Neodredeni integrali

Definicija. Za funkciju F : I R, gde je I interval, kazemo da je primitivnafunkcija funkcije f : I R ako je

F (x) = f(x)

za svako x I.

Teorema 1. Ako je F : I R primitivna funkcija za f : I R, tada je ifunkcija F (x) + C, C R, takode primitivna funkcija funkcije f.

Definicija. Skup svih primitivnih funkcija funkcije f naziva se neodredeniintegral funkcije f i oznacava sa

f(x)dx.

Osnovne osobine neodredenog integrala:

1o(

f(x)dx) = f(x) ,

2oF (x)dx = F (x) + C ,

3of(x)dx =

f(x)dx, R \ {0} ,

4o (f(x) g(x))dx = f(x)dx g(x)dx .

Cuvajmo drvece. Nemojte stampati ovaj materijal, ukoliko to nije neophodno.

1

2Tablica integrala

xndx =

xn+1

n+ 1+ C, n 6= 1

dx

x= ln |x|+ C

dx

1 + x2={

arctanx+ C, arcctg x+ C

dx

x2 1 =12

lnx 1x+ 1

+ Cdx

1 x2 ={

arcsinx+ C, arccosx+ C

dxx2 1 = ln |x+

x2 1|+ C

exdx = ex + Caxdx =

ax

ln a+ C, a > 0, a 6= 1

sinx dx = cosx+ C cosx dx = sinx+ Cdx

sin2 x= cotx+ C

dx

cos2 x= tanx+ C

sinhx dx = coshx+ C

coshx dx = sinhx+ Cdx

sinh2 x= cothx+ C

dx

cosh2 x= tanhx+ C

Osnovne metode integracije:

METOD SMENEAko je

f(x)dx = F (x) +C, tada je

f(u)du = F (u) +C. Uzmimo da je

x = (t), gde je neprekidna funkcija zajedno sa svojim izvodom . Tada

f((t)

)(t)dt = F

((t)

)+ C.

METOD PARCIJALNE INTEGRACIJEAko su u i v diferencijabilne funkcije i funkcija uv ima primitivnu funkciju,

tada je udv = uv

vdu.

3Primeri najcesce koriscenih smena:

dx =1ad(ax+ b), a 6= 0, smena: t = ax+ b, dt = a dx,

xdx =12d(x2), smena: t = x2, dt = 2xdx,

xdx =12a

d(ax2 + b), a 6= 0, smena: t = ax2 + b, dt = 2axdx,

xn1dx =1nd(xn), n 6= 0, smena: t = xn, dt = nxn1dx,

dxx

= 2 d(x), smena: t =

x, 2t dt = dx,

dxax+ b

=2ad(ax+ b), a 6= 0, smena: t = ax+ b, 2

at dt = dx,

dx

x= d(lnx), smena: t = lnx, dt =

dx

x,

dx

x= d(

ln ax), a 6= 0, smena: t = ln ax, dt = dx

x,

exdx = d(ex), smena: t = ex, dt = exdx,

eaxdx = d(

1aeax), a 6= 0, smena: t = eax, dt = aeaxdx,

sinx dx = d(cosx), smena: t = cosx, dt = sinx dx,cosx dx = d(sinx), smena: t = sinx, dt = cosx dx,

dx

cos2 x= d(tanx), smena: t = tanx,

dt

1 + t2= dx,

dx

sin2 x= d(cotx), smena: t = cotx, dt

1 + t2= dx,

sinhx dx = d(coshx), smena: t = coshx, dt = sinhx dx,

coshx dx = d(sinhx), smena: t = sinhx, dt = coshx dx,

dx

cosh2 x= d(tanhx), smena: t = tanhx,

dt

t2 1 = dx,dx

sinh2 x= d(cothx), smena: t = cothx, dt

1 t2 = dx.

4METOD REKURZIVNIH FORMULAOdredivanje integrala In =

fn(x)dx, funkcija koje zavise od celobro-

jnog parametra n, moguce je primenom parcijalne integracije ili nekog drugogmetoda, svesti na izracunavanja integrala Im, 0 m < n, istog tipa. Takverelacije, oblika

In = (In1, . . . , Ink),

zovemo rekurzivne formule. Da bi se na osnovu njih odredila vrednost integralaIn, neophodno je poznavati uzastopnih k integrala In1, . . . , Ink, kao i prvihk integrala I0, . . . , Ik1.

INTEGRALI SA KVADRATNIM TRINOMOM

1. I =

mx+ nax2 + bx+ c

dx

m = 0 : I =

ndx

ax2 + bx+ c=n

a

dx

x2 + bax+ca

=n

a

dx(

x+ b2a)2+ ca b24a2

=t = x+ b2a , h2 = | ca b24a2 | = na

dt

t2 h2 .

m 6= 0 : I = m

xdx

ax2 + bx+ c+ n

dx

ax2 + bx+ c= mI1 + nI2;

I1 =

xdx

ax2 + bx+ c=

12a

2ax dx

ax2 + bx+ c

=12a

2ax+ b bax2 + bx+ c

dx =12a

d(ax2 + bx+ c)ax2 + bx+ c

b2aI2

=12a

ln |ax2 + bx+ c| b2aI2.

Izracunavanje integrala I2 opisano je u prethodnom slucaju (m = 0).

2. I =

mx+ nax2 + bx+ c

dx, analogno tipu integrala pod 1.

3. I=

dx

(mx+ n)ax2+ bx+ c

, smenom mx+n =1t

svodi se na slucaj 2.

4. I =

ax2 + bx+ c dx, resava se dovodenjem na potpun kvadrat izraza

pod korenom

ax2+bx+c =a((x+

b

2a

)2+c

a b

2

4a2

)=a(t2h2), t=x+ b

2a, h2 =

ca b

2

4a2

.

5INTEGRACIJA RACIONALNIH FUNKCIJA

FunkcijaP (x)Q(x)

, P (x) i Q(x) su polinomi, jeste prava racionalna funkcija

ukoliko je stepen polinoma P (x) manji od stepena polinoma Q(x).Prava racionalna funkcija

P (x)Q(x)

=P (x)

ni=1

(x ai)si mi=1

(x2 + pix+ qi

)ki ,

gde su nule kvadratnih trinoma x2 + pix + qi, i = 1, . . . ,m kompleksne, imarazvoj

P (x)Q(x)

=A11x a1 +

A12(x a1)2 + +

A1s1(x a1)s1

+A21x a2 +

A22(x a2)2 + +

A2s2(x a2)s2

+ +

An1x an +

An2(x an)2 + +

Ansn(x an)sn

+B11x+ C11x2 + p1x+ q1

+B12x+ C12

(x2 + p1x+ q1)2+ + B1k1x+ C1k1

(x2 + p1x+ q1)k1

+B21x+ C21x2 + p2x+ q2

+B22x+ C22

(x2 + p2x+ q2)2+ + B2k2x+ C2k2

(x2 + p2x+ q2)k2+ +

Bm1x+ Cm1x2 + pmx+ qm

+Bm2x+ Cm2

(x2 + pmx+ qm)2+ + Bmkmx+ Cmkm

(x2 + pmx+ qm)km.

Konstante Aij , Bij , Cij nalazimo metodom neodredenih koeficijenata.

INTEGRACIJA IRACIONALNIH FUNKCIJA:

1. IntegralR[x,(ax+ bcx+ d

)p1/q1,(ax+ bcx+ d

)p2/q2, . . .

]dx, gde je R neka ra-

cionalna funkcija i pi Z, qi N, uvodenjem smeneax+ bcx+ d

= tn, n = NZS{q1, q2, . . . },

svodi se na integral racionalne funkcije.

62.

Pn(x)dxax2 + bx+ c

= Qn1(x)ax2 + bx+ c+

dx

ax2 + bx+ c,

koeficijente polinoma Qn1(x) i odredujemo metodom neodredenihkoeficijenata nakon diferenciranja gornje jednakosti.

3. Integral

dx

(x+ )nax2 + bx+ c

smenom x+ =1t

svodi se na slucaj

pod 2.

4. IntegralR(x,

ax2 + bx+ c)dx, gde je R neka racionalna funkcija,

primenom Ojlerovih ili trigonometrijskih/hiperbolickih smena svodi sena integral racionalne funkcije.

Ojlerove smene:

I) Ukoliko je a > 0, uvodimo smenuax2 + bx+ c = ta x ;

II) Kada je c 0, smena glasi ax2 + bx+ c = xtc ;III) U slucaju ax2 + bx + c = a(x x1)(x x2), x1, x2 R, koristimo

smenuax2 + bx+ c = t(x x1).

Trigonometrijske i hiperbolicke smene:

(a) ako se u integralu javia2 x2, smena: x = a sin t ili x = a cos t;

(b) ako se u integralu javia2 + x2, smena: x = a tan t ili x = a sinh t;

(c) ako se u integralu javix2 a2, smena: x = a

sin tili x =

a

cos tili

x = a cosh t.

5. Integracija binomnog diferencijala:

I =xr(a+ bxq

)pdx, za a, b R, p, q, r Q

a) U slucaju p Z, q = q1m, r =

r1m, gde su q1, r1 Z i m N,

uvodimo smenu: x = tm, dx = mtm1dt. Polazni integral tada

postaje I = mtr1+m1

(a+ b tq1

)pdt.

b) Kada jer + 1q Z, p = s

m, smena: a+ bxq = tm ili x = t1/q.

c) Za slucajr + 1q

+ p Z, p = sm, smena: x = t1/q ili axq + b = tm.

7INTEGRACIJA TRIGONOMETRIJSKIH FUNKCIJA

Neka je R neka racionalna funkcija. Posmatramo integral oblika

I =R(sinx, cosx)dx.

1. U slucaju

R( sinx, cosx) = R(sinx, cosx), smena: t = cosx; R(sinx, cosx) = R(sinx, cosx), smena: t = sinx; R( sinx, cosx) = R(sinx, cosx), smena: t = tanx.

Inace, smenom t = tanx

2, dx =

2dt1 + t2

, imajuci u vidu transformacije

sinx =2t

1 + t2, cosx =

1 t21 + t2

,

polazni integral postaje integral racionalne funkcije.

2. Transformacije proizvoda u zbir

sin ax sin bx =12(

cos(a b)x cos(a+ b)x),sin ax cos bx =

12(

sin(a b)x+ sin(a+ b)x),cos ax cos bx =

12(

cos(a b)x+ cos(a+ b)x),svode integrale oblika

sin ax sin bx dx,

sin ax cos bx dx,

cos ax cos bx dx,

na elementarne integrale.

3. Za I =

sinm x cosn xdx, gde su m,n Z razlikujemo sledece slucajeve:

n = 2k + 1 (m = 2k + 1 analogno):

I =

sinm x(cos2 x)k cosxdx = t = sinxdt = cosxdx

= tm(1t2)kdt;

8 m = 2k, n = 2l : Snizavamo stepen trigonometrijskih funkcija podintegralom primenom transformacija

sin2 x =1 cos 2x

2, cos2 x =

1 + cos 2x2

;

m = k, n = l, k, l N :

I =

dx

sink x cosl x=

1sink x cosl2 x

dx

cos2 x

=

1sink xcosk x

cosl+k2 x

dx

cos2 x

=

tank x(1 + tan2 x

) l+k22 d(tanx)

4.R(

tanx)dx, smena: t = tanx, dx =

dt

1 + t2, takode

R(

cotx)dx, smena: t = cotx, dx = dt

1 + t2.

