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Neodred¯eniintegrali Definicija. Za funkciju F : I R, gde je I interval, kaˇ zemo da je primitivna funkcija funkcije f : I R ako je F 0 (x)= f (x) za svako x I. Teorema 1. Ako je F : I R primitivna funkcija za f : I R, tada je i funkcija F (x)+ C, C R, takod¯e primitivna funkcija funkcije f. Definicija. Skup svih primitivnih funkcija funkcije f naziva se neodred¯eni integral funkcije f i oznaˇ cava sa Z f (x)dx. Osnovne osobine neodred¯enog integrala: 1 o (R f (x)dx ) 0 = f (x) , 2 o R F 0 (x)dx = F (x)+ C, 3 o R λf (x)dx = λ R f (x)dx, λ R \{0} , 4 o R( f (x) ± g(x) ) dx = R f (x)dx ± R g(x)dx . ˇ Cuvajmo drve´ ce. Nemojte ˇ stampati ovaj materijal, ukoliko to nije neophodno. 1

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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) ,

2o∫F ′(x)dx = F (x) + C ,

3o∫λf(x)dx = λ

∫f(x)dx, λ ∈ 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

2

Tablica 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

ln∣∣∣x− 1x+ 1

∣∣∣+ C

∫dx√

1− x2={

arcsinx+ C,− arccosx+ C

∫dx√x2 ± 1

= ln |x+√x2 ± 1|+ C

∫exdx = ex + C

∫axdx =

ax

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

∫sinx dx = − cosx+ C

∫cosx dx = sinx+ C

∫dx

sin2 x= − cotx+ C

∫dx

cos2 x= tanx+ C

∫sinhx dx = coshx+ C

∫coshx dx = sinhx+ C

∫dx

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

∫f(ϕ(t)

)ϕ′(t)dt = F

(ϕ(t)

)+ C.

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

tada je∫udv = uv −

∫vdu.

3

Primeri 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,

xn−1dx =1nd(xn), n 6= 0, smena: t = xn, dt = nxn−1dx,

dx√x

= 2 d(√x), smena: t =

√x, 2t dt = dx,

dx√ax+ b

=2ad(√ax+ b), a 6= 0, smena: t =

√ax+ b,

2at 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.

4

METOD 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 = Φ(In−1, . . . , In−k),

zovemo rekurzivne formule. Da bi se na osnovu njih odredila vrednost integralaIn, neophodno je poznavati uzastopnih k integrala In−1, . . . , In−k, kao i prvihk integrala I0, . . . , Ik−1.

INTEGRALI SA KVADRATNIM TRINOMOM

1. I =∫

mx+ n

ax2 + bx+ cdx

m = 0 : I =∫

ndx

ax2 + bx+ c=n

a

∫dx

x2 + bax+ c

a

=n

a

∫dx(

x+ b2a

)2+ ca− b2

4a2

=∣∣∣t = x+ b

2a , h2 = | ca − b2

4a2 |∣∣∣ =

n

a

∫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

− b

2aI2

=12a

ln |ax2 + bx+ c| − b

2aI2.

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

2. I =∫

mx+ n√ax2 + 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− b2

4a2

)=a(t2±h2

), t=x+

b

2a, h2 =

∣∣∣ ca− b2

4a2

∣∣∣.

5

INTEGRACIJA 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)

n∏

i=1

(x− ai)si ·m∏

i=1

(x2 + pix+ qi

)ki,

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

P (x)Q(x)

=A11

x− a1+

A12

(x− a1)2+ · · ·+ A1s1

(x− a1)s1

+A21

x− a2+

A22

(x− a2)2+ · · ·+ A2s2

(x− a2)s2+ · · ·+

An1

x− an +An2

(x− an)2+ · · ·+ Ansn

(x− an)sn

+B11x+ C11

x2 + p1x+ q1+

B12x+ C12

(x2 + p1x+ q1)2+ · · ·+ B1k1x+ C1k1

(x2 + p1x+ q1)k1

+B21x+ C21

x2 + p2x+ q2+

B22x+ C22

(x2 + p2x+ q2)2+ · · ·+ B2k2x+ C2k2

(x2 + p2x+ q2)k2

+ · · ·+

Bm1x+ Cm1

x2 + 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. Integral∫R[x,(ax+ b

cx+ d

)p1/q1,(ax+ b

cx+ d

)p2/q2, . . .

]dx, gde je R neka ra-

cionalna funkcija i pi ∈ Z, qi ∈ N, uvodenjem smene

ax+ b

cx+ d= tn, n = NZS{q1, q2, . . . },

svodi se na integral racionalne funkcije.

6

2.∫

Pn(x)dx√ax2 + bx+ c

= Qn−1(x)√ax2 + bx+ c+ λ

∫dx√

ax2 + bx+ c,

koeficijente polinoma Qn−1(x) i λ odredujemo metodom neodredenihkoeficijenata nakon diferenciranja gornje jednakosti.

3. Integral∫

dx

(x+ α)n√ax2 + bx+ c

smenom x+α =1t

svodi se na slucaj

pod 2.

4. Integral∫R(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 smenu√ax2 + bx+ c = t±√a x ;

II) Kada je c ≥ 0, smena glasi√ax2 + bx+ c = xt±√c ;

III) U slucaju ax2 + bx + c = a(x − x1)(x − x2), x1, x2 ∈ R, koristimosmenu

√ax2 + bx+ c = t(x− x1).

Trigonometrijske i hiperbolicke smene:

(a) ako se u integralu javi√a2 − x2, smena: x = a sin t ili x = a cos t;

(b) ako se u integralu javi√a2 + x2, smena: x = a tan t ili x = a sinh t;

(c) ako se u integralu javi√x2 − 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 =q1

m, r =

r1

m, gde su q1, r1 ∈ Z i m ∈ N,

uvodimo smenu: x = tm, dx = mtm−1dt. Polazni integral tada

postaje I = m

∫tr1+m−1

(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 =s

m, smena: x = t1/q ili ax−q + b = tm.

7

INTEGRACIJA 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(1−t2)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 cosl−2 x

dx

cos2 x

=∫

1sink xcosk x

cosl+k−2 x

dx

cos2 x

=∫

tan−k x(1 + tan2 x

) l+k−22 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 − e−x

2, coshx =

ex + e−x

2, tanhx =

sinhxcoshx

,

cosh2x−sinh2x=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 =

1√1− tanh2 x

,

arc sinhx = ln∣∣x+

√x2 + 1

∣∣, arc coshx = ln∣∣x+

√x2 − 1

∣∣,

arc tanhx =12

ln∣∣∣∣x+ 1x− 1

∣∣∣∣, arc cothx =12

ln∣∣∣∣x− 1x+ 1

∣∣∣∣,

arc tanhx

a=

12

ln∣∣∣∣x+ a

x− a

∣∣∣∣, arc cothx

a=

12

ln∣∣∣∣x− ax+ a

∣∣∣∣,

9

Zadaci

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+ b

x+ c

∣∣∣+C,

c)∫

dx

a2 + x2=

1a

arctanx

a+ C, d)

∫dx

a2 − x2=

12a

ln∣∣∣x+ a

x− a∣∣∣+ C,

e)∫

x2

a2 + x2dx = x− a arctan

x

a+ C, f)

∫x2

x2 − a2dx = x− a

2ln∣∣∣x+ a

x− a∣∣∣+ C,

g)∫

dx√a2 − x2

= arcsinx

a+ C, h)

∫dx√x2 ± a2

=ln∣∣x+

√x2 ± a2

∣∣+C,

i)∫

x dx

a2 ± x2= ±1

2ln |a2 ± x2|+ C, j)

∫x dx√a2 ± x2

= ±√a2 ± x2 + C,

k)∫

dx

x(a2±x2)

=1

2a2ln∣∣∣ x2

a2±x2

∣∣∣+C, l)∫

dx

x√a2±x2

=12a

log∣∣∣∣√a2±x2−a√a2±x2+a

∣∣∣∣+C,

m)∫ √

a2 − x2dx =x

2

√a2 − x2 +

a2

2arcsin

x

a+ C,

n)∫ √

x2 ± a2dx =x

2

√x2 ± a2 ± a2

2ln∣∣x+

√x2 ± a2

∣∣+ C.

