Bài tập giải tích

8/11/2019 Bi tp gii tch

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GII TCH 1

HONG HI HBCH KHOA TPHCM

17th June 2013

HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 1 / 85


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1 GII HN DYPhp ton v gii hn dy

BI TP THC HNH2 HM S

Hm s c bn

Gii hn hmVCBV cng lnV. Hm lin tc

3 O HMCc php ton o hmQuy tc LHospitale

CNG THC TAYLORHONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 1 / 85


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4 KHO ST HM SCc tr hm sTim cn

5 TCH PHN BT NH

Phng php tnh v cc dng tch phnTch phn hu tTch phn lng gicTch phn v t

6 TCH PHN SUY RNG

Tp suy rng loi 1HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 2 / 85


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CHNG I: DY S



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I. Cc php ton v gii hn dy

Hai dy an, bnc lim an =a, lim bn =bth:

lim(an bn) =a blim(anbn) =ablim an

bn= a

bnu bn

=0, b

=0

lim anbn =ab nu an >0



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I. Cc php ton v gii hn dy

Hai dy an, bnc lim an =a, lim bn =bth:

lim(an bn) =a blim(anbn) =ablim an

bn= a

bnu bn

=0, b

=0

lim anbn =ab nu an >0CH :

1 a = , a

=0, a

0 = (a =0),a+= +(a>1), a=0(a>1).

2 Cc dng v nh:0

0

,

,

, 0.

,

0, 00, 1.

.HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 3 / 85


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II. Phng php tnh gii hn

VCB-VCL

Dy s{an}l VCL nu lim |an| = +, l VCB nulim an =0.



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II. Phng php tnh gii hn

VCB-VCL

Dy s{an}l VCL nu lim |an| = +, l VCB nulim an =0.Tng ng VCL

Hai VCL{an},{bn}gi l tng ng nu lim anbn

=1.K hiu : an bn

So snh bc VCLVCL{an}c bc nh hn{bn}nu liman

bn=0. K hiu:

an


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So snh bc cc VCL

ln


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Ta c ng thc sau: Nu an, bn, cn, dnln lt l ccVCL, v an

cn, bn

dn, th:lim

an

bn=limcn

dn.



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Gii hn kpNu un

xn

vn, lim

nun = lim

nvn =A. Khi :

limn

xn =A



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Gii hn kpNu un

xn

vn, lim

nun = lim

nvn =A. Khi :

limn

xn =A

V d 1.1a. lim

nsin x

x =0, ( >0)do1

xsin x

x 1

x

b. limn

2008

n

n=0 do 00limxx0

g=1 TH lnf lng


0


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Cng thc tnh gii hn dnh cho dng 00

Nu f , g khi x x0 TH limxx0 fg = limxx0

Cc tng ng thc c bn khi x

01/ ln(1+x) x 2/ ex 1 x

3/ (1+x) 1 x 4/1 cosx x2

2

5/ arctanx x 6/ shx x7/ chx 1 x

2

2



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Tm hm g(x) =Axn sao cho f gkhi x 01. f =sin

2x+

x+x2 x5

2. f =2x2

1

3. f = 4 x4 +x2 24. f =ln(1+sinx) x2

5. f =2sinx tan2x

6. f = x2artanx

x5 +x2 +17. f =ex2+x e2x


p dng tng ng VCB tnh cc gii hn sau


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1. limx x

3 1x 1

2.

limx03

1+3x

5

1+2x1+5x 41+2x

3.

limx0ex

2

1

1+sin2x 1

4.lim

x1+xx x

ln(1+

x2 1)

5. limx1ex

2+x

e2x

cosx

26.

limx0 2x+2arctan3x+3x2

ln(1+3x+sin2x) +xex


Gii hn dng m 1


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1/ limx0

cosx+arctan2x

1arctan2x 2/ lim

x1lnx x+1

x xx

3/ limx

e1/x +1xx

4/ limx+

2 arctanxx

5/ limx0(chx)1

1 cosx


IV. VCL


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NH NGHA(x)c gi l VCL khi x x0nu lim

xx0|(x)| =


IV. VCL


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xx0|(x)| =

So snh cc VCLXt 2 VCL (x), (x)khi x x0v t s : limxx0

=k.


