I = A∫dx +

∫ Bd(sin x + 4 cos x)

sin x + 4 cos x= Ax + B ln (sin x + 4 cos x) + C






sin x + 4 cos xPh¥n t½ch 2 sin x + 3 cos x = A(sin x + 4 cos x) + B(sin x + 4 cos x)′

2 sin x + 3 cos x = (A− 4B) sin x + (4A + B) cos x

{A− 4B = 24A + B = 3

⇔{

A = 1B = −1/4


sin x + 4 cos xdx +


sin x + 4 cos xdx

I = A∫dx +










⇔{

A = 1B = −1/4


sin x + 4 cos xdx +


sin x + 4 cos xdx

I = A∫dx +










⇔{

A = 1B = −1/4


sin x + 4 cos xdx +


sin x + 4 cos xdx

I = A∫dx +







T½ch ph¥n d¤ng I =∫ a1 sin x + b1 cos x + c1

a sin x + b cos x + cdx Ph¥n t½ch

a1 sin x + b1 cos x + c1 =A (a sin x + b cos x + c)′ + B (a sin x + b cos x + c) + C= (Aa + Bb) cos x + (Ab − aB) sin x + (Bc + C )

�çng nh§t hai v¸:

Ab − aB = a1Aa + Bb = b1Bc + C = c1

gi£i t¼m A,B ,C .

I = A ln(a sin x + b cos x + c) + Bx +∫ Cdxa sin x + b cos x + c

T½ch ph¥n cuèi còng t½nh b¬ng c¡ch �êi bi¸n chung t = tanx2





T½ch ph¥n d¤ng I =∫ a1 sin x + b1 cos x + c1

a sin x + b cos x + cdx Ph¥n t½ch

a1 sin x + b1 cos x + c1 =A (a sin x + b cos x + c)′ + B (a sin x + b cos x + c) + C= (Aa + Bb) cos x + (Ab − aB) sin x + (Bc + C )

�çng nh§t hai v¸:

Ab − aB = a1Aa + Bb = b1Bc + C = c1

gi£i t¼m A,B ,C .

I = A ln(a sin x + b cos x + c) + Bx +∫ Cdxa sin x + b cos x + c

T½ch ph¥n cuèi còng t½nh b¬ng c¡ch �êi bi¸n chung t = tanx2





V½ dö. T½nh I =∫ (2 sin x + cos x + 3)dx

3 sin x + 4 cos x + 5Ph¥n t½ch

2 sin x + cos x + 3 = A(3 sin x + 4 cos x + 5)+B(3 sin x + 4 cos x + 5)′+C

2 sin x + cos x + 3 = (3A− 4B) sin x + (4A + 3B) cos x + (5A + C )

⇒

3A− 4B = 24A + 3B = 15A + C = 3

I = A∫dx + B

∫ d(3 sin x + 4 cos x + 5)3 sin x + 4 cos x + 5

+∫ Cdx3 sin x + 4 cos x + 5

I = Ax + ln(3 sin x + 4 cos x + 5) + I1 vîi I1 �¢ t½nh ð v½ dö tr÷îc.







2 sin x + cos x + 3 = A(3 sin x + 4 cos x + 5)+B(3 sin x + 4 cos x + 5)′+C2 sin x + cos x + 3 = (3A− 4B) sin x + (4A + 3B) cos x + (5A + C )

⇒

3A− 4B = 24A + 3B = 15A + C = 3

I = A∫dx + B










2 sin x + cos x + 3 = A(3 sin x + 4 cos x + 5)+B(3 sin x + 4 cos x + 5)′+C2 sin x + cos x + 3 = (3A− 4B) sin x + (4A + 3B) cos x + (5A + C )

⇒

3A− 4B = 24A + 3B = 15A + C = 3

I = A∫dx + B





Education

Toan 1 - Chuong 8