Resenje: a)∫

dx

x+ b=∫d(x+ b)x+ b

= log |x+ b|+ C.

b)∫

dx

(x+b)(x+c)=

1c−b

∫(x+c)−(x+b)(x+b)(x+c)

dx=1c−b

(∫ dx

x+b−∫

dx

x+c

)

1.a)=

1c−b 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 − x2jeste zapravo specijalan slucaj integrala pod b) za

a = a i b = −a. Tako na osnovu rezulata pod b) zakljucujemo da je

I =12a

ln∣∣∣x+ a

x− a∣∣∣+ C.

e)∫

x2

a2+x2dx=

∫a2+x2−a2

a2+x2dx=

∫dx−a2

∫dx

a2+x2

1.c)= x−a arctan

x

a+C.

f)∫

x2

x2−a2dx=

∫x2−a2+a2

x2−a2dx=

∫dx+a2

∫dx

x2−a2

1.d)= x−a

2ln∣∣∣x+ a

x− a∣∣∣+C.

g) I =∫

dx√a2 − x2

=1a

∫dx√

1− x2

a2

=∣∣∣∣t = x

adx = a dt

∣∣∣∣ =∫

dt√1− t2

= arcsin t+ C = arcsinx

a+ C.

h) I =∫

dx√x2 ± a2

=1a

∫dx√x2

a2 ± 1=∣∣∣∣t = x

adx = a dt

∣∣∣∣ =∫

dt√t2 ± 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

a2±x2=∣∣∣∣t=a2±x2

±2xdx=dt

∣∣∣∣=±12

∫dt

t=±1

2ln |t|+C=±1

2ln∣∣a2±x2

∣∣+C.

j) I =∫

xdx√a2±x2

=∣∣∣∣t=a2±x2

±2xdx=dt

∣∣∣∣=±12

∫t−

12dt=±1

2t

12

1/2+C=±

√a2±x2+C.

k) I =∫

dx

x(a2 ± x2)

=1a2

∫a2 ± x2 ∓ x2

x(a2 ± x2

) dx =1a2

∫dx

x∓ 1a2

∫x dx

a2 ± x2

1.i)=

1a2

log |x| − 12a2

log∣∣a2 ± x2

∣∣+ C =1

2a2log∣∣∣ x2

a2 ± x2

∣∣∣+ C.

l) I =∫

dx

x√a2 ± x2

=∫

x dx

x2√a2 ± x2

=

∣∣∣∣∣t =√a2 ± x2, x2 = ±(t2 − a2

)dt = ±x dx√

a2±x2

∣∣∣∣∣

=∫

dt

t2 − a2

1.d)=

12a

log∣∣∣ t− at+ a

∣∣∣+ C =12a

log∣∣∣√a2 ± x2 − a√a2 ± x2 + a

∣∣∣+ C.

11

m) Oblast definisanosti podintegralne funkcije√a2 − 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 ∈ [−π/2, π/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 +

a2

2arcsin

x

a+ C.

n) I =∫ √

x2 ± a2dx =

∣∣∣∣∣u =√x2 ± a2, du = x dx√

x2±a2

dv = dx, v =∫dx = x

∣∣∣∣∣

= x√x2 ± a2 −

∫x2 dx√x2 ± a2

= x√x2 ± a2 −

∫x2 ± a2 ∓ a2

√x2 ± a2

dx

= x√x2 ± a2 −

(∫ √x2 ± a2 dx∓ a2

∫dx√x2 ± 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 ± a2

2ln |x+

√x2 ± a2|+ C.

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

I =∫ √

x2 ± a2dx =∫

x2 ± a2

√x2 ± a2

dx = (c0x+ c1)√x2 ± a2 + λ

∫dx√x2 ± a2

Diferenciranjem ove jednakosti nalazimo

x2 ± a2

√x2 ± a2

= c0

√x2 ± a2 + (c0x+ c1)

x√x2 ± a2

+ λdx√x2 ± a2

.

Posle mnozenja sa√x2 ± a2 i sredivanja izraza, dobijamo

x2 ± a2 = 2c0x2 + c1x± c0a

2 + λ.

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

1 = 2c0, 0 = c1, ±a2 = ±c0a2 + λ,

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

Time trazeni integral postaje

I =x

2

√x2 ± a2 ± a2

2

∫dx√x2 ± a2

1.h)=

x

2

√x2 ± a2 ± a2

2ln |x+

√x2 ± a2|+C.

2. Odrediti sledece integrale:

10

∫ √x(x+ 2)3dx 20

∫x2(1 + x)20dx 30

∫x dx

(x+ 9)10

40

∫dx

x2 + 4x+ 450

∫dx

x2 + x− 260

∫ √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.

20

∫x2(1 + x)20dx=

∣∣∣∣t=1 + xdt = dx

∣∣∣∣ =∫

(t− 1)2t20dt=∫

(t2 − 2t+ 1)t20dt

=∫ (

t22 − 2t21 + t20)dt =

t23

23− 2

t22

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)t−10dt =∫ (

t−9 − 9t−10)dt

=t−8

−8− 9

t−9

−9+ C =

1(x+ 9)9

− 18(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

ln∣∣∣x− 1x+ 2

∣∣∣+ C.

60

∫ √1− 2x2 + x4 dx =

∫ √(1− x2)2 dx =

∫ (1−x2

)dx =

(x−x

3

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 − 220

∫x7dx

(1 + x4)230

∫x8dx

(x3 − 1)3

40

∫x9

x5 + 1dx 50

∫dx

x(x10 + 1

)2 60

∫1− x7

x(1 + x7

) dx

70

∫xdx√1 + x4

80

∫dx

x√x2 + 1

90

∫ (x+

12

)√x2 + x+ 1 dx

Resenje: 10

∫x3dx

x8 − 2=

14

∫d(x4)

(x4)2 − (√

2)2

1.d)=

18√

2ln∣∣∣∣x4 −√2x4 +

√2

∣∣∣∣+ C.

20

∫x7dx

(1 + x4)2=∫

x4

(1 + x4)2x3dx =

∣∣∣∣t = 1 + x4

dt = 4x3dx

∣∣∣∣=14

∫t− 1t2

dt

=14

∫ (t−1 − t−2

)dt =

14(

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

5log∣∣1 + x5

∣∣+ C.