IV. VCL


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xx0|(x)| =

So snh cc VCLXt 2 VCL (x), (x)khi x x0v t s : lim

xx0

=k.

a. k=0


IV. VCL


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xx0|(x)| =


xx0

=k.

a. k=0=o()


IV. VCL


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xx0|(x)| =


xx0

=k.

a. k=0=o()

b. k=


IV. VCL


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xx0|(x)| =


xx0

=k.

a. k=0=o()

b. k= l VCL cp CAO hn .


IV. VCL


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xx0|(x)| =


xx0

=k.

a. k=0=o()

b. k= l VCL cp CAO hn .c. kHH. v l 2 VCL cng cp. Nu k=1, ta c


CC TNH CHT CA V CNG LN TNG NG


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a. Nu f TH f .




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a. Nu f TH f .

b. Tng cc VCL khc cp s T v cng ln cp CAO

nht.




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a. Nu f TH f .


nht.c.

limxx0

f =A TH f A




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a. Nu f TH f .


nht.c.

limxx0

f =A TH f A

d. f g, f, g>0limxx0

g

=1 TH lnf lng



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TH T SO SNH CC VCL HAY GPlnx


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TH T SO SNH CC VCL HAY GPlnx


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f sau:

a. f = x2arctanx

x5 +x2 +1, x0= +

b. f =

x

x+2 2x+1+ x, x0= +

c. f =

x2 +1

x, x0=

d. f =

x2 +1 31 x2, x0= HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 31 / 85

TNH GII HN BNG CCH THAY TNG NGV CNG LN:


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V CNG LN:

1. limx(x2 +8x+3 x2 +4x+3)

2. limx

+

(2x 4x2 7x+4)

3. limx

arcsin(

x2 +x+x)

4. limx

x+

x3

+2x2

x+1



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5. limx+

( 6

x6 +x5 6x6 x5)

6.(*) limx+ ex

(sh

x

2

+x sh

x

2

x)


HM LIN TC


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Hm lin tca. Hm f(x)l lin tc ti x0nu tha mn:f(x0+) =f(x0) =f(x0)


HM LIN TC


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Hm lin tca. Hm f(x)l lin tc ti x0nu tha mn:f(x0+) =f(x0) =f(x0)b. Hm f(x)lin tc trn tp Dnu n lin tc ti miim ca D.


HM LIN TC


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Hm lin tca. Hm f(x)l lin tc ti x0nu tha mn:f(x0+) =f(x0) =f(x0)b. Hm f(x)lin tc trn tp Dnu n lin tc ti miim ca D.c. Cc hm s cp lin tc trn tp xc nh.


HM LIN TC


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Hm lin tca. Hm f(x)l lin tc ti x0nu tha mn:f(x0+) =f(x0) =f(x0)b. Hm f(x)lin tc trn tp Dnu n lin tc ti miim ca D.c. Cc hm s cp lin tc trn tp xc nh.d. Kho st tnh lin tc mt hm s( thng l hm

ghp) trn tp xc nh, ta ch cn kho st ti cc imghp.



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Tm gi tr a, b cc hm sau l hm lin tc trn tp

xc nha. f(x) =

e1x , x=0

a, x=0



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xc nha. f(x) =

e1x , x=0

a, x=0

b. f(x) =

(x

1)3, x

0

ax+b, 0


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xc nha. f(x) =

e1x , x=0

a, x=0

b. f(x) =

(x

1)3, x

0

ax+b, 0


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CHNG III: O HM

HM S


I. Cc php ton o hm v vi phn

o hm hm hp


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o hm hm hpf =f(u), u=u(x)

f

(x) =f

(u)u(x)



o hm hm hp


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f

(x) =f

(u)u(x)

o hm tham sHm tham s c cho bi: x=x(t),y=y(t). y(x) = y(t)

x(t) .