5. Hiperbolicke funkcije analogno, uz napomenu o jednakostima koje vazeza ove funkcije:

sinhx =ex ex

2, coshx =

ex + ex

2, tanhx =

sinhxcoshx

,

cosh2xsinh2x=1, sinh 2x=2 sinhx coshx, cosh 2x=cosh2x+sinh2x,

sinhx

2= sgn(x)

coshx 1

2, cosh

x

2=

coshx+ 1

2,

sinhx =tanhx

1 tanh2 x, coshx =

11 tanh2 x

,

arc sinhx = lnx+x2 + 1 , arc coshx = ln x+x2 1 ,

arc tanhx =12

lnx+ 1x 1

, arc cothx = 12 lnx 1x+ 1

,arc tanh

x

a=

12

lnx+ ax a

, arc coth xa = 12 lnx ax+ a

,

9Zadaci

1. Dokazati za a > 0 i b 6= c :

a)

dx

x+ b= log |x+ b|+ C, b)

dx

(x+b)(x+c)=

1c b ln

x+ bx+ c

+C,c)

dx

a2 + x2=

1a

arctanx

a+ C, d)

dx

a2 x2 =12a

lnx+ ax a

+ C,e)

x2

a2 + x2dx = x a arctan x

a+ C, f)

x2

x2 a2 dx = xa

2lnx+ ax a

+ C,g)

dxa2 x2 = arcsin

x

a+ C, h)

dxx2 a2 =ln

x+x2 a2+C,i)

x dx

a2 x2 = 12

ln |a2 x2|+ C, j)

x dxa2 x2 =

a2 x2 + C,

k)

dx

x(a2x2) =

12a2

ln x2a2x2

+C, l) dxxa2x2 =

12a

loga2x2aa2x2+a

+C,m)

a2 x2dx = x

2

a2 x2 + a

2

2arcsin

x

a+ C,

n)

x2 a2dx = x2

x2 a2 a

2

2lnx+x2 a2+ C.

Resenje: a)

dx

x+ b=d(x+ b)x+ b

= log |x+ b|+ C.

b)

dx

(x+b)(x+c)=

1cb

(x+c)(x+b)(x+b)(x+c)

dx=1cb

( dxx+b

dx

x+c

)1.a)=

1cb log

x+bx+c

+C.c) Kako je I =

dx

a2 + x2=

1a2

dx

1 +(xa

)2 ,uvodenjem smene t =

x

a, dx = a dt, polazni integral postaje tablicni

I =1a

dt

1 + t2=

1a

arctan t+ C.

Vracanjem smene, konacno dobijamo I =1a

arctanx

a+ C.

10

d) Integral I =

dx

a2 x2 jeste zapravo specijalan slucaj integrala pod b) zaa = a i b = a. Tako na osnovu rezulata pod b) zakljucujemo da je

I =12a

lnx+ ax a

+ C.e)

x2

a2+x2dx=

a2+x2a2a2+x2

dx=dxa2

dx

a2+x21.c)= xa arctanx

a+C.

f)

x2

x2a2 dx=x2a2+a2x2a2 dx=

dx+a2

dx

x2a21.d)= xa

2lnx+ ax a

+C.g) I =

dxa2 x2 =

1a

dx

1 x2a2

= t = xadx = a dt

= dt1 t2= arcsin t+ C = arcsin

x

a+ C.

h) I =

dxx2 a2 =

1a

dxx2

a2 1

= t = xadx = a dt

= dtt2 1= ln

t+t2 1+C1 = ln xa

+

x2

a2 1+C1 = ln |x+x2 a2|+C,

gde je C = C1 ln a.

i) I =

xdx

a2x2 = t=a2x22xdx=dt

=12dt

t=1

2ln |t|+C=1

2lna2x2+C.

j) I =

xdxa2x2 =

t=a2x22xdx=dt=12

t

12dt=1

2t

12

1/2+C=

a2x2+C.

k) I =

dx

x(a2 x2) =

1a2

a2 x2 x2x(a2 x2) dx = 1a2

dx

x 1a2

x dx

a2 x2

1.i)=

1a2

log |x| 12a2

loga2 x2+ C = 1

2a2log x2a2 x2

+ C.l) I =

dx

xa2 x2 =

x dx

x2a2 x2 =

t =a2 x2, x2 = (t2 a2)

dt = x dxa2x2

=

dt

t2 a21.d)=

12a

log t at+ a

+ C = 12a

loga2 x2 aa2 x2 + a

+ C.

11

m) Oblast definisanosti podintegralne funkcijea2 x2 integrala

I =

a2 x2 dx

je segment Df =[ a, a]. To opravdava uvodenje smene x = a sin t, pri tom

je dx = a cos t dt, kada t na primer prolazi segment t [pi/2, pi/2]. Za takavizbor vrednosti nove nezavisno promenljive t vazi cos t 0. Tada je

I = a

a2 a2 sin2 t cos t dt = a2

1 sin2 t cos t dt

= a2

cos2 t dt =a2

2

(1 + cos 2t)dt =

a2

2

(dt+

cos 2t dt

)=a2

2

(t+

12

cos 2t d(2t)

)=a2

2

(t+

12

sin 2t)

+ C.

Primetimo da je sin t =x

a, t = arcsin

x

a, kao i

sin 2t = 2 sin t cos t = 2 sin t

1 sin2 t = 2xa

1

(xa

)2=

2xa2

a2 x2.

Tada imamo

I =a2

2

(arcsin

x

a+

x

a2

a2 x2

)+ C =

x

2

a2 x2 + a

2

2arcsin

x

a+ C.

n) I =

x2 a2dx = u =

x2 a2, du = x dx

x2a2dv = dx, v =

dx = x

= x

x2 a2

x2 dxx2 a2 = x

x2 a2

x2 a2 a2

x2 a2 dx

= xx2 a2

( x2 a2 dx a2

dxx2 a2

)1.h)= x

x2 a2 I a2 ln |x+

x2 a2|.

Odatle je2I = x

x2 a2 a2 ln |x+

x2 a2|+ C1,

12

odnosno

I =x

2

x2 a2 a

2

2ln |x+

x2 a2|+ C.

Jos jedan pogodan nacin izracunavanja integrala I dat je u nastavku.

I =

x2 a2dx =

x2 a2x2 a2 dx = (c0x+ c1)

x2 a2 +

dxx2 a2

Diferenciranjem ove jednakosti nalazimo

x2 a2x2 a2 = c0

x2 a2 + (c0x+ c1) x

x2 a2 + dxx2 a2 .

Posle mnozenja sax2 a2 i sredivanja izraza, dobijamo

x2 a2 = 2c0x2 + c1x c0a2 + .

Izjednacavanjem koeficijenata uz iste stepene promenljive x dolazimo do sis-tema jednacina

1 = 2c0, 0 = c1, a2 = c0a2 + ,

cije je resenjec0 = 1/2, c1 = 0, = a2/2.

Time trazeni integral postaje

I =x

2

x2 a2 a

2

2

dxx2 a2

1.h)=

x

2

x2 a2 a

2

2ln |x+

x2 a2|+C.

2. Odrediti sledece integrale:

10

x(x+ 2)3dx 20x2(1 + x)20dx 30

x dx

(x+ 9)10

40

dx

x2 + 4x+ 450

dx

x2 + x 2 60

1 2x2 + x4 dx

Resenje: 10

x(x+ 2)3dx =x1/2

(x3 + 6x2 + 12x+ 8

)dx

13

= (

x7/2 + 6x5/2 + 12x3/2 + 8x1/2)dx

=x(2

9x4 +

127x3 +

245x2 +

163x)

+ C.

20x2(1 + x)20dx=

t=1 + xdt = dx = (t 1)2t20dt= (t2 2t+ 1)t20dt

= (

t22 2t21 + t20)dt = t2323 2 t

22

22+t21

21+ C

=(1 + x)23

23 (1 + x)

22

11+

(1 + x)21

21+ C.

30

x dx

(x+ 9)10= t = x+ 9dt = dx

= (t 9)t10dt = (t9 9t10)dt=t8

8 9t9

9 + C =1

(x+ 9)9 1

8(x+ 9)8+ C.

40

dx

x2 + 4x+ 4=d(x+ 2)(x+ 2)2

=1x+ 2

+ C.

50

dx

x2 + x 2 =

dx

(x+ 2)(x 1)1.b)=

13

lnx 1x+ 2

+ C.60

1 2x2 + x4 dx =

(1 x2)2 dx =

(1x2) dx = (xx3

3

)+C,

za 1 x2 0, tj. x [1, 1].Za x (,1) ili x (1,+) vazi

1 2x2 + x4 dx =

(1 x2) dx = (x x3

3

)+ C1.

14

3. Metodom smene odrediti:

10

x3dx

x8 2 20

x7dx

(1 + x4)230

x8dx

(x3 1)3

40

x9

x5 + 1dx 50

dx

x(x10 + 1

)2 60 1 x7x(1 + x7) dx70

xdx1 + x4

80

dx

xx2 + 1

90 (

x+12

)x2 + x+ 1 dx

Resenje: 10

x3dx

x8 2 =14

d(x4)

(x4)2 (2)21.d)=

18

2lnx4 2x4 +2

+ C.

20

x7dx

(1 + x4)2=

x4

(1 + x4)2x3dx =

t = 1 + x4dt = 4x3dx= 14

t 1t2

dt

=14

(t1 t2)dt = 1

4(

ln |t|+ 1/t)+ C=

14

(ln(1 + x4

)+

11 + x4

)+ C.

30

x8dx

(x3 1)3 =

(x3)2

(x3 1)3x2dx =

t = x3 1dt = 3x2dx= 13

(t+ 1)2

t3dt

=13

(1t

+2t2

+1t3

)dt =

13

(ln |t| 2

t 1

2t2)

+ C

=13

(ln |x3 1| 2

x3 1 1

2(x3 1)2)

+ C.

40

x9

x5 + 1dx=

15

x5d(x5)

x5 + 1=

15

x5 + 1 1x5 + 1

d(x5)

=15

(d(x5) d(x5 + 1)

x5 + 1

)=x5

5 1

5log1 + x5+ C.

15

50

dx

x(x10 + 1

)2 = x9dxx10(x10 + 1

)2 = t = x10dt = 10x9dx = 110

dt

t(t+ 1)2

=110

t+ 1 tt(t+ 1)2

dt =110

dt

t(t+ 1) 1

10

d(t+ 1)(t+ 1)2

1b)=

110

log tt+1

+ 110(t+1)

+C=110

log x10x10+1

+ 110(x10+1

)+C.

60

1x7x(1+x7

) dx= 1x7x7(1+x7

) x6dx= t = x7dt = 7x6dx= 17

1tt(1+t)

dt

=17

1 + t 2tt(1 + t)

dt =17

log|x7|(

x7 + 1)2 + C.

70

xdx1 + x4

=12

d(x2)

1 + (x2)2=

12

lnx2 +1 + x4 + C.