15

50

∫dx

x(x10 + 1

)2 =∫

x9dx

x10(x10 + 1

)2 =∣∣∣∣t = x10

dt = 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∣∣∣ t

t+1

∣∣∣+ 110(t+1)

+C=110

log∣∣∣ x10

x10+1

∣∣∣+ 110(x10+1

)+C.

60

∫1−x7

x(1+x7

) dx=∫

1−x7

x7(1+x7

) x6dx=∣∣∣∣t = x7

dt = 7x6dx

∣∣∣∣=17

∫1−tt(1+t)

dt

=17

∫1 + t− 2tt(1 + t)

dt =17

log|x7|(

x7 + 1)2 + C.

70

∫xdx√1 + x4

=12

∫d(x2)√

1 + (x2)2=

12

ln∣∣x2 +

√1 + x4

∣∣+ C.

80

∫dx

x√x2 + 1

=∫

x dx

x2√x2 + 1

=

∣∣∣∣∣t =√x2 + 1, x2 = t2 − 1

dt = x dx√x2+1

∣∣∣∣∣

=∫

dt

t2 − 11.d)=

12

ln∣∣∣∣t− 1t+ 1

∣∣∣∣+ C2 =12

ln∣∣∣∣√x2 + 1− 1√x2 + 1 + 1

∣∣∣∣+ C2.

90

∫ (x+

12

)√x2 + x+ 1 dx =

∣∣∣∣t = x2 + x+ 1dt = (2x+ 1)dx = 2

(x+ 1

2

)dx

∣∣∣∣

=12

∫ √t dt =

12t3/2

3/2+ C =

13

√(x2 + x+ 1)3 + C.

16

4. Metodom smene odrediti:

10

∫dx√

5x− 220

∫x√

1− 4xdx 30

∫4√

1− 4x dx

40

∫dx√

x+ 1 +√x− 1

50

∫dx√

x+ 3−√x− 2

60

∫x5

√1− x2

dx 70

∫x dx

(1 + x2

)3/2 80

∫x3dx(

x2 + 1)3√

x2 + 1

90

∫2

(2− x)23

√2− x2 + x

dx

Resenje: 10

∫dx√

5x− 2=

25

∫d(√

5x− 2) =25√

5x− 2 + C.

20

∫x√

1− 4xdx =

∣∣∣∣t =√

1− 4x, x = 1−t24

dx = −12 t dt

∣∣∣∣ =18

∫(t2 − 1)dt

=18

( t33− t)

+ C =−112√

1− 4x (1 + 2x) + C.

30

∫4√

1− 4x dx=∣∣∣∣t = 4√

1− 4x, t4 = 1− 4xx = 1−t4

4 , dx = −t3dt

∣∣∣∣=−∫t4dt=− t

5

5+ C

= 4√

1− 4x4x− 1

5+ C.

40

∫dx√

x+1+√x−1

=∫ √

x+1−√x−1x+ 1−(x− 1)

dx=12

∫ (√x+1−√x−1

)dx

=12

∫ √x+ 1 d(x+ 1)− 1

2

∫ √x− 1 d(x− 1)

=12

(x+1)3/2

3/2− 1

2(x−1)3/2

3/2+C=

13((x+1)3/2−(x−1)3/2

)+C.

17

50

∫dx√

x+3−√x−2=∫ √

x+3+√x−2

(x+3)−(x−2)dx =

15

∫ (√x+ 3 +

√x− 2

)dx

=215((x+ 3)3/2 + (x− 2)3/2

)+ C.

60

∫x5

√1− x2

dx =∫

x4

√1− x2

xdx =

∣∣∣∣∣t =√

1− x2, x2 = 1− t2dt = − x dx√

1−x2

∣∣∣∣∣

= −∫

(1− t2)2dt = −√

1− x2(1

5(1− x2)2 − 2

3(1− x2) + 1

)+ C.

70

∫x dx

(1+x2

)3/2 =

∣∣∣∣∣t= 1√

1+x2

dt=− x dx(1+x2)3/2

∣∣∣∣∣=−∫dt=−t+C=− 1√

1+x2+C.

80

∫x3dx(

x2 + 1)3√

x2 + 1=

∣∣∣∣∣t =√x2 + 1, x2 = t2 − 1

dt = x dx√x2+1

∣∣∣∣∣ =∫t2 − 1t6

dt

=∫ (t−4−t−6

)dt=

t−3

−3− t−5

−5+C=

15√

(x2+1)5− 1

3√

(x2+1)3+C.

90

∫2

(2− x)23

√2− x2 + x

dx=

∣∣∣∣∣∣∣

t = 3

√2+x2−x

dt = 4/3(2−x)2

3

√(2−x2+x

)2dx

∣∣∣∣∣∣∣=

32

∫t dt

=34t2 + C =

34

3

√(2 + x

2− 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)− lnx

x(1 + x)dx 60

∫xx(1 + log x)dx

70

∫x(

ln(1 + x) + ln(1− x))2

x2 − 1dx

Resenje: 10

∫dx

x lnx=∫d(lnx)

lnx= ln | lnx|+ C.

20

∫ √x+ lnxx

dx=∫x−1/2dx+

∫lnx

dx

x=x1/2

1/2+∫

lnx d(lnx)

=2√x+

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)−lnx

x(1+x)dx=

∫ln(1+ 1

x

)

x(1+x)dx=

∣∣∣∣∣∣t=ln

(1+ 1

x

)

dt= 11+1x

−dxx2 =− dx

x(1+x)

∣∣∣∣∣∣

=−∫t dt = − t

2

2+ C = −1

2ln2(

1 +1x

)+ C.

60∫xx(1 + log x)dx =

∣∣∣∣∣t=xx, log t=x log xdtt = (1 + log x)dx

∣∣∣∣∣=∫dt= t+ C=xx + C.

19

70

∫x(

ln(1+x)+ln(1−x))2

x2−1dx=

∫ln2(1−x2

) x dx

x2−1=

∣∣∣∣∣t=ln

(1−x2

)

dt= −2x dx1−x2

∣∣∣∣∣

=12

∫t2dt =

16t3 + C =

16

ln3(1− x2

)+ C.

6. Metodom smene odrediti:

10

∫dx

ex + 120

∫dx√

1 + e2x30

∫ √ex − 1

1 + 3e−xdx

40

∫e−x

2−1x dx 50

∫x e−(1+x2)−1/2

√(1 + x2)3

dx 60

∫x+ 1

x(1 + x ex

) dx

Resenje: 10

∫dx

ex+1=∫

dx

ex(1+e−x

)=−∫d(1+e−x)

1+e−x=− ln

(1+e−x

)+C

= lnex

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

∫dx√

1 + e2x=∫

dx

ex√e−2x + 1

=∣∣∣∣t = e−x

dt = −e−xdx∣∣∣∣ = −

∫dt√t2 + 1

= − ln |t+√t2 + 1|+ C = − ln(e−x +

√e−2x + 1) + C

= lnex

1 +√

1 + e2x+ C = x− ln

(1 +

√1 + e2x

)+ C.

