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Томский Государственный Университет Систем Управления и Радиоэлектроники

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Математический анализ

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<<< < Предыдущая 12 / 52 3 4 5 > Следующая >>>

1: Z

2: Z

3: Z

4: Z

5: Z

x dx

d(x2)

d(3+x2)

3+x2 = Z

3+x2 = 2 Z

3+x2

= ln juj + C

GDE

u = (3 + x2)

2 ln jx2 + 3j + C:

x dx

d(x2)

2 Z

= aarctg a

+ C

3+x4

3+(x2)2

a2 +u2

GDE

u = x

2 Z

3)2 +u2 =

arctgp

+C:

d(;5x)

; 5x)

p3;5x

;5 Z p3;5x

p3;5x

= 2 p

+ C

GDE

u = (3

5x)

5x + C:

Z pu

;

d(2x)

2 Z

= j u = (2x)

;

4x2

;

(2x)2

5)2

;

Z p

= arcsin a +C

= 2 arcsin p

2 arcsin p

+C:

a2 ;u2

x2 dx

d(x3)

1 1 d(3

;

5x3)

3 Z

5 Z

5x3

;

(3;5x3);1=5 d(3;5x3) =

u;1=5 du = 4(u)4=5 + C =

= ;15

1 5

4=5

;15

GDE u = (3 ; 5x

)

4(3;5x )

= ;12 q(3;5x )

+ C:

2 d(p

)

6: Z p

= Z

(1 + 4p

)2 =

(1 + 4p

d (4p

+1)

GDE u = (1+4px) =

= 2 4 Z

(1+4px)2

Z u2

= ;u

+ C

+ C:

= ;2

1 + 4p

7: Z

dx = Z e

= Z e

2 d(

x) = 2 Z e

x) =

eu du = eu

GDE u = p

= 2 ep

+ C:

+ C

x3 dx = 4 Z

8: Z

x3e1;x

dx = Z e1;x

e1;x

d(x4) =

= ;4 Z e1;x

d(;x4) = ;4 Z

e1;x

d(1 ; x4) = ;

4 e1;x

+ C:

2 ln x+5

9: Z

dx = Z

p2 ln x+5

= Z (2 ln x+5)1=2 d(ln x) =

= 2 Z (2 ln x+5)1=2 d(2 ln x+5) =

(u)1=2 d(u) =

3(u)3=2 + C

3=2

j =

3(2 ln x+5)

GDE

u = (2 ln x + 5)

2 3

(2 ln x+5)

+ C:

sin 3x

d(3x)

d(cos 3x)

10: Z

dx =

cos2 3x

= ;3

cos2 3x

;

cos2 3x

;

11:

u ; a

+ C

1 ln

cos 3x ; 2

+ C:

Z u2 ; a2

ju + aj

cos 3x + 2

sin(2=x)

dx = Z sin(2=x) x2

= Z sin(2=x) d ;x! =

1			2						1
= ;2 Z sin(2=x) d			x!= Z sin u du= ;cos u + C = 2 cos(2=x)+C:
	dx		1				dx		d(arcsin x)
12: Z p			= Z			p		= Z		=
		arcsin x		arcsin x
	1;x2						1;x2		arcsin x
= j Z duu = ln juj + C GDE					u = arcsin x j = ln j arcsin xj + C:

ex dx

= Z

d(ex)

d(2ex)

13: Z p

2 Z

4e2x+3

(2ex)2 +3

x 2

q(2e )

= j Z

= ln ju+pu2

+a2j + Cj= 2 ln j2ex+p4e2x+3 j+C:

u2 +a2

14:

15:

16:

d(3x=2)

= Z

= 2 Z

1 ; cos 3x

2 sin2

sin2

d(3x=2)

= 2

Z sin2

= ;ctgu + C

;3 ctg

+ C:

sin2 u

d(tgx)

= Z

cos2 x (2 + 5tgx)

2 + 5tgx

cos2 x

2 + 5tgx

d(5tgx + 2)

2 + 5tgx

u = ln juj + C

= 5 ln j2 + 5tgxj + C:

= Z

(1 + 9x2) arctg53x

arctg53x

(1 + 9x2)

d(3x)

d(arctg3x)

= 3 Z

3 Z

arctg53x

(1 + (3x)2)

arctg53x

= Z

= Z (u);5

u;4

du =

4 = ;

+ C = ;

+ C:

;

4u4

12 arctg4(3x)

(x+3) dx

x dx

3 dx

d(x2)