o hm cp 2: y(x) =[y(x)]

t

x(

t)



o hm hm hp


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f(x) =f(u)u(x)

o hm tham sHm tham s c cho bi: x=x(t),y=y(t). y(x) = y(t)

x(t) .

o hm cp 2: y(x) =[y(x)]

t

x(

t)

o hm cp cao

Cng thc Lebnitz:(fg)(n) =n=

Ckn f(k)gnk



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Cc cng thc o hm cp cao c bn

1. f =ax, f(n) =axlnna2. f = 1

x+a, f(n) = (1)n n!

(x+a)n+1

3. f =ln(x+a),

f(n) =

1x+a

(n1)

= (1)n1(n 1)!(x+a)n

.

4. f =sin(ax),

f(n) =ansin(ax+n

2

)


Tnh o hm cc hm s sau:


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Tnh o hm cc hm s sau:

1. y=aarcsin(1+x2)

x , a>0

2.y

= (ln

(1+x

))

sinx

3. y=

xnsin1

x

, x

=0, n

N

0 x=0

4. y=arctan(2x)

5. y=x|x2

5x+6

|6. x=ln(1+t2),y=t arctant



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Gi s f(x), g(x)c o hm vi mi x. Tnh ohm ca:a. y=f(x)g(x), f(x)>0

b. y=f(ex)ef(x)


Vi phn

Vi phn cp n ca hm y = f (x) ti x0


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Vi phn cp n ca hm y=f(x)ti x0

dn

f(x0) =fn

(x0)dxn


Vi phn

Vi phn cp n ca hm y = f (x) ti x0


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Vi phn cp n ca hm y f(x)ti x0

dn

f(x0) =fn

(x0)dxn

Quy tc vi phn

d(fg) =fdg+gdf

d(f

g) =

gdf fdgg2

d(f

) =f

1

dfd(af) =aflnadf, d(lnf) =df/f

Trong ,f,g l hm ph thuc x. CH : d2x=0.HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 41 / 85

Tnh df(x0):

a f =1 + ln x 1 x0 = 1


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a. f =x

+ln

x , x0= 1

b. f = x22xxx

, x0=2

Vit biu thc dytheo du v dv:

a. y=euvb. y= uv

u2 +v2

Tnh vi phn cp 2 ti x0ca hm:f =arctan

2+x22 x2


Tnh d2ytheo dx2 cc hm tham s:t2 t3


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1. x= t

1+t3

,y= t

1+t3

2. x= (t2 +1)et,y=t2e2t ti t=0

Tnh cc o hm cp cao sau: yn(

x0)

1. y= x2

1 x, n=8, x0=0.

2. y= (x2

2x)cos3x, n=101, x0=0.3. y=x2lnx, n=100, x0=1


Quy tc LHospitale

0


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Xt gii hn dng 0

0

hoc

Nu limxx0

f(x)g(x)

tn ti th:

limxx0 f(x)g(x)= limxx0 f(x)

g(x)

CH :

i vi cc dng v nh cn li, mun dng LHopitale,phi bin i v hai dng trn trc.


Tnh cc gii hn sau:1 lim xarcsinx

2

2 lim (x+1)ln(x+1) x


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1. limx0 xcosx

sinx

2. limx0 ex

x

1

3. limx0+

3+lnx2 3lnsinx 4. limx+

3xln(lnx)3

2x+3

lnx

5. limx

0+(xx

1)lnx 6. lim

x

+

(

2arctan

x)

x

7. limx0

1sinx

1x

8. lim

x0

1x 1

ex 1

9. limx1 x 1x1

10. limx0+(arcsinx)

tanx

11. limx

+

(3x2 +3x)1x 12. lim

x

0+

xxx1


II. KHAI TRIN TAYLOR


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nh nghaHm f(x)lin tctrn[a, b]v kh vi n cp n-1. Tnti f(n)(x0)vi x0 [a, b]. Khai trin Taylor ca f(x)ncp n ti x0l :

f(x) =n

k=0

f(n)(x0)

k! (x x0)k +o((x x0)n).