30

∫ √ex − 1

1 + 3e−xdx =

∫ex − 1ex + 3

ex dx√ex − 1

=

∣∣∣∣∣t =√ex − 1, ex = t2 + 1

dt = 12

ex dx√ex−1

∣∣∣∣∣

=2∫

t2

t2+4dt

1.e)= 2t−4 arctan

t

2+C=2

√ex−1−4 arctan

√ex−12

+C.

20

40

∫e−x

2−1x dx =∣∣∣∣t=−x2−1dt=−2x dx

∣∣∣∣= −12

∫etdt = −1

2e−x

2−1 + C.

50

∫xe−(1+x2)−

12

√(1+x2)3

dx=

∣∣∣∣∣∣∣

t=e−(1+x2)−12

dt= e−(1+x2)− 1

2 x dx

(1+x2)32

∣∣∣∣∣∣∣=∫dt= t+C=e−(1+x2)−

12+C.

60

∫x+ 1

x(1 + x ex

) dx=∫

(x+ 1)ex

xex(1 + x ex

) dx =∣∣∣∣t = xex

dt = (1 + x)exdx

∣∣∣∣

=∫

dt

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

∣∣∣ t

t+ 1

∣∣∣+ C = ln∣∣∣ xex

xex + 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 = ln

∣∣∣1 + sinxcosx

∣∣∣+ C

= log∣∣∣∣cos x2 + sin x

2

cos x2 − sin x2

∣∣∣∣+C = log∣∣∣∣1 + tan x

2

1− 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

∣∣ tanx

2

∣∣+ C.

60

∫dx

1 + sinx=∣∣∣x = t+ π/2, dx = dt

∣∣∣ =∫

dt

1 + sin(t+ π/2)=∫

dt

1+cos t7.30

= tan t/2+C=tanx− π

2

2+C=

tan x2−tan π

4

1+tan x2 tan π

4

+C

=tan x

2 − 1tan x

2 + 1+ C = − cosx

1 + sinx+ C =

21 + 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 =

∫ (1−sin2 x

)d(sinx) = sinx− sin3 x

3+

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

)2

dx=∫

1+2 cos 2x+cos22x4

dx

=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 2x

4 cos2 x+ 9 sin2 xdx 110

∫sinx cos3 x

1 + cos2 xdx

Resenje: 10 Za x ∈ (− π2 + 2kπ, π2 + 2kπ

), k ∈ Z, imamo cosx > 0. Tada je

∫dx

cos3 x=∫

1cosx

dx

cos2 x=

∣∣∣∣∣∣t=tanx, dt= dx

cos2 x

cosx= 1√1+t2

∣∣∣∣∣∣=∫ √

1+t2 dt

1.n)=( t

2

√1+t2 +

12

ln∣∣t+√

1+t2∣∣)

+C

=( tanx

2 cosx+

12

ln∣∣∣ tanx+

1cosx

∣∣∣)

+ C

=sinx

2 cos2 x+

12

ln∣∣∣sinx+ 1

cosx

∣∣∣+ C =sinx

2 cos2 x+

12

ln∣∣∣cos x2 + sin x

2

cos x2 − sin x2

∣∣∣+ C.

23

Integral, takode mozemo odrediti na sledeci nacin:∫

dx

cos3 x=∫

sin2 x+ cos2 x

cos3 xdx =

∫sin2 x dx

cos3 x+∫

dx

cosx

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

cos3 xv = 1

2 cos2 x

∣∣∣∣ =sinx

2 cos2 x− 1

2

∫dx

cosx+∫

dx

cosx

=sinx

2 cos2 x+

12

∫dx

cosx7.20

=sinx

2 cos2 x+

12

log∣∣∣∣1 + tan x

2

1− 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 x

sinx cos2 xdx =

∫sinx dxcos2 x

+∫

dx

sinx

7.50

= −∫d(cosx)cos2 x

+ ln∣∣ tan

x

2

∣∣ =1

cosx+ ln

∣∣∣ tanx

2

∣∣∣+ C.

40

∫dx

sin3 x=

18

( 1cos2 x

2

− 1sin2 x

2

)+

12

log∣∣ tan

x

2

∣∣+ C.

50

∫dx

sin4 x= − cotx− 1

3cot3 x+ C.

60

∫dx

sin2 x cosx= log

∣∣∣∣cos x2 + sin x

2

cos x2 − sin x2

∣∣∣∣−1

sinx+ C.

70

∫tan2x dx=

∫sin2x

cos2xdx=

∫1−cos2x

cos2 xdx=

∫dx

cos2x−∫dx=tanx−x+C.

80

∫tan3 x dx=

∫tanx tan2 x dx=

∫tanx

1− cos2 x

cos2 xdx

=∫

tanx d(tanx)−∫

tanx dx 7.10

=12

tan2 x+ ln | cosx|+ C.

24

90

∫tan4 x dx =

∫tan2x

1−cos2 x

cos2 xdx=

∫tan2 x d(tanx) dx−

∫tan2 x dx

9.70

=tan3 x

3− tanx+ x+ C.

100

∫sin 2x

4 cos2 x+9 sin2 xdx=

∣∣∣∣∣t=4 cos2 x+9 sin2 x

dt=5 sin 2x dx

∣∣∣∣∣=15

∫dt

t=

15

ln |t|+ C

=15

ln(4 cos2 x+9 sin2 x

)+ C.

110

∫sinx cos3 x

1 + cos2 xdx=

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

∣∣∣∣ =12

∫1− tt

dt

=12

(∫ dt

t−∫dt)

=12

ln |t| − 12t+ C

=12

ln(1 + cos2 x

)− 1 + cos2 x

2+ C.

10. Metodom smene odrediti:

10

∫x−√arctan 2x

1 + 4x2dx 20

∫sin 2x√sinx+ 1

dx 30

∫ √1− sin 2x dx

40

∫cos

1x

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 cosx

100

∫tanx

(1 + cosx)2dx 110

∫dx

(3 + cosx) sinx120

∫sin 3x+ cos 3xsin 3x cos 3x

dx

25

Resenje: 10

∫x−√arctan 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 2x√sinx+1

dx=2∫

sinx cosx√sinx+1

dx=

∣∣∣∣∣t=√

sinx+1, sinx= t2−1dt= 1

2cosx√sinx+1

dx

∣∣∣∣∣

=4∫ (

t2 − 1)dt =

43t3 − 4t+ C =

43t(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

1x

+ C.

50

∫ √sinx cos2 x dx =

∫ √sinx cosx dx=

∫ √sinx d(sinx)=

(sinx)3/2

3/2+C,

za x ∈ (2kπ, 2kπ + π/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 x

2 dx = 2dt1+t2

sinx = 2t1+t2

cosx = 1−t21+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 x

2 , dx= 2dt1+t2

cosx = 1−t21+t2

∣∣∣∣∣=∫

2dtt2+5

1.c)=

2√5

arctantan x

2√5

+C.

90

∫dx

4− 3 cosx=

∣∣∣∣∣t = tan x

2 , dx = 2dt1+t2

cosx = 1−t21+t2

∣∣∣∣∣ =∫

2dt1 + 7t2

1.c)=

2√7

arctan(√

7 tanx

2)

+ C.