80

dx

xx2 + 1

=

x dx

x2x2 + 1

=

t =x2 + 1, x2 = t2 1

dt = x dxx2+1

=

dt

t2 11.d)=

12

ln t 1t+ 1

+ C2 = 12 lnx2 + 1 1x2 + 1 + 1

+ C2.

90 (

x+12

)x2 + x+ 1 dx =

t = x2 + x+ 1dt = (2x+ 1)dx = 2(x+ 12)dx

=12

t dt =

12t3/2

3/2+ C =

13

(x2 + x+ 1)3 + C.

16

4. Metodom smene odrediti:

10

dx5x 2 2

0

x

1 4x dx 30

4

1 4x dx

40

dxx+ 1 +

x 1 5

0

dx

x+ 3x 2

60

x51 x2 dx 7

0

x dx(

1 + x2)3/2 80 x3dx(x2 + 1)3x2 + 1

90

2(2 x)2

3

2 x2 + x

dx

Resenje: 10

dx5x 2 =

25

d(

5x 2) = 25

5x 2 + C.

20

x1 4x dx =

t = 1 4x, x = 1t24dx = 12 t dt = 18

(t2 1)dt

=18

( t33 t)

+ C =112

1 4x (1 + 2x) + C.

30

4

1 4x dx= t = 41 4x, t4 = 1 4xx = 1t44 , dx = t3dt

= t4dt= t55 + C= 4

1 4x 4x 15

+ C.

40

dxx+1+

x1 =

x+1x1

x+ 1(x 1) dx=12

(x+1x1 ) dx

=12

x+ 1 d(x+ 1) 1

2

x 1 d(x 1)

=12

(x+1)3/2

3/2 1

2(x1)3/2

3/2+C=

13((x+1)3/2(x1)3/2)+C.

17

50

dxx+3x2 =

x+3+

x2

(x+3)(x2) dx =15

(x+ 3 +

x 2 ) dx

=215((x+ 3)3/2 + (x 2)3/2)+ C.

60

x51 x2 dx =

x4

1 x2 xdx = t =

1 x2, x2 = 1 t2

dt = x dx1x2

=

(1 t2)2dt =

1 x2

(15

(1 x2)2 23

(1 x2) + 1)

+ C.

70

x dx(1+x2

)3/2 = t=

11+x2

dt= x dx(1+x2)3/2

=dt=t+C= 1

1+x2+C.

80

x3dx(x2 + 1

)3x2 + 1

=

t =x2 + 1, x2 = t2 1

dt = x dxx2+1

=t2 1t6

dt

= (t4t6)dt= t33 t55 +C= 15(x2+1)5 13(x2+1)3 +C.

90

2(2 x)2

3

2 x2 + x

dx=

t = 3

2+x2x

dt = 4/3(2x)2

3

(2x2+x

)2dx

=32

t dt

=34t2 + C =

34

3

(2 + x2 x

)2+ C.

18

5. Metodom smene izracunati:

10

dx

x lnx20

x+ lnxx

dx 30

ln 2xx ln 4x

dx

40

dx

x

lnx50

ln(1 + x) lnxx(1 + x)

dx 60xx(1 + log x)dx

70x(

ln(1 + x) + ln(1 x))2x2 1 dx

Resenje: 10

dx

x lnx=d(lnx)

lnx= ln | lnx|+ C.

20

x+ lnxx

dx=x1/2dx+

lnx

dx

x=x1/2

1/2+

lnx d(lnx)

=2x+

12

ln2 x+ C.

30 S obzirom da je d(ln 4x)=dx

xi ln 4x=ln 2x+ln 2, uvodimo smenu t=ln 4x :

ln 2xx ln 4x

dx=t ln 2

tdt =

dt ln 2

dt

t= t ln 2 ln |t|+ C

= ln 4x ln 2 ln | ln 4x|+ C.

40

dx

x

lnx=

(lnx)1/2d(lnx) = 2

lnx+ C.

50

ln(1+x)lnxx(1+x)

dx=

ln(1+ 1x

)x(1+x)

dx=

t=ln

(1+ 1x

)dt= 1

1+1x

dxx2

= dxx(1+x)

=

t dt = t

2

2+ C = 1

2ln2(

1 +1x

)+ C.

60xx(1 + log x)dx =

t=xx, log t=x log x

dtt = (1 + log x)dx

= dt= t+ C=xx + C.

19

70x(

ln(1+x)+ln(1x))2x21 dx=

ln2(1x2) x dx

x21 = t=ln

(1x2)

dt= 2x dx1x2

=

12

t2dt =

16t3 + C =

16

ln3(1 x2)+ C.

6. Metodom smene odrediti:

10

dx

ex + 120

dx1 + e2x

30

ex 11 + 3ex

dx

40ex

21x dx 50x e(1+x2)1/2

(1 + x2)3dx 60

x+ 1

x(1 + x ex

) dxResenje: 10

dx

ex+1=

dx

ex(1+ex

)= d(1+ex)1+ex

= ln (1+ex)+C= ln

ex

1 + ex+ C = x ln (1 + ex)+ C.

Isti rezultat mozemo da dobijemo i na sledeci nacin.dx

ex + 1=

1 + ex exex + 1

dx=dx

exdx

ex + 1=x

d(ex + 1

)ex + 1

= x ln (ex + 1)+ C.

20

dx1 + e2x

=

dx

exe2x + 1

= t = exdt = exdx

= dtt2 + 1= ln |t+

t2 + 1|+ C = ln(ex +

e2x + 1) + C

= lnex

1 +

1 + e2x+ C = x ln (1 +1 + e2x)+ C.

30

ex 11 + 3ex

dx =ex 1ex + 3

ex dxex 1 =

t =ex 1, ex = t2 + 1

dt = 12ex dxex1

=2

t2

t2+4dt

1.e)= 2t4 arctan t

2+C=2

ex14 arctan

ex12

+C.

20

40ex

21x dx = t=x21dt=2x dx

= 12etdt = 1

2ex

21 + C.

50xe(1+x2)

12(1+x2)3

dx=

t=e(1+x2)

12

dt= e(1+x2)

12 x dx

(1+x2)32

=dt= t+C=e(1+x

2)12+C.

60

x+ 1x(1 + x ex

) dx= (x+ 1)exxex(1 + x ex

) dx = t = xexdt = (1 + x)exdx

=

dt

t(t+ 1)1.b)= ln

tt+ 1

+ C = ln xexxex + 1

+ C.7. Metodom smene odrediti:

10

tanx dx 20

dx

cosx30

dx

1 + cosx

40

cotx dx 50

dx

sinx60

dx

1 + sinx

Resenje: 10

tanx dx =

sinx dxcosx

= d(cosx)

cosx= ln | cosx|+ C.

20

dx

cosx=

cosx dxcos2 x

=

d(sinx)1 sin2 x

1.b)=

12

ln1 + sinx1 sinx + C.

=12

ln(1 + sinx)2

1 sin2 x + C = ln1 + sinx

cosx

+ C= log

cos x2 + sin x2cos x2 sin x2+C = log 1 + tan x21 tan x2

+C.30

dx

1 + cosx=

dx

2 cos2 x2=

d(x2

)cos2 x2

= tanx

2+ C.

40

cotx dx = log | sinx|+ C.

21

50

dx

sinx= log

tan x2

+ C.

60

dx

1 + sinx=x = t+ pi/2, dx = dt = dt

1 + sin(t+ pi/2)=

dt

1+cos t7.30= tan t/2+C=tan

x pi22

+C=tan x2tan pi4

1+tan x2 tanpi4

+C

=tan x2 1tan x2 + 1

+ C = cosx1 + sinx

+ C =2

1 + cot x2+ C1.

8. Metodom smene odrediti:

10

cos2 x dx 20

cos3 x dx 30

cos4 x dx

40

sin2 x dx 50

sin3 x dx 60

sin4 x dx

Resenje: 10

cos2 x dx =

1 + cos 2x2

dx =12

dx+

14

cos 2x d(2x)

=x

2+

sin 2x4

+ C.

20

cos3 x dx =

cos2 x cosx dx = (

1sin2 x)d(sinx) = sinx sin3 x3

+

C.

Drugi oblik primitivne funkcije moze se dobiti snizavanjem stepena trigo-nometrijske funkcije pod integralom.

cos3 x dx=

cos2 x cosx dx =

12

(1 + cos 2x

)cosx dx

=12

(cosx+cos 2x cosx

)dx=

12

(cosx+

12

(cos 3x+cosx))dx

=14

(3 cosx+cos 3x

)dx=

34

sinx+112

sin 3x+C.

22

30

cos4x dx= (

cos2x)2dx=

(1+cos 2x

2

)2dx=

1+2 cos 2x+cos22x

4dx

=14

dx+

14

cos 2x d(2x) +

18

(1 + cos 4x)dx

=38x+

14

sin 2x+132

sin 4x+ C.

40

sin2 x dx =x

2 1

4sin 2x+ C.

50

sin3 x dx = 34

cosx+112

cos 3x+ C = cosx+ cos3 x

3+ C.

60

sin4 x dx =38x 1

4sin 2x+

132

sin 4x+ C.

9. Metodom smene odrediti:

10

dx

cos3 x20

dx

cos4 x30

dx

sinx cos2 x

40

dx

sin3 x50

dx

sin4 x60

dx

sin2 x cosx

70

tan2 x dx 80

tan3 x dx 90

tan4 x dx

100

sin 2x4 cos2 x+ 9 sin2 x

dx 110

sinx cos3 x1 + cos2 x

dx

Resenje: 10 Za x ( pi2 + 2kpi, pi2 + 2kpi), k Z, imamo cosx > 0. Tada jedx

cos3 x=

1cosx

dx

cos2 x=

t=tanx, dt= dx

cos2 x

cosx= 11+t2

=

1+t2 dt

1.n)=( t

2

1+t2 +

12

lnt+1+t2 )+C

=( tanx

2 cosx+

12

ln tanx+ 1

cosx

)+ C=

sinx2 cos2 x

+12

lnsinx+ 1

cosx

+ C = sinx2 cos2 x

+12

lncos x2 + sin x2cos x2 sin x2

+ C.

23

Integral, takode mozemo odrediti na sledeci nacin:dx

cos3 x=

sin2 x+ cos2 xcos3 x

dx =

sin2 x dxcos3 x

+

dx

cosx

= u = sinx du = cosx dxdv = d(cosx)

cos3 xv = 1

2 cos2 x

= sinx2 cos2 x 12

dx

cosx+

dx

cosx

=sinx

2 cos2 x+

12

dx

cosx7.20=

sinx2 cos2 x

+12

log1 + tan x21 tan x2

+ C.

20

dx

cos4 x=

1cos2 x

dx

cos2 x=

t = tanx, dt = dx

cos2 x

cos2 x = 11+tan2 x

= 11+t2

=

(1 + t2)dt = t+t3

3+ C = tanx+

13

tan3 x+ C.

30

dx

sinx cos2 x=

sin2 x+ cos2 xsinx cos2 x

dx =

sinx dxcos2 x

+

dx

sinx

7.50= d(cosx)cos2 x

+ ln tan x

2

= 1cosx

+ ln tan x

2

+ C.40

dx

sin3 x=

18

( 1cos2 x2

1sin2 x2

)+

12

log tan x

2

+ C.50

dx

sin4 x= cotx 1

3cot3 x+ C.