Trong : o((x

x0)n)l phn d Peano.


II. KHAI TRIN TAYLOR


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nh nghaHm f(x)lin tctrn[a, b]v kh vi n cp n-1. Tnti f(n)(x0)vi x0 [a, b]. Khai trin Taylor ca f(x)ncp n ti x0l :

f(x) =n

k=0

f(n)(x0)

k! (x x0)k +o((x x0)n).

Trong : o((x

x0)n)l phn d Peano.

Vix0=0 : Khai trin MACLAURINT.


Khai trin Maclaurint cc hm s cp n cp n:1 ex = 1 + x + x

2

+x3

+ +xn

+ o(xn)


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1. e =1+x+ 2! + 3! +...+ n! +o(x )

2. ln(1+x) =x x2

2 +x3

3! +...+ (1)n1x

n

n +o(xn)

3. (1+x) =1+x+( 1)2! x2 +...+( 1)...( n+1)

n! xn +o(xn)

4. sinx=x x3

3! +x5

5! +...+ (1)n x

2n+1

(2n+1)!+o(x2n+1)



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Khai trin TAYLOR trong ln cn x0n cp ncc


82/124

hm s sau:

1. y=ln(2x+1), x0=12

2. y=

x, x0=1

3. y= x2 +4x+4

x2 +10x+25 , x0= 2

4. y=e2x2

+2x1


THC HIN KHAI TRIN MACLAURINT N CP n:


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1. y=tanx, n=3

2. y=

1 2x+x3 31 3x+x2, n=3

3. y= 1

x2 +3x+2 , n=44. y=e

1+2x, n=2

5. y=sin(arctanx), n=4


Tnh o hm cp n:f(n)(0)


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1. y=ex2, n=6

2. y= 11+x+x2 , n=32

3. y= 11 x4 , n=60

4. y= (x2 +2x)cos(x2

x)


Tm bc cc v cng b sau khi x 0


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Tm bc cc v cng b sau khi x 0

1. y= (1+x2)sinx arctanx

2. y=ln(

1+2x ln(1+x))

3. y=ln(1+sinx) tanx

4. y=tanx xex2


P DNG KHAI TRIN MACLAURINT TNH GIIHN HM S:f (x)


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Dng thng gp tnh l I

=limx0f(x)

g(x)vi dng vnh00

Khai trin Maclaurint c t v mun bc b nht:f(x) =axn +o(xn), g(x) =bxm +o(xm).Xy ra cc trng hp sau:

1

Nu n=mTH I = a

b.2 Nu nmTH I =0


TNH CC GII HN SAU:

1 i 2

1 i 2


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limx0

1+sin2x

1

sin2x

ln(1+sinx) tanx

limx0

arctanx arcsinxtanx

sinx

limx0

ex2+x 1+2x x2ex

2 1limx x x

2ln(x+1

x

)limx0

3

1 2x2 xcotxxsinx


li ( )cot2x


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limx

0(cosx)cot x

limx0

(x+1)x+1 xln(cosx)

limx0 1+xcosx 1+2xln(1+x) sinx

limx

0

ecosx e 31 4x2

(1/x)arcsin2x 2coshx2



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CHNG IV: KHO ST

HM S

HONG HI H (HBK TPHCM) GII TCH 1 17th J 2013 56 / 85

CC TR HM S

1. TX.2 Gii phng trnh y=0 + tm nhng im y


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2. Gii phng trnh y =0 + tm nhng imy

KHNG TN TI. Tp hp nhng im ny gi l IMNG.3. Lp BBT kho st cc tr: Sp cc IM NG trn

BBT v xt du y(x0). Nu qua x0, y i du t + sang- th ti hm t cc i. Ngc li, hm t cc tiu.Qua x0y khng i du th hm khng c cc tr ti .Hoc:Xt du y(x0). Nu y(x0)>0 hm t CT. Nu

y(x0)