100

∫tanx

(1 + cosx)2dx =

∫sinx dx

cosx(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

∣∣∣+1

1 + cosx+ C.

110

∫dx

(3+cosx) sinx=−

∫d(cosx)

(3+cosx)(1−cos2 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)− 1

2

∫dt

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

18

log∣∣∣ t−1t+3

∣∣∣+ 14

log∣∣∣ t+3t+1

∣∣∣+C=18

log∣∣∣(1−cosx)(3+cosx)

(1+cosx)2

∣∣∣+C

27

120

∫sin 3x+ cos 3xsin 3x cos 3x

dx =∫

dx

cos 3x+∫

dx

sin 3x=∣∣∣∣t = 3xdx = dt

3

∣∣∣∣

=13

∫dt

cos t+

13

∫dt

sin t7.20,7.50

=13

(log∣∣∣1+tan t

2

1−tan t2

∣∣∣+log∣∣∣ tan

t

2

∣∣∣)

+C

=13

log

∣∣∣∣∣

(1 + tan 3x

2

)tan 3x

2

1− tan 3x2

∣∣∣∣∣+ C.

11. Metodom smene odrediti:

10

∫dx

sinhx20

∫sinhx√cosh 2x

dx

30

∫sinhx coshx√

sinh4 x+ cosh4 xdx 40

∫dx

cosh2 x3√

tanh2 x

Resenje: 10

∫dx

sinhx=

12

∫dx

sinh x2 cosh x

2

=12

∫dx

tanh x2 cosh2 x

2

=∫d(

tanh x2

)

tanh x2

= ln∣∣ tanh

x

2

∣∣+ C.

20

∫sinhx√cosh 2x

dx =∫

sinhx dx√2 cosh2 x− 1

=1√2

∫d(coshx)√cosh2 x− 1

2

1.h)=

1√2

ln∣∣∣ coshx+

√cosh2 x− 1

2

∣∣∣+ C.

30

∫sinhx coshx√

sinh4 x+ cosh4 xdx =

12

∫sinh 2x dx√

(sinh2 x)2 + (cosh2 x)2

=12

∫sinh 2x dx√(

cosh 2x−12

)2 +(

cosh 2x+12

)2 =∫

sinh 2x dx√2(cosh2 2x+ 1)

=∣∣∣∣t=cosh 2xdt=2 sinh 2x dx

∣∣∣∣=1

2√

2

∫dt√t2+1

=1

2√

2ln∣∣t+√t2+1

∣∣+C

=1

2√

2ln(

cosh 2x+√

cosh2 2x+ 1)

+ C.

28

40

∫dx

cosh2 x3√

tanh2 x=∫

tanh−2/3 x d(tanhx) = 3 3√

tanhx+ C.

12. Primenom metoda parcijalne integracije odrediti:

10

∫dx

(a2 + x2)220

∫x2dx

(x2 − a2)2

30

∫x3

√a2 + x2

dx 40

∫x3

√a2 − x2

dx

50

∫dx(√

x2 − a2)5

Resenje: 10

∫dx

(a2 + x2)2=

1a2

∫a2 + x2 − x2

(a2 + x2)2dx

=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 = 1

2

∫d(a2 + x2)(a2 + x2)2

= − 12(a2 + x2)

∣∣∣∣∣∣

=1a3

arctanx

a− 1a2

(− x

2(a2 + x2)+

12

∫dx

a2 + x2

)

=1

2a3arctan

x

a+

x

2a2(a2 + x2)+ C.

20

∫x2dx

(x2−a2)2=

∣∣∣∣∣u=x dv= x dx

(x2−a2)2

du=dx v=− 12(x2−a2)

∣∣∣∣∣=−x

2(x2−a2)+

12

∫dx

x2−a2

1.d)=

x

2(a2 − x2)+

14a

log∣∣∣x− ax+ a

∣∣∣+ C.

29

30

∫x3

√a2 + x2

dx =

∣∣∣∣∣u = x2 du = 2x dx

dv = x dx√a2+x2

v1.j)=√a2 + x2

∣∣∣∣∣

=x2√a2+x2 −2

∫x√a2+x2 dx=x2

√a2+x2−

∫√a2+x2 d

(a2+x2

)

=x2√a2+x2− 2

3

√(a2+x2

)3+C=x2−2a2

3

√a2+x2 +C.

Drugi nacin odredivanja ovog integrala podrazumeva koriscenje rezultata∫

Pn(x)dx√ax2 + bx+ c

= Qn−1(x)√ax2 + bx+ c+ λ

∫dx√

ax2 + bx+ c.

Tada je I =∫

x3

√a2 + x2

dx =(c0x

2 + c1x + c2

)√a2 + x2 + λ

∫dx√a2 + x2

,

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

x3

√a2 + x2

= (2c0x+ c1)√a2 + x2 +

(c0x

2 + c1x+ c2

) x√a2 + x2

+λ√

a2 + x2.

Nakon mnozenja sa√a2 + x2 i sredivanja, izraz postaje

x3 = 3c0x3 + 2c1x

2 + (2a2c0 + c2)x+ a2c1 + λ.

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

c0 =13, c1 = 0, c2 = −2a2

3, λ = 0,

sto ponovo daje vrednost integrala I =x2 − 2a2

3

√a2 + x2 + C.

40

∫x3

√a2 − x2

dx =

∣∣∣∣∣u = x2 du = 2x dx

dv = x dx√a2−x2

v1.j)= −√a2 − x2

∣∣∣∣∣

=−x2√a2− x2 +2

∫x√a2−x2 dx=−x2

√a2−x2−

∫√a2−x2 d

(a2−x2

)

=−x2√a2−x2− 2

3

√(a2−x2

)3+C=−x2+2a2

3

√a2−x2+C.

30

Drugi nacin odredivanja :

I =∫

x3

√a2 − x2

dx =(c0x

2 + c1x+ c2

)√a2 − x2 + λ

∫dx√a2 − x2

.

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

c0 = −13, c1 = 0, c2 = −2a2

3, λ = 0,

tj., I = −x2 + 2a2

3

√a2 − x2 + C.

50

∫dx(√

x2−a2)5 =

1a2

∫x2−(x2−a2)

(x2−a2)52

dx=1a2

∫x2

(x2−a2)52

dx− 1a2

∫dx

(x2−a2)32

=

∣∣∣∣∣∣

u = x dv = x dx(x2−a2)5/2

du = dx v = 12

∫ d(x2−a2)

(x2−a2)5/2 = − 13(x2−a2)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

(x2−a2)3/2− 2

3a4

(∫x2

(x2−a2)3/2dx−

∫dx√x2−a2

)

=

∣∣∣∣∣∣

u = x dv = x dx(x2−a2)3/2

du = dx v = 12

∫ d(x2−a2)

(x2−a2)3/2 = − 1√x2−a2

∣∣∣∣∣∣

=−13a2

x

(x2−a2)3/2− 2

3a4

( −x√x2−a2

+∫

dx√x2−a2

−∫

dx√x2−a2

)

=2

3a4

x√x2 − a2

− 13a2

x(√x2 − a2

)3 + C =x

3a4

2x2 − 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 x

x2dx 80

∫x2 log

(x+

√a2 + x2

)dx

30

∫x2 log2 x dx 60

∫x log

2 + x

2− x dx 90

∫x log(x+

√1 + x2)√

1 + x2dx

Resenje: 10

∫log x dx =

∣∣∣∣u = log x du = dx

xdv = dx v = x

∣∣∣∣ = x log x−∫dx

= x log x− x+ C = x(log x− 1) + C = x logx

e+ C.