60

dx

sin2 x cosx= log

cos x2 + sin x2cos x2 sin x2 1sinx + C.

70

tan2x dx=

sin2xcos2x

dx=

1cos2xcos2 x

dx=

dx

cos2xdx=tanxx+C.

80

tan3 x dx=

tanx tan2 x dx=

tanx1 cos2 x

cos2 xdx

=

tanx d(tanx)

tanx dx 7.10

=12

tan2 x+ ln | cosx|+ C.

24

90

tan4 x dx =

tan2x1cos2 x

cos2 xdx=

tan2 x d(tanx) dx

tan2 x dx

9.70=tan3 x

3 tanx+ x+ C.

100

sin 2x4 cos2 x+9 sin2 x

dx=

t=4 cos2 x+9 sin2 x

dt=5 sin 2x dx

= 15dt

t=

15

ln |t|+ C

=15

ln(4 cos2 x+9 sin2 x

)+ C.

110

sinx cos3 x1 + cos2 x

dx= t = 1 + cos2 xdt = 2 cosx sinx dx

= 12

1 tt

dt

=12

( dttdt)

=12

ln |t| 12t+ C

=12

ln(1 + cos2 x

) 1 + cos2 x2

+ C.

10. Metodom smene odrediti:

10xarctan 2x

1 + 4x2dx 20

sin 2xsinx+ 1

dx 30

1 sin 2x dx

40

cos1x

dx

x250

sinx cos2 x dx 60

cos2 x sin 2x dx

70

dx

sinx+ 2 cosx+ 380

dx

2 cosx+ 390

dx

4 3 cosx100

tanx

(1 + cosx)2dx 110

dx

(3 + cosx) sinx120

sin 3x+ cos 3xsin 3x cos 3x

dx

25

Resenje: 10xarctan 2x

1 + 4x2dx =

x dx

1 + 4x2

arctan 2x1 + 4x2

dx

=18

d(1 + 4x2)

1 + 4x2 1

2

arctan 2x d(arctan 2x)

=18

ln(1 + 4x2) 12

arctan3/2 2x3/2

+ C

=18

ln(1 + 4x2) 13

arctan3/2 2x+ C.

20

sin 2xsinx+1

dx=2

sinx cosxsinx+1

dx=

t=

sinx+1, sinx= t21dt= 12

cosxsinx+1

dx

=4 (

t2 1)dt = 43t3 4t+ C = 4

3t(t2 3) + C

=43

sinx+ 1 (sinx 2) + C.

30

1 sin 2x dx=

sin2 x+ cos2 x 2 sinx cosx dx

=

(sinx+ cosx)2 dx =

(sinx+ cosx)dx

=(

sinx cosx)+ C, za sinx+ cosx 0.40

cos

1x

dx

x2=

cos

1xd(1x

)= sin 1

x+ C.

50

sinx cos2 x dx =

sinx cosx dx=

sinx d(sinx)=(sinx)3/2

3/2+C,

za x (2kpi, 2kpi + pi/2), k Z, gde je sinx, cosx 0.

60

cos2 x sin 2x dx=2

cos3 x sinx dx = 2

cos3 x d(cosx)

=12

cos4 x+ C.

26

70

dx

sinx+2 cosx+3=

t = tan x2 dx = 2dt1+t2sinx = 2t1+t2

cosx = 1t2

1+t2

= 2

dt

t2+2t+5

= 2

d(t+ 1)(t+ 1)2 + 4

= arctant+ 1

2+ C = arctan

tan x2 + 12

+ C.

80

dx

2 cosx+3=

t=tan x2 , dx= 2dt1+t2cosx = 1t21+t2

=

2dtt2+5

1.c)=

25

arctantan x2

5+C.

90

dx

4 3 cosx = t = tan x2 , dx = 2dt1+t2cosx = 1t2

1+t2

=

2dt1 + 7t2

1.c)=

27

arctan(

7 tanx

2)

+ C.

100

tanx(1 + cosx)2

dx =

sinx dxcosx(1 + cosx)2

= t = cosxdt = sinx dx

=

dt

t(1 + t)2=

1 + t tt(1 + t)2

dt =

dt

t(1 + t)d(1 + t)(1 + t)2

1.b)= log

1 + cosxcosx

+ 11 + cosx

+ C.

110

dx

(3+cosx) sinx=

d(cosx)

(3+cosx)(1cos2 x) = t = cosxdt = sinx dx

=

dt

(3 + t)(t2 1) =12

t+ 1 (t 1)

(3 + t)(t 1)(t+ 1) dt

=12

dt

(3 + t)(t 1) 12

dt

(3 + t)(t+ 1)1.b)=

18

log t1t+3

+ 14

log t+3t+1

+C= 18

log(1cosx)(3+cosx)

(1+cosx)2

+C

27

120

sin 3x+ cos 3xsin 3x cos 3x

dx =

dx

cos 3x+

dx

sin 3x= t = 3xdx = dt3

=

13

dt

cos t+

13

dt

sin t7.20,7.50

=13

(log1+tan t21tan t2

+log tan t2

)+C=

13

log

(1 + tan 3x2

)tan 3x2

1 tan 3x2

+ C.11. Metodom smene odrediti:

10

dx

sinhx20

sinhxcosh 2x

dx

30

sinhx coshxsinh4 x+ cosh4 x

dx 40

dx

cosh2 x 3

tanh2 x

Resenje: 10

dx

sinhx=

12

dx

sinh x2 coshx2

=12

dx

tanh x2 cosh2 x

2

=d(

tanh x2)

tanh x2= ln

tanh x2

+ C.

20

sinhxcosh 2x

dx =

sinhx dx2 cosh2 x 1

=12

d(coshx)cosh2 x 12

1.h)=

12

ln coshx+cosh2 x 1

2

+ C.

30

sinhx coshxsinh4 x+ cosh4 x

dx =12

sinh 2x dx

(sinh2 x)2 + (cosh2 x)2

=12

sinh 2x dx(

cosh 2x12

)2 + ( cosh 2x+12 )2 =

sinh 2x dx2(cosh2 2x+ 1)

= t=cosh 2xdt=2 sinh 2x dx

= 122

dtt2+1

=1

2

2lnt+t2+1 +C

=1

2

2ln(

cosh 2x+

cosh2 2x+ 1)

+ C.

28

40

dx

cosh2 x 3

tanh2 x=

tanh2/3 x d(tanhx) = 3 3

tanhx+ C.

12. Primenom metoda parcijalne integracije odrediti:

10

dx

(a2 + x2)220

x2dx

(x2 a2)2

30

x3a2 + x2

dx 40

x3a2 x2 dx

50

dx(x2 a2 )5

Resenje: 10

dx

(a2 + x2)2=

1a2

a2 + x2 x2(a2 + x2)2

dx

=1a2

dx

a2+x2 1a2

x2

(a2+x2)2dx

1.c)=

1a3

arctanx

a 1a2

x2

(a2+x2)2dx

=

u = x du = dx

dv = x dx(a2+x2)2

v = 12

d(a2 + x2)(a2 + x2)2

= 12(a2 + x2)

=

1a3

arctanx

a 1a2

( x

2(a2 + x2)+

12

dx

a2 + x2

)=

12a3

arctanx

a+

x

2a2(a2 + x2)+ C.

20

x2dx

(x2a2)2 = u=x dv=

x dx(x2a2)2

du=dx v= 12(x2a2)

= x2(x2a2) + 12

dx

x2a21.d)=

x

2(a2 x2) +14a

logx ax+ a

+ C.

29

30

x3a2 + x2

dx =

u = x2 du = 2x dx

dv = x dxa2+x2

v1.j)=a2 + x2

=x2

a2+x2 2

xa2+x2 dx=x2

a2+x2

a2+x2 d

(a2+x2

)=x2

a2+x2 2

3

(a2+x2

)3+C=x22a23

a2+x2 +C.

Drugi nacin odredivanja ovog integrala podrazumeva koriscenje rezultataPn(x)dxax2 + bx+ c

= Qn1(x)ax2 + bx+ c+

dx

ax2 + bx+ c.

Tada je I =

x3a2 + x2

dx =(c0x

2 + c1x + c2)

a2 + x2 +

dxa2 + x2

,

gde c0, c1, c2 i odredujemo metodom neodredenih koeficijenata.Diferenciranjem poslednje jednakosti po x dobijamo

x3a2 + x2

= (2c0x+ c1)a2 + x2 +

(c0x

2 + c1x+ c2) x

a2 + x2+

a2 + x2

.

Nakon mnozenja saa2 + x2 i sredivanja, izraz postaje

x3 = 3c0x3 + 2c1x2 + (2a2c0 + c2)x+ a2c1 + .

Jednacenjem koeficijenata uz iste stepene od x na levoj i desnoj strani jed-nakosti, dobijamo

c0 =13, c1 = 0, c2 = 2a

2

3, = 0,

sto ponovo daje vrednost integrala I =x2 2a2

3

a2 + x2 + C.

40

x3a2 x2 dx =

u = x2 du = 2x dx

dv = x dxa2x2 v

1.j)= a2 x2

=x2

a2 x2 +2

xa2x2 dx=x2

a2x2

a2x2 d(a2x2)

=x2a2x2 2

3

(a2x2)3+C=x2+2a2

3

a2x2+C.

30

Drugi nacin odredivanja :

I =

x3a2 x2 dx =

(c0x

2 + c1x+ c2)

a2 x2 +

dxa2 x2 .

Diferenciranjem po x i sredivanjem koeficijenata uz iste stepene od x do-bijamo

c0 = 13 , c1 = 0, c2 = 2a2

3, = 0,

tj., I = x2 + 2a2

3

a2 x2 + C.

50

dx(x2a2 )5 = 1a2

x2(x2a2)

(x2a2) 52dx=

1a2

x2

(x2a2) 52dx 1

a2

dx

(x2a2) 32

=

u = x dv = x dx

(x2a2)5/2

du = dx v = 12 d(x2a2)

(x2a2)5/2 = 13(x2a2)3/2

=

1a2

( x3(x2 a2)3/2 +

13

dx

(x2 a2)3/2) 1a2

dx

(x2 a2)3/2

= 13a2

x

(x2 a2)3/2 2

3a2

dx

(x2 a2)3/2

= 13a2

x

(x2 a2)3/2 2

3a4

x2 (x2 a2)(x2 a2)3/2 dx

= 13a2

x

(x2a2)3/22

3a4

(x2

(x2a2)3/2 dx

dxx2a2

)

=

u = x dv = x dx

(x2a2)3/2

du = dx v = 12 d(x2a2)

(x2a2)3/2 = 1x2a2

=13a2

x

(x2a2)3/22

3a4

( xx2a2 +

dxx2a2

dxx2a2

)=

23a4

xx2 a2

13a2

x(x2 a2 )3 + C = x3a4 2x

2 3a2(x2 a2 )3 + C.

31

13. Primenom metoda parcijalne integracije odrediti:

10

log x dx 40

log xx2

dx 70

log(x+

a2 + x2

)dx

20

log2 x dx 50

log2 xx2

dx 80x2 log

(x+

a2 + x2

)dx

30x2 log2 x dx 60

x log

2 + x2 x dx 9

0

x log(x+

1 + x2)

1 + x2dx

Resenje: 10

log x dx = u = log x du = dxxdv = dx v = x

= x log x dx= x log x x+ C = x(log x 1) + C = x log x

e+ C.