17: Z

2;7x2

= Z 2;7x2 +Z 2;7x2

dx= Z 2;7x2

;3 Z 7x2 ;2

d(2 7x2)

d(p

1 1

= ;2

2 ;7x2

x)2

2)2

;

= ln u

+ C

u ; a

+ C

Z u2 ; a2

j j

u + a

= ;

ln j2 ; 7x2j ;

p7x +; p2

+ C:

V dU:

1.2.3. iNTEGRIROWANIE PO ^ASTQM

w OSNOWE METODA INTEGRIROWANIQ PO ^ASTQM LEVIT FORMULA{SHEMA

R U dV = UV ; R

mETOD ISPOLXZUETSQ PRI INTEGRIROWANII NEKOTORYH TRANSCENDENT- NYH FUNKCIJ I PROIZWEDENIQ RAZNORODNYH FUNKCIJ. tOLXKO PO ^AS- TQM BERUTSQ INTEGRALY SLEDU@]IH TIPOW

tIP	Z		x				Z	x
		Pn(x) e dx						Pn(x) a dx
I
	Z	Pn(x) cos x dx					Z	Pn(x) sin x dx:
tIP	Z			Z	n		Z
		ln x dx,			Qr(x) ln	x dx		arcsin x dx
II
	Z	arctgx dx			Z Qr(x)	arcsin x dx	Z	Qr(x) arctgx dx:
tIP III	Z	e x cos x dx			Z e x sin x dx:

zDESX Pn(x) { MNOGO^LEN CELOJ STEPENI OTNOSITELXNO x DLQ INTEGRA- LOW I-OJ GRUPPY. dLQ INTEGRALOW II-OJ GRUPPY Qr(x) MOVET BYTX KAK CELOJ TAK I IRRACIONALXNOJ ALGEBRAI^ESKOJ FUNKCIEJ.

sHEMA INTEGRIROWANIQ PO ^ASTQM PREDPOLAGAET PREDWARITELXNOE RAZBIENIE PODYNTEGRALXNOGO WYRAVENIQ NA PROIZWEDENIE DWUH SO- MNOVITELEJ U I dV . pRI \TOM OSNOWNYM KRITERIEM PRAWILXNOSTI RAZBIENIQ SLUVIT TO, ^TO INTEGRAL W PRAWOJ ^ASTI SHEMY R U dV DOLVEN BYTX PRO]E ILI, PO KRAJNEJ MERE, NE SLOVNEE ISHODNOGO IN- TEGRALA R V dU, A FUNKCIQ V PO EE DIFFERENCIALU dV OPREDELQLASX DOSTATO^NO LEGKO.

pRIMENQQ METOD INTEGRIROWANIQ PO ^ASTQM, SLEDUET RUKOWODSTWO- WATXSQ SLEDU@]IM PRAWILOM:

1.eSLI W PODYNTEGRALXNOE WYRAVENIE WHODIT PROIZWEDENIE MNO- GO^LENA NA POKAZATELXNU@ ILI TRIGONOMETRI^ESKU@ FUNKCI@, TO W KA^ESTWE FUNKCII U BERETSQ MNOGO^LEN (INTEGRALY I-GO TIPA.)

2.zA U WSEGDA BERUTSQ LOGARIFMI^ESKAQ I OBRATNYE TRIGO- NOMETRI^ESKME FUNKCII (INTEGRALY II-GO TIPA.)

w INTEGRALAH III-EGO TIPA, KOTORYE NAZYWA@TSQ CIKLI^ESKIMI, WSE RAWNO, ^TO WZQTX ZA U; POKAZATELXNU@ ILI TRIGONOMETRI^ESKU@ FUNKCI@. sLEDUET U^ESTX, ^TO PRI NE- OBHODIMOSTI SHEMU INTEGRIROWANIQ PO ^ASTQM MOVNO PRIMENQTX POWTORNO.

rEKOMENDACII PO RAZBIENI@ PODYNTEGRALXNOGO WYRAVENIQ NA MNOVITELI PRIWEDENY W

TABLICE 1.4.

rEALIZACIQ SHEMY INTEGRIROWANIQ PO ^ASTQM tABLICA 1.4.