20

∫log2 x dx =

∣∣∣∣u=log2 x du=2 log x dx

xdv=dx v=x

∣∣∣∣=x log2 x−2∫

log x dx

13.10

= x log2 x− 2x(log x− 1) + C = x(

log2 x

e+ 1)

+ C.

30

∫x2 log2 x dx=

∣∣∣∣∣∣u=log2 x du= 2 log x dx

x

dv=x2dx v= x3

3

∣∣∣∣∣∣=x3

3log2 x− 2

3

∫x2 log x dx

=

∣∣∣∣∣u = log x du = dx

x

dv = x2dx v = x3

3

∣∣∣∣∣ =x3

3log2 x− 2

3

(x3

3log x− 1

3

∫x2dx

)

=13x3 log2 x− 2

9x3 log x+

227x3+C=

x3

27

(9 log2 x−6 log x+2

)+C

=x3

27((3 log x− 1)2 + 1

)+ C.

40

∫log xx2

dx=

∣∣∣∣∣u = log x du = dx

x

dv = dxx2 v = − 1

x

∣∣∣∣∣=−1x

log x+∫dx

x2=−1

xlog x− 1

x+C

=−1x

(log x+ 1) + C = −1x

log ex+ C.

32

50

∫log2 x

x2dx =

∣∣∣∣∣∣u=log2 x dv= dx

x2

du= 2 log x dxx v=− 1

x

∣∣∣∣∣∣=−1

xlog2 x+2

∫log xx2

dx

13.40

= −1x

log2 x− 2x

log ex+ C = −1x

(log2 ex+ 1

)+ C.

60

∫x log

2+x2−x dx=

∣∣∣∣∣∣u=log 2+x

2−x dv=x dx

du= 4 dx4−x2 v= x2

2

∣∣∣∣∣∣=x2

2log

2+x2−x+2

∫x2

x2−4dx

1.f)=x2

2log

2+x2−x+2x−2log

2+x2−x+C=

x2−42

log2+x2−x+2x+C.

70

∫log(x+√a2 + x2

)dx =

∣∣∣∣∣∣u = log

(x+√a2 + x2

)dv = dx

du = dx√a2+x2

v = x

∣∣∣∣∣∣

= x log(x+√a2+x2

)−∫

x dx√a2+x2

1.j)= x log

(x+√a2+x2

)−√a2+x2+C.

80

∫x2 log

(x+

√a2 + x2

)dx =

∣∣∣∣∣∣u = log

(x+√a2 + x2

)dv = x2dx

du = dx√a2+x2

v = x3

3

∣∣∣∣∣∣

=x3

3log(x+

√a2 + x2

)− 13

∫x3

√a2 + x2

dx

12.30

=x3

3log(x+

√a2 + x2

)− x2 − 2a2

9

√a2 + x2 + C.

90

∫x log(x+

√1 + x2)√

1 + x2dx =

∣∣∣∣∣∣u = log

(x+√

1 + x2)

dv = x dx√1+x2

du = dx√1+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:

10∫x3 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: 10∫x3 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.

20

∫x sin2 x dx =

∣∣∣∣u = x dv = sin2 x dx = 1−cos 2x

2 dxdu = dx v = x

2 − 14 sin 2x

∣∣∣∣

=x2

2−x

4sin 2x−

∫ (x2− 1

4sin 2x

)dx

=x2

2−x

4sin 2x−x

2

4− 1

8cos 2x+C=

x2

4−x

4sin 2x− 1

8cos 2x+C.

30

∫x2 cos 2x dx =

∣∣∣∣u = x2 du = 2x dxdv = cos 2x dx v = 1

2 sin 2x

∣∣∣∣

=x2

2sin 2x−

∫x sin 2x dx =

∣∣∣∣u = x du = dxdv = sin 2x dx v = −1

2 cos 2x

∣∣∣∣

=x2

2sin 2x+

x

2cos 2x− 1

2

∫cos 2x dx

=x2

2sin 2x+

x

2cos 2x− 1

4sin 2x+C=

2x2−14

sin 2x+x

2cos 2x+C.

34

40

∫dx

cos5 x=∫

sin2 x+ cos2 x

cos5 xdx =

∫sin2 x

cos5 xdx+

∫dx

cos3 x

=∣∣∣∣u=sinx du=cosx dxdv= sinx dx

cos5 xv= 1

4 cos4 x

∣∣∣∣=sinx

4 cos4x+

34

∫dx

cos3 x

9.10

=sinx

4 cos4 x+

38

sinxcos2 x

+38

log∣∣∣cos x2 + sin x

2

cos x2 − sin x2

∣∣∣+ C.

50

∫tan7 x dx =

∫sin7 x

cos7 xdx =

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

cos7 xv = 1

6 cos6 x

∣∣∣∣

=16

tan6 x−∫

sin5 x

cos5 xdx=

∣∣∣∣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−∫

tan3x1−cos2x

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.

60

∫x sin

√x dx =

∣∣∣∣t =√x, x = t2

dx = 2t dt

∣∣∣∣ = 2∫t3 sin t dt

14.10

= 2(− t3 cos t+ 3t2 sin t+ 6t cos t− 6 sin t

)+ C

= −2√x3 cos

√x+ 6x sin

√x+ 12

√x cos

√x− 12 sin

√x+ C.

35

70

∫(2x− 1) sin 3x dx =

∣∣∣∣u = 2x− 1 dv = sin 3x dxdu = 2dx v = −1

3 cos 3x

∣∣∣∣

= −2x−13

cos 3x+23

∫cos 3x dx=

1−2x3

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=−1

2 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 = 1

cosx−sinx du = − sinx+cosx(cosx−sinx)2 dx

dv = d(sinx+cosx)(sinx+cosx)4 v = −1

3(sinx+cosx)3

∣∣∣∣∣∣

=−1

3(sinx+cosx)3(cosx−sinx)− 1

3

∫dx

(cosx−sinx)2(sinx+cosx)2

=−1

3(sinx+ cosx)2(cos2 x− sin2 x)− 1

3

∫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

∫arccos

x

2dx 30

∫x2 arccosx dx

40

∫arcsin2 x dx 50

∫x arcsin2 x dx 60

∫arcsinxx2

dx

70

∫x arcsinx√

1− x2dx 80

∫arcsinx arccosx dx

Resenje: 10

∫arcsinx dx=

∣∣∣∣∣u=arcsinx dv=dx

du= dx√1−x2

v=x

∣∣∣∣∣=x arcsinx−∫

x dx√1−x2

1.j)= x arcsinx+

√1− x2 + C.

20

∫arccos

x

2dx =

∣∣∣∣∣u=arccos x2 dv=dx

du=− dx√4−x2

v=x

∣∣∣∣∣=x arccosx

2+∫

x dx√4−x2

1.j)= x arccos

x

2−√

4− x2 + C.