20

log2 x dx = u=log2 x du=2 log x dxxdv=dx v=x

=x log2 x2 log x dx13.10= x log2 x 2x(log x 1) + C = x

(log2

x

e+ 1)

+ C.

30x2 log2 x dx=

u=log2 x du= 2 log x dxx

dv=x2dx v= x3

3

= x3

3log2 x 2

3

x2 log x dx

=

u = log x du =dxx

dv = x2dx v = x3

3

= x33 log2 x 23(x33 log x 13x2dx

)=

13x3 log2 x 2

9x3 log x+

227x3+C=

x3

27

(9 log2 x6 log x+2

)+C

=x3

27((3 log x 1)2 + 1)+ C.

40

log xx2

dx=

u = log x du =dxx

dv = dxx2

v = 1x

=1x log x+dx

x2=1

xlog x 1

x+C

=1x

(log x+ 1) + C = 1x

log ex+ C.

32

50

log2 xx2

dx =

u=log2 x dv= dx

x2

du= 2 log x dxx v= 1x

=1x log2 x+2

log xx2

dx

13.40= 1x

log2 x 2x

log ex+ C = 1x

(log2 ex+ 1

)+ C.

60x log

2+x2x dx=

u=log 2+x2x dv=x dx

du= 4 dx4x2 v=

x2

2

= x2

2log

2+x2x+2

x2

x24 dx

1.f)=x2

2log

2+x2x+2x2log

2+x2x+C=

x242

log2+x2x+2x+C.

70

log(x+a2 + x2

)dx =

u = log

(x+a2 + x2

)dv = dx

du = dxa2+x2

v = x

= x log

(x+a2+x2

) x dxa2+x2

1.j)= x log

(x+a2+x2

)a2+x2+C.

80x2 log

(x+

a2 + x2

)dx =

u = log

(x+a2 + x2

)dv = x2dx

du = dxa2+x2

v = x3

3

=x3

3log(x+

a2 + x2

) 13

x3

a2 + x2dx

12.30=x3

3log(x+

a2 + x2

) x2 2a29

a2 + x2 + C.

90x log(x+

1 + x2)

1 + x2dx =

u = log

(x+

1 + x2)

dv = x dx1+x2

du = dx1+x2

v =

1 + x2

=

1+x2 log(x+

1+x2) dx=1+x2 log (x+1+x2)x+C.

33

14. Primenom metoda parcijalne integracije odrediti:

10x3 sinx dx 20

x sin2 x dx 30

x2 cos 2x dx

40

dx

cos5 xdx 50

tan7 x dx 60

x sin

x dx

70

(2x 1) sin 3x dx 80

sin3 x cos3 x dx 90

dx

(sinx+ cosx)4

Resenje: 10x3 sinx dx =

u = x3 du = 3x2dxdv = sinx dx v = cosx

=x3 cosx+ 3 x2 cosx dx = u = x2 du = 2x dxdv = cosx dx v = sinx

=x3 cosx+ 3(x2 sinx 2 x sinx dx) = u = x du = dxdv = sinx dx v = cosx

=x3 cosx+ 3x2 sinx 6

( x cosx+ cosx dx)

=x3 cosx+ 3x2 sinx+ 6x cosx 6 sinx+ C.

20x sin2 x dx =

u = x dv = sin2 x dx = 1cos 2x2 dxdu = dx v = x2 14 sin 2x

=x2

2x

4sin 2x

(x2 1

4sin 2x

)dx

=x2

2x

4sin 2xx

2

4 1

8cos 2x+C=

x2

4x

4sin 2x 1

8cos 2x+C.

30x2 cos 2x dx =

u = x2 du = 2x dxdv = cos 2x dx v = 12 sin 2x

=x2

2sin 2x

x sin 2x dx =

u = x du = dxdv = sin 2x dx v = 12 cos 2x

=x2

2sin 2x+

x

2cos 2x 1

2

cos 2x dx

=x2

2sin 2x+

x

2cos 2x 1

4sin 2x+C=

2x214

sin 2x+x

2cos 2x+C.

34

40

dx

cos5 x=

sin2 x+ cos2 xcos5 x

dx =

sin2 xcos5 x

dx+

dx

cos3 x

=u=sinx du=cosx dxdv= sinx dx

cos5 xv= 1

4 cos4 x

= sinx4 cos4x+ 34

dx

cos3 x

9.10=sinx

4 cos4 x+

38

sinxcos2 x

+38

logcos x2 + sin x2cos x2 sin x2

+ C.

50

tan7 x dx =

sin7 xcos7 x

dx = u = sin6 x du = 6 sin5 x cosx dxdv = sinx dx

cos7 xv = 1

6 cos6 x

=

16

tan6 x

sin5 xcos5 x

dx= u=sin4 x du=4 sin3 x cosx dxdv= sinx dx

cos5 xv= 1

4 cos4 x

=

16

tan6 x 14

tan4 x+

tan3 x dx

9.80=16

tan6 x 14

tan4 x+12

tan2 x+ log | cosx|+ C.Integral se moze takode izracunati i primenom metoda smene.

tan7 x dx=

tan5 x tan2 x dx =

tan5 x1 cos2 x

cos2 xdx

=

tan5x d(tanx)

tan5x dx=16

tan6x

tan3x1cos2x

cos2 xdx

=16

tan6 x 14

tan4 x+

tan3 x dx

9.80=16

tan6 x 14

tan4 x+12

tan2 x+ log | cosx|+ C.

60x sin

x dx =

t = x, x = t2dx = 2t dt = 2 t3 sin t dt

14.10= 2( t3 cos t+ 3t2 sin t+ 6t cos t 6 sin t)+ C

= 2x3 cos

x+ 6x sin

x+ 12

x cos

x 12 sinx+ C.

35

70

(2x 1) sin 3x dx = u = 2x 1 dv = sin 3x dxdu = 2dx v = 13 cos 3x

= 2x1

3cos 3x+

23

cos 3x dx=

12x3

cos 3x+29

sin 3x+C.

80

sin3 x cos3 x dx=18

sin3 2x dx=

u=sin2 2x du=4 sin 2x cos 2x dxdv=sin 2x dx v=12 cos 2x

=12

sin2 2x cos 2x+ 2

sin 2x cos2 2x dx

=12

sin2 2x cos 2x

cos2 2x d(cos 2x)

=12

sin2 2x cos 2x 13

cos3 2x+ C.

90

dx

(sinx+ cosx)4=

d(sinx+ cosx)(cosx sinx)(sinx+ cosx)4

=

u = 1cosxsinx du = sinx+cosx(cosxsinx)2 dxdv = d(sinx+cosx)

(sinx+cosx)4v = 1

3(sinx+cosx)3

=

13(sinx+cosx)3(cosxsinx)

13

dx

(cosxsinx)2(sinx+cosx)2

=1

3(sinx+ cosx)2(cos2 x sin2 x) 13

dx

(cos2 x sin2 x)2

= 13(1 + sin 2x) cos 2x

13

dx

cos2 2x

= 13(1 + sin 2x) cos 2x

16

tan 2x+ C.

36

15. Primenom metoda parcijalne integracije odrediti:

10

arcsinx dx 20

arccosx

2dx 30

x2 arccosx dx

40

arcsin2 x dx 50x arcsin2 x dx 60

arcsinxx2

dx

70x arcsinx

1 x2 dx 80

arcsinx arccosx dx

Resenje: 10

arcsinx dx=

u=arcsinx dv=dxdu= dx1x2 v=x

=x arcsinx

x dx1x2

1.j)= x arcsinx+

1 x2 + C.

20

arccosx

2dx =

u=arccos x2 dv=dxdu= dx4x2 v=x

=x arccos x2 +

x dx4x2

1.j)= x arccos

x

2

4 x2 + C.

30x2arccosx dx =

u=arccosx dv=x2dxdu= dx1x2 v=

x3

3

= x33 arccosx+ 13

x31x2 dx

12.40=x3

3arccosx x

2 + 29

1 x2 + C.

40

arcsin2 x dx =

u=arcsin2 x dv=dxdu=2 arcsinx dx1x2 v=x

= x arcsin2 x 2

arcsinx

x dx1 x2 =

u=arcsinx du= dx

1x2

dv= x dx1x2 v

1.j)= 1x2

=x arcsin2 x+ 2

1 x2 arcsinx 2

dx

= x arcsin2 x+ 2

1 x2 arcsinx 2x+ C.

37

50x arcsin2 x dx==

u = arcsin2 x dv = x dxdu = 2 arcsinx dx1x2 v =

x2

2

=x2

2arcsin2 x

arcsinx

x2dx1 x2

=

u = x arcsinx dv = x dx

1x2

du =(

arcsinx+ x1x2

)dx v

1.j)= 1 x2

=x2

2arcsin2 x+x

1x2 arcsinx

1x2 arcsinx dx

x dx

=

u = arcsinx du =dx1x2

dv =

1 x2 dx v 1.m)= x2

1 x2 + 12 arcsinx

=x2 1

2arcsin2 x+

x

2

1x2 arcsinx x

2

2+

12

x dx

+12

arcsinx d(arcsinx)

=2x2 1

4arcsin2 x+

x

2

1 x2 arcsinx x

2

4+ C.

60

arcsinxx2

dx =

u = arcsinx dv =dxx2

du = dx1x2 v =

1x

= arcsinxx +

dx

x

1 x2

1.l)= 1

xarcsinx+

12

log

1 x2 11 x2 + 1

+ C.

70x arcsinx

1 x2 dx=u=arcsinx dv= x dx

1x2

du= dx1x2 v

1.j)=1x2

=

1x2 arcsinx+dx

= x

1 x2 arcsinx+ C.

38

80

arcsinx arccosx dx=

u=arccosx dv=arcsinx dxdu= dx1x2 v

15.10= x arcsinx+

1x2

=x arcsinx arccosx+

1x2 arccosx+

arcsinx

x dx1x2 +

dx

15.70= x arcsinx arccosx+

1 x2( arccosx arcsinx)+ 2x+ C.Bez koriscenja rezultata 15.10 i 15.70, integral se moze izracunati na sledecinacin.

arcsinx arccosx dx=

u=arcsinx arccosx dv=dx

du= arccosxarcsinx1x2 dx v=x

= x arcsinx arccosx

(arccosx arcsinx) x dx

1 x2

=

u=arccosx arcsinx dv= x dx

1x2

du= 2dx1x2 v=

1 x2

= x arcsinx arccosx+

1 x2(arccosx arcsinx) + 2

dx

= x arcsinx arccosx+

1 x2(arccosx arcsinx) + 2x+ C.

16. Primenom metoda parcijalne integracije odrediti:

10x arctanx dx 20

x2 arctanx dx 30

1x2

arctanx

3dx

40

arctanx dx 50

x2 arctanx

1 + x2dx 60

arctan

2x1 x2 dx

70

x earctanx

(1 + x2)3/2dx

Resenje: 10x arctanx dx =

u = arctanx du =dx

1+x2

dv = x dx v = x2

2

=x2

2arctanx 1

2

x2

1 + x2dx

1.e)=

x2 + 12

arctanx x2

+ C.