Z (x + 2) sin 3xdx

u = x + 2

du = dx

v = ;

dv = sin 3x dx

3 cos 3x

Z (5x ; 1)e;x=2 dx

u = 5x ;

du = 5dx

dv = e;x=2 dx v = ;2e;x=2

u = x2

du = 2xdx

Z x2 cos 3 dx

dv = cos 3 dx

v = 3 sin 3

Z ln xdx

u = ln x

du = dxx

dv = dx

v = x

ln2 x

2 ln x

Z p

u = ln2 x

du =

dv =

v = 2

ln(x2 + 5)dx

u = ln(x2 + 5)

du =

x2 + 5

dv = dx

v = x

arcsin x dx

u = arcsin x

du =

1 ; x2

dv = dx

v = x

Z x arctg x dx

u = arctg x

du =

1 + x2

dv = xdx

v =

1: Z (2x + 3) cos 5x dx=			U = 2x + 3				dV = cos 5x dx
			dU = 2	dx V =		Z		cos 5x dx=			1 sin 5x =
										5
	1				2x + 3					2
		2	Z sin 5x dx =
= (2x + 3)	5 sin 5x ;								sin 5x +				cos 5x + C:
		5				5					25
	1				1				1
2: Z x sin2 x dx= 2 Z x(1;cos 2x) dx=					2 Z	x dx;2 Z x cos 2x dx=
1	1			1		1				1
= 4 x2	; 2 Z x cos 2x dx = 4 x2				;	4	(x sin 2x +			2		cos 2x) + C:

wTOROJ INTEGRAL WY^ISLQLSQ PO ^ASTQM ANALOGI^NO PRIMERU 1.

x dx

U = x

dU = dx

sin2 x

dV =

V = Z

= ;ctgx

sin2 x

= ;x ctgx + R ctgx dx = ;x ctgx

+ ln j sin xj + C:

U = x

dU = dx

x sin x dx

sin x dx

V = Z

sin x dx

dV = cos3 x

cos3 x

;d(cos x) =

; Z

Z dxcos

2 cos

= x

;

2 tgx + C:

2 cos2 x

5: Z

x tg2x dx =

U = x

dU = dx

x =

dV = tg2x dx V =

tg2x dx = tgx

;

(iNTEGRAL

tg2x dx UVE BYL RE[EN RANEE.)

= x (tgx ; x) ; Z (tgx ; x) dx = x (tgx ; x) + ln j cos xj +

2 + C:

6: Z

x e;x dx=

U = x

dU = dx

dV = e;x dx V =

e;x dx =

;

e;x

= ;x e;x;Z

;e;x dx = ;x e;x;Z e;x d(;x) =;x e;x;e;x+C:

7: Z

dx= Z x2 ex

x dx= 2 Z x2 ex

d(x2) =

U = x2

dU = 2x dx

dV = ex2 d(x2) V = Z ex2 dx2 = ex2

d(x2) =

2 x2

2x dx = 2 x2 ex ;Z

2 x2

;ex

2 ex

(x2 ; 1) + C:

8: Z

U = x2 +2x;1

dU = (2x+2) dx

+2x;1) e

dx=

dV = e3x dx

V =

1 e3x

U = x+1

dU = dx

3(x2 +2x;1) e3x;3 Z (x+1)e3x dx =

dV = e3xdx V

e3x

(x+1) e3x ;

3(x2

+2x;1) e3x ; 3

3 Z e3x dx! =

e3x

2e3x

e3x

(x2 +2x;1) ;

(x+1) +

27 (9x2 +12x;13) + C:

ln x

U = ln x

dU = dx

9: 3

dx =

Z px2

dV =

V =

(x);2=3 dx

= 3 x1=3

px2

= 3 x1=3 ln x ; 3 Z x1=3 dxx = 3 x1=3 ln x ; 3 Z (x);2=3 dx =

= 3 x1=3 ln x ; 9x1=3 dx = 3p

(ln x

; 3) + C:

10: Z x2 ln2 x dx =

= ln2 x

dU = 2 ln x dx

V =

1 x3

= x

3 x

2 ln x dx

3x3 ln2 x ; Z

3x3 ln2 x ;

3 Z x2 ln x dx =

= ln x dU

= dxx

ln x

= x

;Z 3

dV = x2 dx V

x A

2x3

ln2 x ; 3

0 3 ln x ;

3 Z x2

dx1 = 3 ln2 x ;

ln x +

9 ln2 x ; 6 ln x + 2 + C:

11:

x2 ln(1 + x3) dx = Z

ln(1 + x3) x2 dx =

ln(1 + x3) d(x3 + 1) =

ln t dt =

U = ln t

dU = dtt

3 Z

dV = dt

V = t

(t ln t ; t) =

t ln t ; Z t t ! =

t ln t ; Z dt = 3

(ln t ; 1) =

1 + x3

ln(1 + x3) ; 1 + C:

arctg 2x dx = U = arctg 2x dU =

2dx

12:

1+4x2

dV = dx

V = x

= x arctg 2x ; Z

2x dx

= x arctg

2x ; Z

d(x2)

1+4x2

d(4x2 + 1)

= x arctg 2x ; 4 Z

1+4x2

= x arctg 2x ;

4 ln(1+4x2) + C:

xarctg x

13: Z p1 + x2 dx= = px2 +1 arctg

U = arctg x

dU =

1+x2

dV =

x dx

p1 + x2

d(x2

+ 1)

= 2 Z

= px2 + 1

1 + x2

+1 dx

x;Z

= px2

+ 1 arctg x;Z p

1+x2

x2 +1

= p

arctg x ; ln jx + p

j + C:

x2 + 1

U = arctgp

2x ; 1

dV = dx

14: Z arctgp

2 dx

dx= dU =

1+2x ; dx1

2x ; 1

V = x

2x ; 1

x dx

= x arctgp2x ; 1;Z

2x p

= x arctgp2x ; 1;

Z p

2x ; 1

arctgp

d(2x ; 1)

= x

;

p2x ; 1

= x arctg

1 p

2x ; 1 = x arctg

2x ; 1+C:

2x ; 1; 4

2x ; 1; 2

U = arcsin x

dU =

1;x2

arcsin x

V = Z

15: Z p1 + x dx= dV = p

= =

1 + x

d(x + 1)

= Z

= 2p1 + x

1 + x

= 2p1 + x arcsin x ; 2 Z

p1 + x p

1 ; x2

= 2p1 + x arcsin x ; 2 Z

p1 + x p

1 ; x

1 + x

= 2p

arcsin x+2

d(1;x)

1+x

arcsin x

1+x

Z p1 ; x

;

p1;x

= 2p

arcsin x + 4p

+ C:

1 + x

1 ; x

cIKLI^ESKIE INTEGRALY

16: Z e2x sin 3x dx =

U = e2x

dU = 2e2x

dV = sin 3x dx V =

;

1 cos 3x

= ;3e2x cos 3x + 3 Z

e2x cos 3x dx =

K INTEGRALU

Z e2x cos 3x dx TAKVE PRIMENIM FORMULU INTEGRIRO-

WANIQ PO ^ASTQM

=	U = e2x		dU = 2e2x			=
	dV = cos 3x dx	V =		1 sin 3x
	1	2	1	3		2

= ;3e2x cos 3x + 3			3e2x sin 3x ; 3				Z e2x sin 3x dx! =
	1	2			4	Z e2x sin 3x dx
= ;3e2x cos 3x +		9e2x sin 3x ; 9

tAKIM OBRAZOM, MY WNOWX PRI[LI K ISHODNOMU INTEGRALU I, ESLI OTBROSITX WSE PROMEVUTO^NYE WYKLADKI, TO MOVNO ZAPISATX RAWEN- STWO

Z e2x sin 3x dx =	e2x	4	Z e2x sin 3x dx
	9 (2 sin 3x ; 3 cos 3x) ; 9

KOTOROE MOVNO RASSMATRIWATX KAK URAWNENIE OTNOSITELXNO ISKOMOGO INTEGRALA. rE[AEM \TO URAWNENIE

4				e2x
1 + 9! Z e2x sin 3x dx =				9 (2 sin 3x ; 3 cos 3x):
	13	Z		e2x
oTS@DA POLU^AEM	9	Z	e2x sin 3x dx = 9 (2 sin 3x ; 3 cos 3x)
			e2x
Z e2x sin 3x dx = 13 (2 sin 3x ; 3 cos 3x) + C:
17: Z cos(ln x) dx =		U = cos(ln x) dU = ; sin(ln x)dxx			=
			dV = dx	V = x
				dx
= x cos(ln x) + Z			x sin(ln x)	x = x cos(ln x) + Z sin(ln x) dx =

K INTEGRALU Z sin(ln x) dx TAKVE PRIMENIM FORMULU INTEGRIROWA- NIQ PO ^ASTQM

= U = sin(ln x)		dU = cos(ln x)dxx	=
	dV = dx	V = x
			dx	! =
= x cos(ln x) +		x sin(ln x) ; Z x cos(ln x) x

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