30

∫x2arccosx dx =

∣∣∣∣∣u=arccosx dv=x2dx

du=− dx√1−x2

v= x3

3

∣∣∣∣∣=x3

3arccosx+

13

∫x3

√1−x2

dx

12.40

=x3

3arccosx− x2 + 2

9

√1− x2 + C.

40

∫arcsin2 x dx =

∣∣∣∣∣u=arcsin2 x dv=dx

du=2 arcsinx dx√1−x2

v=x

∣∣∣∣∣

= x arcsin2 x− 2∫

arcsinxx dx√1− x2

=

∣∣∣∣∣∣u=arcsinx du= dx√

1−x2

dv= x dx√1−x2

v1.j)= −√1−x2

∣∣∣∣∣∣

=x arcsin2 x+ 2√

1− x2 arcsinx− 2∫dx

= x arcsin2 x+ 2√

1− x2 arcsinx− 2x+ C.

37

50

∫x arcsin2 x dx==

∣∣∣∣∣u = arcsin2 x dv = x dx

du = 2 arcsinx dx√1−x2

v = x2

2

∣∣∣∣∣

=x2

2arcsin2 x−

∫arcsinx

x2dx√1− x2

=

∣∣∣∣∣∣u = x arcsinx dv = x dx√

1−x2

du =(

arcsinx+ x√1−x2

)dx v

1.j)= −√1− x2

∣∣∣∣∣∣

=x2

2arcsin2 x+x

√1−x2 arcsinx−

∫ √1−x2 arcsinx dx−

∫x dx

=

∣∣∣∣∣u = arcsinx du = dx√

1−x2

dv =√

1− x2 dx v1.m)= x

2

√1− x2 + 1

2 arcsinx

∣∣∣∣∣

=x2 − 1

2arcsin2 x+

x

2

√1−x2 arcsinx− x2

2+

12

∫x dx

+12

∫arcsinx d(arcsinx)

=2x2 − 1

4arcsin2 x+

x

2

√1− x2 arcsinx− x2

4+ C.

60

∫arcsinxx2

dx =

∣∣∣∣∣u = arcsinx dv = dx

x2

du = dx√1−x2

v = − 1x

∣∣∣∣∣ = −arcsinxx

+∫

dx

x√

1− x2

1.l)= −1

xarcsinx+

12

log∣∣∣∣√

1− x2 − 1√1− x2 + 1

∣∣∣∣+ C.

70

∫x arcsinx√

1− x2dx=

∣∣∣∣∣∣u=arcsinx dv= x dx√

1−x2

du= dx√1−x2

v1.j)=−√1−x2

∣∣∣∣∣∣=−

√1−x2 arcsinx+

∫dx

= x−√

1− x2 arcsinx+ C.

38

80

∫arcsinx arccosx dx=

∣∣∣∣∣u=arccosx dv=arcsinx dx

du=− dx√1−x2

v15.10

= x arcsinx+√

1−x2

∣∣∣∣∣

=x arcsinx arccosx+√

1−x2 arccosx+∫

arcsinxx dx√1−x2

+∫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= arccosx−arcsinx√1−x2

dx v=x

∣∣∣∣∣∣

= x arcsinx arccosx−∫

(arccosx− arcsinx)x dx√1− x2

=

∣∣∣∣∣∣u=arccosx− arcsinx dv= x dx√

1−x2

du=− 2dx√1−x2

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:

10

∫x arctanx dx 20

∫x2 arctanx dx 30

∫1x2

arctanx

3dx

40

∫arctan

√x dx 50

∫x2 arctanx

1 + x2dx 60

∫arctan

2x1− x2

dx

70

∫x earctanx

(1 + x2)3/2dx

Resenje: 10

∫x 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− x

2+ C.

39

20

∫x2arctanx dx=

∣∣∣∣∣u=arctanx du= dx

1+x2

dv=x2 dx v= x3

3

∣∣∣∣∣=x3

3arctanx− 1

3

∫x3

1+x2dx

=∣∣∣∣t=1+x2

dt=2x dx

∣∣∣∣=x3

3arctanx−1

6

∫t−1tdt=

x3

3arctanx−1

6(t−log|t|)+C

=x3

3arctanx− 1

6(1 + x2 − log(1 + x2)

)+ C.

30

∫1x2

arctanx

3dx =

∣∣∣∣∣∣u=arctan x

3 dv= dxx2

du= 3 dx9+x2 v=− 1

x

∣∣∣∣∣∣=−1

xarctan

x

3+∫

3 dxx(9+x2

)

1.k)= −1

xarctan

x

3+

16

log∣∣∣∣

x2

x2 + 9

∣∣∣∣+ C.

40

∫arctan

√x dx =

∣∣∣∣t =√x, t2 = x

2t dt = dx

∣∣∣∣ = 2∫t arctan t dt

16.10

=(t2 + 1

)arctan t− t+ C = (x+ 1) arctan

√x−√x+ C.

50

∫x2 arctanx

1 + x2dx =

∣∣∣∣∣u = arctanx du = dx

1+x2

dv = x2 dx1+x2 v

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

∫arctan

2x1− x2

dx =∣∣∣∣u = arctan 2x

1−x2 du = 2dx1+x2

dv = dx v = x

∣∣∣∣

= x arctan2x

1−x2−2∫x dx

1+x2

1.i)= x arctan

2x1−x2

−log(1+x2

)+C.

40

70 I =∫

x earctanx

(1 + x2)3/2dx =

∣∣∣∣∣∣u = x√

1+x2dv = earctan xdx

1+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 xdx1+x2 v = earctanx

∣∣∣∣∣∣

=xearctanx

√1 + x2

− earctanx

√1 + x2

−∫

x earctanx

(1 + x2)3/2dx =

(x− 1) earctanx

√1 + x2

− I.

Odatle je I =earctanx(x− 1)

2√

1 + x2+ C.

17. Primenom metoda parcijalne integracije odrediti:

10

∫x3e2x dx 20

∫x2ex

(x+ 2)2dx 30

∫e√xdx

40

∫eax cos bx dx 50

∫eax sin bx dx

60

∫e2x sin2 x dx 70

∫e2x cos2 x dx

Resenje: 10 I =∫x3e2xdx =

∣∣∣∣u = x3 du = 3x2dxdv = e2xdx v = 1

2e2x

∣∣∣∣

=x3

2e2x − 3

2

∫x2e2xdx =

∣∣∣∣u = x2 du = 2x dxdv = e2xdx v = 1

2e2x

∣∣∣∣

=x3

2e2x − 3

2

(x2

2e2x −

∫xe2xdx

)=x3

2e2x − 3x2

4e2x +

32

∫xe2xdx

=∣∣∣∣u = x du = dxdv = e2xdx v = 1

2e2x

∣∣∣∣ =x3

2e2x − 3x2

4e2x +

32

(x2e2x − 1

2

∫e2xdx

)

=x3

2e2x− 3x2

4e2x+

3x4e2x− 3

8e2x+C=

e2x

8(4x3−6x2+6x−3

)+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 = − x2

x+ 2ex + xex − ex+C

=x− 2x+ 2

ex + C.