39

20x2arctanx dx=

u=arctanx du=dx

1+x2

dv=x2 dx v= x3

3

=x33 arctanx 13

x3

1+x2dx

= t=1+x2dt=2x dx

=x33 arctanx16t1tdt=

x3

3arctanx1

6(tlog|t|)+C

=x3

3arctanx 1

6(1 + x2 log(1 + x2))+ C.

30

1x2

arctanx

3dx =

u=arctan x3 dv=

dxx2

du= 3 dx9+x2

v= 1x

=1x arctan x3 +

3 dxx(9+x2

)1.k)= 1

xarctan

x

3+

16

log x2x2 + 9

+ C.

40

arctanx dx =

t = x, t2 = x2t dt = dx = 2 t arctan t dt

16.10=(t2 + 1

)arctan t t+ C = (x+ 1) arctanxx+ C.

50x2 arctanx

1 + x2dx =

u = arctanx du =dx

1+x2

dv = x2 dx

1+x2v

1.e)= x arctanx

= x arctanx arctan2 x

x dx

1 + x2+

arctanx d(arctanx)

1.i)= x arctanx 1

2arctan2 x 1

2log(1 + x2

)+ C.

60

arctan2x

1 x2 dx = u = arctan 2x1x2 du = 2dx1+x2dv = dx v = x

= x arctan

2x1x22

x dx

1+x21.i)= x arctan

2x1x2log

(1+x2

)+C.

40

70 I =

x earctanx

(1 + x2)3/2dx =

u = x

1+x2dv = e

arctan xdx1+x2

du = dx(1+x2)3/2

v = earctanx

=xearctanx

1+x2

earctanx

(1+x2)3/2dx=

u= 1

1+x2du= x dx

(1+x2)3/2

dv= earctan xdx

1+x2v = earctanx

=xearctanx

1 + x2 e

arctanx

1 + x2

x earctanx

(1 + x2)3/2dx =

(x 1) earctanx1 + x2

I.

Odatle je I =earctanx(x 1)

2

1 + x2+ C.

17. Primenom metoda parcijalne integracije odrediti:

10x3e2x dx 20

x2ex

(x+ 2)2dx 30

exdx

40eax cos bx dx 50

eax sin bx dx

60e2x sin2 x dx 70

e2x cos2 x dx

Resenje: 10 I =x3e2xdx =

u = x3 du = 3x2dxdv = e2xdx v = 12e2x

=x3

2e2x 3

2

x2e2xdx =

u = x2 du = 2x dxdv = e2xdx v = 12e2x

=x3

2e2x 3

2

(x2

2e2x

xe2xdx

)=x3

2e2x 3x

2

4e2x +

32

xe2xdx

= u = x du = dxdv = e2xdx v = 12e2x

= x32 e2x 3x24 e2x + 32(x2 e2x 12e2xdx

)=x3

2e2x 3x

2

4e2x+

3x4e2x 3

8e2x+C=

e2x

8(4x36x2+6x3

)+C.

41

20

x2ex

(x+ 2)2dx=

u = x2ex du = x(x+ 2)exdxdv = dx(x+2)2 v = 1x+2

= x2

x+ 2ex +

xexdx =

u = x du = dxdv = exdx v = ex

= x2

x+ 2ex+xex

exdx = x

2

x+ 2ex + xex ex+C

=x 2x+ 2

ex + C.

30exdx=

xexdxx

=

u=x dv= e

xdxx

du= dx2x

v=2ex

=2xexexdxx

= 2xex 2e

x + C = 2e

x(x 1)+ C.

40 I =eax cos bx dx =

u = cos bx du = b sin bx dxdv = eaxdx v = 1aeax

=1aeax cos bx+

b

a

eax sin bx dx =

u = sin bx du = b cos bx dxdv = eaxdx v = 1aeax

=1aeax cos bx+

b

a

(1aeax sin bx b

a

eax cos bx dx

)=eax

a2(a cos bx+ b sin bx

) b2a2I.

Odatle jea2 + b2

a2I =

eax

a2(a cos bx+ b sin bx

)+ C1, odnosno

I =eax

a2 + b2(a cos bx+ b sin bx

)+ C.

50eax sin bx dx =

u = sin bx du = b cos bx dxdv = eaxdx v = 1aeax

=1aeaxsin bx b

a

eaxcos bx dx17.4

0

=eax

a2+b2(a sin bxb cos bx)+C.

42

60e2x sin2 x dx =

u = sin2 x du = 2 sinx cosx dx = sin 2x dxdv = e2xdx v = 12e2x

=12

(e2xsin2x

e2xsin2x dx

)17.50=

e2x

8(2cos2xsin2x)+C.

70e2x cos2 x dx =

e2x(1 sin2 x) dx = e2x dx e2x sin2 x dx

17.60=e2x

8(2 + cos 2x+ sin 2x

)+ C.

18. Odrediti rekurentnu formulu za izracunavanje integrala In za n N, n 2i a, b 6= 0.

10 In =

x logn x dx 20 In = (

1 x2)n/2 dx30 In =

dx

(x2 + a2)n40 In =

dx

(ax2 + bx+ c)n

50 In =x2nx2 a2 dx 60 In =

x2n+1

x2 a2 dx

70 In =

sinn x dx 80 In =

dx

cosn x

90 In =

tann x dx 100 In =

cotn x dx

110 In =

dx

(a sinx+ b cosx)n120 In =

dx

(a+ b cosx)n

Resenje: 10 In =

x logn x dx = u = logn x du = n logn1x dxxdv = x dx v = 23xx

=

23x3/2 logn x 2n

3

x logn1 x dx =

23x3/2 logn x 2n

3In1.

43

Iskoristimo izvedenu relaciju za izracunavanje, na primer, integrala I3. Kakoje

I1 =

x log x dx =

u = log x du =dxx

dv =x v = 23x

3/2

= 23x3/2 log x 23

x dx

=23x3/2 log x 4

9x3/2 + C =

29x3/2

(3 log x 2)+ C,

na osnovu izvedene rekurentne relacije imamo

I2 =23x3/2 log2 x 4

3I1 =

227x3/2

(9 log2 x 12 log x+ 8)+ C,

I3 =23x3/2 log3 x 2I2 = 227x

3/2(9 log3 x 18 log2 x+ 24 log x 16)+ C.

20 In = (

1 x2)n/2dx = (1 x2)(1 x2)(n2)/2dx= In2

x2(1x2)n22 dx= u=x dv=x

(1x2)n22 dx

du=dx v= 1n(1x2)n/2

= In2+

x

n

(1x2)n/2 1

n

(1x2)n/2dx= 1

nIn+In2+

x

n

(1x2)n/2.

Dakle,n+ 1n

In = In2 +x

n

(1 x2)n/2, pa je

In =n

n+ 1In2 +

x

n+ 1(1 x2)n/2.

Znamo da je I1 =

1 x2 dx 1.m)= x2

1 x2 + 1

2arcsinx + C. Na osnovu

izvedene rekurentne relacije dobijamo

I3 =34I1 +

x

4(1 x2)3/2 = 3

8arcsinx+

x

8

1 x2(5 2x2)+ C.

Kako je I2 = (

1 x2)dx = x x33

+ C, mozemo odrediti

I4 =45I2 +

x

5(1 x2)2 + C = x 2

3x3 +

15x5 + C.

44

30 In =

dx

(x2+a2)n=

1a2

x2+a2x2(x2+a2)n

dx=1a2In1 1

a2

x2

(x2+a2)ndx

=

u = x dv =x dx

(x2+a2)n

du = dx v = 1(22n)(x2+a2)n1

=

2n 3(2n 2)a2 In1 +

1(2n 2)a2

x

(x2 + a2)n1.

Primetimo da su ovaj integral i dobijena rekurentna relacija samo specijalanslucaj integrala i relacije iz narednog primera (a = 1, b = 0, c = a2).

40 In =

dx

(ax2 + bx+ c)n=

ax2 + bx+ c(ax2 + bx+ c)n+1

dx

=

d(ax2 + bx+ c) = (2ax+ b)dx

ax2 + bx+ c = 14a((2ax+ b)2 + 4ac b2)

=

(2ax+b)2+4acb24a(ax2+bx+c)n+1

dx=14a

(2ax+b)2

(ax2+bx+c)n+1dx+

4acb24a

In+1

=

u = 2ax+ b du = 2a dxdv = d(ax2+bx+c)(ax2+bx+c)n+1

v = 1n(ax2+bx+c)n

=

14a

( (2ax+ b)n(ax2 + bx+ c)n

+2an

dx

(ax2 + bx+ c)n

)+

4ac b24a

In+1

= 2ax+ b4an(ax2 + bx+ c)n

+1

2nIn +

4ac b24a

In+1.

4ac b2

4aIn+1 = In 12nIn +

2ax+ b4an

(ax2 + bx+ c

)n In+1 = 2a4ac b2

2n 1n

In +2ax+ b

n(4ac b2)(ax2 + bx+ c)n .

45

50 In =x2nx2 a2 dx =

x2n2

(x2 a2 a2)x2 a2 dx

=x2n2

(x2a2) 32dxa2In1 =

u=(x2a2) 32 dv=x2n2dx

du=3x2a2xdx v= x2n12n1

=

x2n1

2n 1(x2 a2) 32 3

2n 1In a2In1

2n+ 22n 1In =

x2n1

2n 1(x2 a2) 32 a2In1

In = x2n1

2n+ 2(x2 a2) 32 2n 1

2n+ 2a2In1.

60 In =x2n+1

x2 a2 dx =

u = x2n du = 2nx2n1dxdv = x2 a2xdx v = 13(x2 a2) 32

=13x2n(x2 a2) 32 2n

3

x2n1

(x2 a2) 32dx

=13x2n(x2 a2) 32 2n

3(In a2In1

) 2n+ 3

3In =

13x2n(x2 a2) 32 2n

3a2In1

In = 12n+ 3x2n(x2 a2) 32 2na2

2n+ 3In1.

70 In=

sinn x dx=

sinn2 x(1cos2 x)dx=In2

cos2 x sinn2 x dx

= u = cosx dv = sinn2 x d(sinx)du = sinx dx v = 1n1 sinn1 x

= In2 1

n 1 sinn1x cosx 1

n 1In

nn 1In = In2

1n 1 sin

n1 x cosx

In = n 1n

In2 1n

sinn1 x cosx.

46

80 In =

dx

cosn x=

sin2 x+ cos2 xcosn x

dx =

sin2 xcosn x

dx+

dx

cosn2 x

=

sin2 xcosn x

dx+ In2 =

u = sinx du = cosx dxdv = d(cosx)cosn x v = 1(n1) cosn1 x

=sinx

(n 1) cosn1 x 1

(n 1)In2 + In2

=sinx

(n 1) cosn1 x +n 2n 1 In2.

90 In =

tann x dx =

tann2 x1 cos2 x

cos2 xdx =

tann2 x

dx

cos2 x In2

=

tann2 x d(tanx) In2 = 1n 1 tan

n1 x In2.

100 In =

cotn x dx =

cotn2 x1 sin2 x

sin2 xdx =

cotn2 x

dx

sin2 x In2

=

cotn2 x d(cotx) In2 = 1n 1 cot

n1 x In2.