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TM TIM CN HM Sy

=f

(x

)

TIM CN NG : x=al TC nu limxa

=


TIM CN


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TM TIM CN HM Sy

=f

(x

)


=

TIM CN NGANG: y=bl TCN nu limxy=b


TIM CN


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TM TIM CN HM S y=f(x)


=

TIM CN NGANG: y=bl TCN nu limxy=bTIM CN XIN: y=ax+bl TCX nu

a= limxy

xb= limx(y ax)

HONG HI H (HBK TPHCM) GII TCH 1 17 h J 2013 58 / 85

Tnh li lm, im un


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Tm im unGii nghim y hoc im y khng tn ti.Xt du y qua cc im trn, nu qua x0, y i duth l im un.

Tnh li lmHm y=f(x) li trn khong (a,b) nu trn , f(x)


95/124

1. y=x+ 4x2

+1

2. y= (x+3)e1x

3. y= x2

|x| +4

4. y=

1+1xx

5. y=1 xe1

|x| 1x

6. y=xx1

x

x

7. y=x

x

x+4

8. y=1+xe3x


TM CC TR CC HM S SAU:

1 3


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1. y= 5 x2 4x 2. y= 1+3x4+x2

3. y= 3

x2 x 4. y= |x 5|(x 3)3

TM GTLN-GTNN ca hm y=f(x)trn khong [a,b]:y=x 2x, x [0, 5]

y= |x2

+2x 3|, x [1/2, 2]


KHO ST V V CC HM SAU


97/124

y=x+

3 x

y=e

1

x

x(x+2)y=

lnxx

HONG HI H (HBK TPHCM) GII TCH h J 6 /


98/124

CHNG I: TCH PHN

BT NH

O G ( C ) G C /

I. Phng php tnh v cc dng tch phn


99/124

i binXt tp:

f(x)dx. t x=(u). Khi , tp tr thnh:

f(x)dx= f((u))(u)du

( ) /

I. Phng php tnh v cc dng tch phn


100/124

i binXt tp:

f(x)dx. t x=(u). Khi , tp tr thnh:

f(x)dx= f((u))(u)duTng phn

udv=uv vdu

Tnh cc tp bt nh sau


101/124

1. 1

1+ 3

x+1dx

2.

dxex +e

x

3. arcsinx

x2 dx

4.

dx(x2 +4)2

II. Tch phn hu t

Xt tp hu t: P(x)Q(x)

dx, vi degP(x)


102/124

1. Phn tch mu thnh tch cc nhn t cha cc nhthc v tam thc bc 2 v nghim.

2. Gi s ta c: Q(x) = (x

a)(x

b)s1(ax2 +bx+c).

3. Khi :


II. Tch phn hu t

Xt tp hu t: P(x)Q(x)

dx, vi degP(x)


103/124

1. Phn tch mu thnh tch cc nhn t cha cc nhthc v tam thc bc 2 v nghim.

2. Gi s ta c: Q(x) = (x

a)(x

b)s1(ax2 +bx+c).

3. Khi :P(x)

Q(x)=

A

x

a+

B1

x

b+...+

Bs1(x

b)s1

+ Cx+D

ax2 +bx+c

4. Tnh Cx+D

ax2 +bx+cdx


Tnh Cx+D

ax2 +bx+cdx.

Phn tch: Cx+Dax2 + bx + c

= C

2a 2ax+b

ax2 + bx + c+


104/124

ax +bx+c 2a ax +bx+c

+(D bC2a) 1

a

(x b2a )

2

+4ac b2

4a2

.

Khi , tch phn cn tnh l:C

2aln(ax2

+bx+c) +

2Da

bC

2a21Karctan

x

b/2a

K

. Vi K2 =4ac b2

4a2HONG HI H (HBK TPHCM) GII TCH 1 17th June 2013 67 / 85

Tnh cc tch phn hu t sau


105/124

1.

xdx2x2 x+1

2. dx(x+2)(4x2 +8x+7)

3.