47

110 In =

dx

(a sinx+ b cosx)n=

a sinx+ b cosx(a sinx+ b cosx)n+1

dx

=

u=1

(a sinx+b cosx)n+1du=(n+ 1) a cosxb sinx

(a sinx+b cosx)n+2dx

dv=(a sinx+b cosx)dx v=(a cosxb sinx)

= a cosx b sinx

(a sinx+ b cosx)n+1 (n+ 1)

(a cosx b sinx)2

(a sinx+ b cosx)n+2dx

= a cosxb sinx(a sinx+b cosx)n+1

(n+ 1)a2+b2(a sinx+b cosx)2

(a sinx+b cosx)n+2dx

= a cosx b sinx(a sinx+ b cosx)n+1

(n+ 1)((a2 + b2

)In+2 In

) In+2 = n(n+ 1)(a2 + b2)In 1(n+ 1)(a2 + b2) a cosx b sinx(a sinx+ b cosx)n+1 .

120 In =

dx

(a+ b cosx)n=

a+ b cosx(a+ b cosx)n+1

dx

= a

dx

(a+b cosx)n+1+b

cosx dx(a+b cosx)n+1

=aIn+1+b

cosx dx(a+b cosx)n+1

=

u = 1(a+b cosx)n+1 du = (n+ 1)b sinx dx(a+b cosx)n+2dv = cosx dx v = sinx

=aIn+1+b(

sinx(a+b cosx)n+1

(n+1)b

sin2x dx(a+b cosx)n+2

)= aIn+1 +

b sinx(a+ b cosx)n+1

+ (n+ 1)b2

1 cos2 x(a+ b cosx)n+2

dx

=b sinx

(a+b cosx)n+1+aIn+1+(n+1)b2In+2

(n+1)b2 cos2 x(a+b cosx)n+2

dx

=b sinx

(a+b cosx)n+1+aIn+1(n+1)b2In+2(n+1)b2

(a+b cosxab

)2dx

(a+ b cosx)n+2

=b sinx

(a+b cosx)n+1(n+1)In+a(2n+3)In+1(n+1)

(a2+b2

)In+2.

48

(n+ 1)(a2 + b2)In+2 = b sinx(a+ b cosx)n+1 (n+ 2)In + a(2n+ 3)In+1 In+2 = b(n+ 1)(a2 + b2) sinx(a+ b cosx)n+1 n+ 2(n+ 1)(a2 + b2)In

+(2n+ 3)a

(n+ 1)(a2 + b2

)In+1.

19. Razlaganjem racionalne funkcije, izracunati sledece integrale

10

x3 + 6x3 5x2 + 6x dx 2

0

dx

(x 1)2(x2 + 1)2

30

dx

x(x+ 1)(x2 + x+ 1)340

x

x3 3x+ 2 dx

50

dx

x3 + 160

x+ 1x3 1 dx 7

0

dx

x4 + 1

80

x2dx(x2 + 2x+ 2

)2 90 2x2 5x4 5x2 + 6 dx100

x3 1x4 + x2

dx 110

x+ 2x3 + 1

dx 120

x+ 1x(x2 + 2x+ 2)

dx

Resenje: 10 I =

x3 + 6x3 5x2 + 6x dx; integral najpre svedemo na integral

prave racionalne funkcije.

x3 + 6x3 5x2 + 6x =

x3 5x2 + 6x+ 5x2 6x+ 6x3 5x2 + 6x = 1 +

5x2 6x+ 6x3 5x2 + 6x.

Kako je x3 5x2 + 6x = x(x2 5x+ 6) = x(x 2)(x 3), to je

I =dx+

5x2 6x+ 6x(x 2)(x 3) dx.

Nova racionalna podintegralna funkcija dozvoljava razvoj:

5x2 6x+ 6x(x 2)(x 3) =

A

x+

B

x 2 +C

x 3 , (1)

gde je koeficijente A,B,C potrebno odrediti.

49

Metod neodredenih koeficijenata podrazumeva sledece:pomnozimo jednakost (1) sa x(x 2)(x 3). Ona postaje jednakost dva poli-noma u razvijenom obliku

5x2 6x+ 6 =A(x 2)(x 3) +Bx(x 3) + Cx(x 2)= x2(A+B + C) + x(5A 3B 2C) + 6A.

Izjednacavanjem koeficijenata uz iste stepene promenljive x dolazimo do sis-tema jednacina

5 = A+B + C, 6 = 5A 3B 2C, 6 = 6A.Resavanjem ovog sistema nalazimo A = 1, B = 7 i C = 11.

Time polazni integral svodimo na zbir cetiri elementarna integrala:

I =dx+

dx

x 7

dx

x 2 + 11

dx

x 3= x+ log |x| 7 log |x 2|+ 11 log |x 3|+ C.

20 I =

dx

(x 1)2(x2 + 1)2 , podintegralna funkcija dozvoljava razvoj:

1(x 1)2(x2 + 1)2 =

A1x 1 +

A2(x 1)2 +

B1x+ C1x2 + 1

+B2x+ C2(x2 + 1

)2 . (2)OdredimoAj , Bj , Cj metodom neodredenih koeficijenata. Nakon mnozenja

(2) sa (x 1)2(x2 + 1)2 dolazimo do jednakosti1 = A1(x 1)

(x2 + 1

)2 +A2(x2 + 1)2 + (B1x+ C1)(x 1)2(x2 + 1)+(B2x+ C2)(x 1)2

= x5(A1 +B1) + x4(A1 +A2 2B1 + C1) + x3(2A1 + 2B1 +B2 2C1)+x2(2A1 + 2A2 2B1 2B2 + 2C1 + C2)+x(A1 +B1 +B2 2C1 2C2)A1 +A2 + C1 + C2.

Izjednacavanjem koeficijenata dva polinoma uz odgovarajuce stepene od x,dolazimo do sistema jednacina

0 =A1 +B10 =A1 +A2 2B1 + C10 = 2A1 + 2B1 +B2 2C10 =2A1 + 2A2 2B1 2B2 + 2C1 + C20 =A1 +B1 +B2 2C1 2C21 =A1 +A2 + C1 + C2,

50

cijim resavanjem nalazimo vrednosti

A1 = 1/2, A2 = 1/4, B1 = 1/2, C1 = 1/4, B2 = 1/2, C2 = 0.Polazni integral tada postaje

I =12

dx

x 1 +14

dx

(x 1)2 +14

2x+ 1x2 + 1

dx+12

x dx(

x2 + 1)2

=12

log |x 1| 14(x 1) +

14

d(x2 + 1

)x2 + 1

+14

dx

x2 + 1+

14

d(x2 + 1

)(x2 + 1

)2=

14

(log

x2 + 1(x 1)2

x2 + x(x 1)(x2 + 1) + arctanx)+ C.

30 I =

dx

x(x+ 1)(x2 + x+ 1)3; primetimo sledeci razvoj podintegralne funkcije

1x(x+ 1)(x2 + x+ 1)3

=x2 + x+ 1 x(x+ 1)x(x+ 1)(x2 + x+ 1)3

=1

x(x+ 1)(x2 + x+ 1)2 1

(x2 + x+ 1)3

=x2 + x+ 1 x(x+ 1)x(x+ 1)(x2 + x+ 1)2

1(x2 + x+ 1)3

=1

x(x+ 1)(x2 + x+ 1) 1

(x2 + x+ 1)2 1

(x2 + x+ 1)3

=1

x(x+ 1) 1x2 + x+ 1

1(x2 + x+ 1)2

1(x2 + x+ 1)3

.

Time polazni integral postaje

I =

dx

x(x+ 1)

dx

x2 + x+ 1

dx

(x2 + x+ 1)2

dx

(x2 + x+ 1)3

1.b)= log

xx+ 1

d(x+ 12)(x+ 12

)2 + 34

d(x+ 12)((x+ 12

)2 + 34)2

d(x+ 12)((x+ 12

)2 + 34)3 = t = x+ 12

1.c)= log

xx+ 1

23

arctan2x+ 1

3

dt(t2 + 34

)2 dt(t2 + 34

)312.10

18.30= log xx+1

143

3arctan

2x+13 2

32x+1

x2+x+1 2x+1

6(x2+x+1

)2 +C.

51

40

x

x3 3x+ 2 dx =29

log1 x2 + x

+ 13(1 x) + C.

50

dx

x3 + 1=

13

arctan2x 1

3+

16

log(x+ 1)2

x2 x+ 1 + C

=13

arctan2x 1

3+

16

log(x+ 1)3x3 + 1

+ C.60

x+ 1x3 1 dx =

13

log (x 1)2x2 + x+ 1

+ C = 13

log(x 1)3x3 1

+ C.

70

dx

x4+1=

dx

(x2+1)22x2 =1

2

2

(arctan(1+

2x)arctan(1

2x)

)+

14

2logx2 +2x+ 1x2 2x+ 1

+ C.80

x2dx(x2 + 2x+ 2

)2 = arctan(x+ 1) + 1x2 + 2x+ 2 + C.90

2x2 5x4 5x2 + 6 dx =

12

2logx2x+2

+ 123 logx3x+3

+ C.100

x3 1x4 + x2

dx =1x

+12

log(1 + x2) + arctanx+ C.

110

x+ 2x3 + 1

dx =

3 arctan2x 1

3+

16

log(x+ 1)3x3 + 1

+ C.120

x+ 1

x(x2 + 2x+ 2)dx =

12

arctan(x+ 1) +14

logx2

x2 + 2x+ 2+ C.

20. Odrediti sledece integrale iracionalnih funkcija

10

dxx+ 2+ 3

x+ 2

20

1x+ 11+ 3x+ 1

dx 30

dx6x 1+ 3x 1+x 1

40

dx

(x+1)5x2+2x

50

x+1x1 dx 6

0

dx

(x 1)(2 x)

52

Resenje: 10

dxx+ 2 + 3

x+ 2

= t6 = x+ 2dx = 6t5dt

= 6 t5dtt3 + t2 = 6

t3dt

t+ 1

= 6t3 + 1 1t+ 1

dt = 6

(t2 t+ 1)dt 6

dt

t+ 1

= 2t3 3t2 + 6t 6 log |t+ 1|+ C

= 2x+ 2 3 3x+ 2 + 6 6x+ 2 6 log 6x+ 2 + 1+ C.

20

1x+ 11 + 3x+ 1

dx = t6 = x+ 1dx = 6t5dt

= 6 (1 t3)t51 + t2 dt= 6

(t6 + t4 + t3 t2 t+ 1)dt+ 6

t 1t2 + 1

dt

= 67t7 +

65t5 +

32t4 2t3 3t2 + 6t+ 3

2t dtt2 + 1

6

dt

t2 + 1

= 67t7 +

65t5 +

32t4 2t3 3t2 + 6t+ 3 log(t2 + 1) 6 arctan t+ C

= 67(

6x+ 1

)7 + 65(

6x+ 1

)5 + 32(

3x+ 1

)2 2x+ 13 3x+ 1 + 6 6x+ 1 + 3 log( 3x+ 1 + 1) 6 arctan 6x+ 1 + C.

30

dx6x 1 + 3x 1 +x 1 =

t6 = x 1dx = 6t5dt = 6 t5t+ t2 + t3 dt

= 6

(t2 t)dt+ 6

t dt

t2 + t+ 1= 2t3 3t2 + 3

2t+ 1 1t2 + t+ 1

dt

= 2t3 3t2 + 3d(t2 + t+ 1)t2 + t+ 1

dt 3

dt

t2 + t+ 1

= 2t3 3t2 + 3 log |t2 + t+ 1| 3

dt

(t+ 12)2 + 34

1.c)= 2t3 3t2 + 3 log

t3 1t 1

33/2

arctant+ 12

3/2+ C

= 2x13 3x1+3 log

x116x1123 arctan 2 6x1+13 +C.

53

40 I =

dx

(x+1)5x2+2x

=

x+1=1t , x, t > 0

dx=dtt2

=

t4dt1t2

=(a0t

3 + a1t2 + a2t+ a3)

1 t2 +

dt1 t2 .

Diferenciranjem po t i mnozenjem sa

1 t2 dobijamot4 = 4a0t4 3a1t3 + (3a0 2a2)t2 + (2a1 a3)t+ a2 + .

Izjednacavanjem koeficijenata uz stepene od t formiramo sistem jednacina

1 = 4a0 0 = 2a1 a30 = 3a1 0 = a2 + 0 = 3a0 2a2

cije je resenje

a0 = 1/4, a1 = 0, a2 = 3/8, a3 = 0, = 3/8.Nakon sredivanja nalazimo da je polazni integral

I =x2 + 2x

( 14(x+ 1)4

+3

8(x+ 1)2) 3

8arcsin

1x+ 1

+ C.

Za x < 2 integral se izracunava analognim postupkom, vodeci racuna da jet2 = t.

Ovaj integral takode se moze odrediti primenom Ojlerove smene.

I=

dx

(x+ 1)5x2 + 2x

=

x2 + 2x = t x, x = 2

t21dx = 4t dt

(t21)2

=4

(t2 1)4(t2 + 1)5

dt = 4

(t2 + 1 2)4(t2 + 1)5

dt

=4

(t2 + 1)4 8(t2 + 1)3 + 24(t2 + 1)2 32(t2 + 1) + 16(t2 + 1)5

dt

=4

dt

t2 + 1+ 32

dt

(t2 + 1)2 96

dt

(t2 + 1)3+ 128

dt

(t2 + 1)4

64

dt

(t2 + 1)5.

54

Koristeci izvedenu rekurentnu relaciju iz zadatka 18.30 nalazimo

I = 32

arctan t+5t7 3t5 + 3t3 5t

2(t2 + 1)4+ C.

Vracanjem smene konacno dobijamo

I =x2 + 2x

3x2 + 6x+ 54(x+ 1)4

32

arctanx2 + 2xx

+ C.

50

x+ 1x 1 dx =

t =

x+1x1

dx = 4t dt(t21)2

= 4

t2

(t2 1)2 dt

12.20=2t

t2 1 + log t+ 1t 1

+ C=

(x+ 1)(x 1) + log x+(x+ 1)(x 1)+ C.

60

dx(x 1)(2 x) =

x (1, 2) x = 1 + sin2 t, t (0, pi/2)dx = 2 sin t cos t dt,

(x 1)(2 x) =

sin2 t(1 sin2 t)

=2dt = 2t+ C = 2 arcsin

x 1 + C.

Integral mozemo resiti i primenom trece Ojlerove smene.

I =

dx(x 1)(2 x) =

(x 1)(2 x) = t(x 1)x = t

2+2t2+1

dx = 2t dt(t2+1)2

= 2

dt

t2 + 1

= 2 arctan t+ C = 2 arctan

2 xx 1 + C.

21. Svodenjem na integral racionalne funkcije, izracunati sledece integrale

10

dx

x+x2 + x+ 1

20

dx

1 +

1 2x x2 30

xx2 + 3x+ 2x+x2 + 3x+ 2

dx

40

x3 + x4 dx 50

x(1 + 3x)2 dx 60 33x x3 dx

55

Resenje: 10 I =

dx

x+x2 + x+ 1

. Potkorena funkcija u integralu odgo-

vara za primenu prve Ojlerove smene:x2 + x+ 1 = x+ t.

Znak ispred x biramo tako da pogoduje sredivanju podintegralne funkcije.Kako je

t = x+x2 + x+ 1,

posle poredenja sa podintegralnom funkcijom uzimamox2 + x+ 1 = x+t.

Tada je

x2 + x+ 1 = x2 2xt+ t2 x+ 1 = t2 2xt,

tj. x =t2 11 + 2t

, pa je dx = 2t2 + t+ 1(2t+ 1)2

dt. Integral onda glasi

I = 2

t2 + t+ 1t(2t+ 1)2

dt.

Rastavimo racionalnu funkciju

2t2 + t+ 1t(2t+ 1)2

=A1t

+A2

2t+ 1+

A3(2t+ 1)2

.

Metodom neodredenih koeficijenata nalazimo da je

A1 = 2, A2 = 3, A3 = 3.

Integral se transformise u elementarne integrale

I = 2dt

t3

dt

2t+1 3

2

d(2t+1)(2t+1)2

=2 log |t| 32

log |2t+1|+ 32(2t+1)

+C1

= 2 logx+x2+x+1 3

2log2x+2x2+x+1+1x+x2+x+1+C,

gde je C = C1 1/2.

56

20 I =

dx

1+

12xx2 =

12xx2 =xt1dx=2t

2+2t+1(t2+1)2

dt

= t2+2t+1t(t1)(t2+1) dt.

t2 + 2t+ 1t(t 1)(t2 + 1) =

A1t

+A2t 1 +

Bt+ Ct2 + 1

A1 = 1, A2 = 1, B = 0, C = 2

I =

dt

t 1 dt

t 2

dt

t2 + 1= log

t 1t

2 arctan t+ C= log

1 x+1 2x x21 +

1 2x x2 2 arctan 1 +1 2x x2

x+ C.

30 I=xx2+3x+2x+x2+3x+2

dx=

x2+3x+2=

(x+2)(x+1)= t(x+1)

x = 2t2

t21 , dx = 2tdt(t21)2

=2

t2 + 2t

(t 2)(t 1)(t+ 1)3 dt.

2 t2 + 2t

(t 2)(t 1)(t+ 1)3 = 16

27(t 2)+3

4(t 1)17

108(t+ 1)+

518(t+ 1)2

+1

3(t+ 1)3

I=1627

log |t 2|+ 34

log |t 1| 17108

log |t+ 1| 518(t+ 1)

16(t+ 1)2

+ C

=1627

log

x+ 2x+ 1

2+ 34 log

x+ 2x+ 1

1 17108 log

x+ 2x+ 1

+ 1

5

18(

x+ 2x+ 1

+ 1) 1

6(

x+ 2x+ 1

+ 1)2 + C.

57

40 I=

x3 + x4 dx =x2(1 + x1)1/2 dx =

1 + x1 = t2dx = 2t dt(t21)2

=2

t2dt

(t2 1)4 = 2t2 1 + 1(t2 1)4 dt

=2

dt

(t2 1)3 2

dt

(t2 1)4 = 2I3 2I4,

gde je In =

dt

(t2 1)n . Rekurentnu relaciju za izracunavanje In

In =t

2(n 1)(t2 1)n1 2n 3

2(n 1)In1

mozemo dobiti, na primer, iz 18.40 za a = 1, b = 0 i c = 1, imajuci u viduda je

I1 =

dt

t2 1 =12

log t 1t+ 1

+ C1.Tada je

I2 = t2(t2 1) 14

log t 1t+ 1

+ C2.I3 =

3t3 5t8(t2 1)2 +

316

log t 1t+ 1

+ C3.I4 = 15t

5 40t3 + 33t48(t2 1)3

532

log t 1t+ 1

+ C4.I = 3t

5 8t3 3t24(t2 1)3

116

log t 1t+ 1

+ C=

124

x2 + x(8x2 + 2x 3) + 1

8logx+1 + x + C.

58

50

x(1+ 3x)2 dx= x= t6dx=6t5dt

=6 t8 dt(t2+1)2= 6

(t4 2t2 + 3)dt 6

4t2 + 3

(t2 + 1)2dt

=65t5 4t3 + 18t 6

4(t2 + 1) 1

(t2 + 1)2dt

=65t5 4t3 + 18t 24

dt

t2 + 1+ 6

dt

(t2 + 1)2

12.10=65t5 4t3 + 18t 21 arctan t+ 3 t

t2 + 1+ C

=6x

5(1 + 3x )(6x 14 3

x2 + 70 3

x+ 105

) 21 arctan 6x+ C.

60

3

3xx3 dx=x(1+3x2)1/3dx=

1+3x2 = t3, x=

3

t3+1

dx= 3

3t2dt2(t3+1)3/2

= 9

2

t3dt

(t3+1)2=

u= t dv= t2dt

(t3+1)2

du=dt v= 13(t3+1)

= 3t2(t3+1) 32

dt

t3+1

19.50=3t

2(t3 + 1)

32

arctan2t 1

3 1

4log(t+ 1)3t3 + 1

+ C=

118

log

(3

3x x3 + x)3

3x+x2

9+

19

3arctan

2 3

3x x3 x3x

+ C.

22. Svesti sledece integrale na integral racionalne funkcije.

10

3 tanx+ 13 tanx dx 2

0

dx

4 + 3 tanx

30

1 2 cosx+ 3 sinx2 + 3 cosx 4 sinx dx 4

0

dx

(sinx+ 2 cosx)3

50

dx

sin2 x cos4 x60

dx(sinx+

2cosx

)270

2x 13 x dx 8

0

3

x+ 22x 1 dx

59

Resenje: 10

3 tanx+ 13 tanx dx =

t = tanxdx = dt1+t2

= 3t+ 1(3 t)(1 + t2) dt .20

dx

4 + 3 tanx= t = tanxdx = dt

1+t2

= dt(4 + 3t)(1 + t2) .

30

1 2 cosx+ 3 sinx2 + 3 cosx 4 sinx dx =

t = tan x2 dx = 2dt1+t2sinx = 2t1+t2

cosx = 1t2

1+t2

= 2

3t2 + 6t 1

(5 8t t2)(1 + t2) dt .

40

dx

(sinx+2 cosx)3=

t=tan x2 dx= 2dt1+t2sinx= 2t1+t2

cosx= 1t2

1+t2

= 12

(1+t2)dt(1+tt2)3 .

50

dx

sin2 x cos4 x=

1sin2 x cos2 x

dx

cos2 x=

t=tanx dt=dx

cos2 x

sin2 x= t2

1+t2cos2 x= 1

1+t2

=

(1 + t2)2

t2dt.

60

dx(sinx+

2cosx

)2 = t=tan x2 dx= 2dt1+t2sinx= 2t

1+t2cosx= 1t

2

1+t2

= 12

(1+t2)(1t2)2dt(t(1t2)+(1+t2)2)2 .

70

2x 13 x dx =

t =

2x13x

dx = 10t dt(2+t2)2

= 10

t2dt

(2 + t2)2.

80

3

x+ 22x 1 dx =

t =3

x+22x1

dx = 15t2dt

(2t31)2

= 15

t3dt

(2t3 1)2 .