4x2 +4x 11(2x

1)(2x+3)2


III. TCH PHN HM LNG GIC


106/124

R(sinx, cosx)dx

t t=tan x

2

cho TH TQ.


III. TCH PHN HM LNG GIC


107/124

R(sinx, cosx)dx

t t=tan x

2

cho TH TQ.

Trng hp c bit1 R(sinx, cosx) = R(sinx, cosx), t=cosx2 R(sinx,cosx) = R(sinx, cosx), t=sinx3 R(

sinx,

cosx) =R(sinx, cosx),

t=tanx


Dng c bit


108/124

A1sinx+B1cosx+C1Asinx+Bcosx+C

dx.

Tch:A1sinx+B1cosx+C1=(Asinx+Bcosx) +(Acosx Bsinx) +


Tnh cc TPLG sau


109/124

1.

cosxdxsin3x+cos3x

2. dxsin3xcos5x

3.

dx

3 2sinx+cosx

4.

cosx 2sinxcosx+sinx

dx

5.

tan3xdx


IV. Tch phn v t c bn


110/124

1.

dxa2 x2 =arcsin xa +C

2. dxx2 a2 =ln|x+

x2

a2

|+C

3. dx

x2 +a2=ln|x+ x2 +a2| +C



111/124

R(x,

ax+b

cx+d

p1q1 ,

ax+b

cx+d

p2q2 )dx.

t

ax+b

cx+d =ts

vi sl BCNN ca q1, q2.


V d:

4 3 x 6 x


112/124

1

x

x

x( 6x+1)dx

x 1 1x+1+1 x 3x+2

x+ 3

x+2 x

4

x3(4 x)


TP EULER

R(x,

ax2 +bx+c)dx


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Phng php 1: Lng gic.

aR(x,

ax2 +bx+c)v mt trong 3 dng sau:

R(u,

u2 A2). t u= Asint

hocu=Acht(Acosht)

R(u, u2

+A2

). t u=AtantR(u,

A2 u2). t u=Asint


Phng php 2: Euler tng qut.


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1 t

ax2 +bx+c=x

a+tnu a>02

ax2 +bx+c=xt+

cnu c>0

3 ax2 +bx+c= (x x1)tnu x1l nghim ca tamthc bc 2.


Tnh cc tch phn v t sau:5

dx


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3/4

2+3x 2x2

1

0

dx

2x2 +2x+51

0

x

2

+2x+5dx


(x+2)dx


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3x2 +2x+2 (x+2)

2x2 x+1dx

21

e2xe2x +ex


Tch phn dng: dx2

.


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(x x0)

ax +bx+c

t t= 1x x0 . V du:

(3x+2)dx(x+1)

x2 +3x+3



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dx

(x+2)

3x2 +5

dxx1 ln2x



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CHNG VI. TCH PHN

SUY RNG


I. Tch phn suy rng loi 1

im k di


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im x0c gi l im k d ca hm f(x) nu:limxx0

f(x) =



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Cng thc tnh


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+a

f(x)dx= F(x)|+a =F(+) F(a).

Tp c gi l hi t nu gi tr ca n hu hn, ngcli tp phn k.


Tiu chun so snh

Xt hai tch phn sr loi 1: I=+

a

f(x)dx, J=+

a

g(x)dx,


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trong f(x), g(x) 0, x [x0, +), x0 a. Xt:k= lim

x+f(x)

g(x).

Nu k hu hn th I,J cng bn cht(THBk=1,I J)k=0, ch kt lun c I nu J hi t, khi I hi t.

k=+, ch kt lun c I nu J phn k, khi Iphn k.


Ch


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1. +a>0

dx

xhi t khi >1.

2. +

a

dxxlnxhi t khi >1 hoc =1, >1


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