30

∫e√xdx=

∫ √xe√xdx√x

=

∣∣∣∣∣u=√x dv= e

√xdx√x

du= dx2√x

v=2e√x

∣∣∣∣∣=2√xe√x−∫e√xdx√x

= 2√xe√x − 2e

√x + C = 2e

√x(√x− 1

)+ C.

40 I =∫eax cos bx dx =

∣∣∣∣u = cos bx du = −b sin bx dxdv = eaxdx v = 1

aeax

∣∣∣∣

=1aeax cos bx+

b

a

∫eax sin bx dx =

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

aeax

∣∣∣∣

=1aeax cos bx+

b

a

(1aeax sin bx− b

a

∫eax cos bx dx

)

=eax

a2

(a cos bx+ b sin bx

)− b2

a2I.

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.

50

∫eax sin bx dx =

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

aeax

∣∣∣∣

=1aeaxsin bx− b

a

∫eaxcos bx dx17.40

=eax

a2+b2(a sin bx−b cos bx

)+C.

42

60

∫e2x sin2 x dx =

∣∣∣∣u = sin2 x du = 2 sinx cosx dx = sin 2x dxdv = e2xdx v = 1

2e2x

∣∣∣∣

=12

(e2xsin2x−

∫e2xsin2x dx

)17.50

=e2x

8(2−cos2x−sin2x

)+C.

70

∫e2x 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

dx

30 In =∫

dx

(x2 + a2)n40 In =

∫dx

(ax2 + bx+ c)n

50 In =∫x2n√x2 ± 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 logn−1x dx

xdv =

√x dx v = 2

3x√x

∣∣∣∣

=23x3/2 logn x− 2n

3

∫ √x logn−1 x dx =

23x3/2 logn x− 2n

3In−1.

43

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

I1 =∫ √

x log x dx =

∣∣∣∣∣u = log x du = dx

x

dv =√x v = 2

3x3/2

∣∣∣∣∣ =23x3/2 log x− 2

3

∫ √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 =

227x3/2

(9 log3 x− 18 log2 x+ 24 log x− 16

)+ C.

20 In =∫ (

1− x2)n/2

dx =∫ (

1− x2)(

1− x2)(n−2)/2

dx

= In−2−∫x2(1−x2

)n−22 dx=

∣∣∣∣∣u=x dv=x

(1−x2

)n−22 dx

du=dx v=− 1n(1−x2)n/2

∣∣∣∣∣

= In−2+x

n

(1−x2

)n/2− 1n

∫ (1−x2

)n/2dx=− 1

nIn+In−2+

x

n

(1−x2

)n/2.

Dakle,n+ 1n

In = In−2 +x

n

(1− x2

)n/2, pa je

In =n

n+ 1In−2 +

x

n+ 1(1− x2

)n/2.

Znamo da je I1 =∫ √

1− x2 dx1.m)=

x

2

√1− x2 +

12

arcsinx + C. Na osnovuizvedene rekurentne relacije dobijamo

I3 =34I1 +

x

4(1− x2

)3/2 =38

arcsinx+x

8

√1− x2

(5− 2x2

)+ C.

Kako je I2 =∫ (

1− x2)dx = x− x3

3+ C, mozemo odrediti

I4 =45I2 +

x

5(1− x2

)2 + C = x− 23x3 +

15x5 + C.

44

30 In =∫

dx

(x2+a2)n=

1a2

∫x2+a2−x2

(x2+a2)ndx=

1a2In−1− 1

a2

∫x2

(x2+a2)ndx

=

∣∣∣∣∣u = x dv = x dx

(x2+a2)n

du = dx v = 1(2−2n)(x2+a2)n−1

∣∣∣∣∣

=2n− 3

(2n− 2)a2In−1 +

1(2n− 2)a2

x

(x2 + a2)n−1.

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+1dx

=

∣∣∣∣∣d(ax2 + bx+ c) = (2ax+ b)dx

ax2 + bx+ c = 14a

((2ax+ b)2 + 4ac− b2)

∣∣∣∣∣

=∫

(2ax+b)2+4ac−b24a(ax2+bx+c)n+1

dx=14a

∫(2ax+b)2

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

4ac−b24a

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+ b

4an(ax2 + bx+ c)n+

12nIn +

4ac− b24a

In+1.

⇒ 4ac− b24a

In+1 = In − 12nIn +

2ax+ b

4an(ax2 + bx+ c

)n

⇒ In+1 =2a

4ac− b22n− 1n

In +2ax+ b

n(4ac− b2)(ax2 + bx+ c

)n .

45

50 In =∫x2n√x2 ± a2 dx =

∫x2n−2

(x2 ± a2 ∓ a2

)√x2 ± a2 dx

=∫x2n−2

(x2±a2

) 32dx∓a2In−1 =

∣∣∣∣∣u=(x2±a2

) 32 dv=x2n−2dx

du=3√x2±a2xdx v= x2n−1

2n−1

∣∣∣∣∣

=x2n−1

2n− 1(x2 ± a2

) 32 − 3

2n− 1In ∓ a2In−1

⇒ 2n+ 22n− 1

In =x2n−1

2n− 1(x2 ± a2

) 32 ∓ a2In−1

⇒ In =x2n−1

2n+ 2(x2 ± a2

) 32 ∓ 2n− 1

2n+ 2a2In−1.

60 In =∫x2n+1

√x2 ± a2 dx =

∣∣∣∣∣u = x2n du = 2nx2n−1dx

dv =√x2 ± a2xdx v = 1

3

(x2 ± a2

) 32

∣∣∣∣∣

=13x2n(x2 ± a2

) 32 − 2n

3

∫x2n−1

(x2 ± a2

) 32dx

=13x2n(x2 ± a2

) 32 − 2n

3(In ± a2In−1

)

⇒ 2n+ 33

In =13x2n(x2 ± a2

) 32 ∓ 2n

3a2In−1

⇒ In =1

2n+ 3x2n(x2 ± a2

) 32 ∓ 2na2

2n+ 3In−1.

70 In=∫

sinn x dx=∫

sinn−2 x(1−cos2 x)dx=In−2−∫

cos2 x sinn−2 x dx

=∣∣∣∣u = cosx dv = sinn−2 x d(sinx)du = − sinx dx v = 1

n−1 sinn−1 x

∣∣∣∣

= In−2 − 1n− 1

sinn−1x cosx− 1n− 1

In

⇒ n

n− 1In = In−2 − 1

n− 1sinn−1 x cosx

⇒ In =n− 1n

In−2 − 1n

sinn−1 x cosx.

46

80 In =∫

dx

cosn x=∫

sin2 x+ cos2 x

cosn xdx =

∫sin2 x

cosn xdx+

∫dx

cosn−2 x

=∫

sin2 x

cosn xdx+ In−2 =

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

cosn x v = 1(n−1) cosn−1 x

∣∣∣∣∣

=sinx

(n− 1) cosn−1 x− 1

(n− 1)In−2 + In−2

=sinx

(n− 1) cosn−1 x+n− 2n− 1

In−2.

90 In =∫

tann x dx =∫

tann−2 x1− cos2 x

cos2 xdx =

∫tann−2 x

dx

cos2 x− In−2

=∫

tann−2 x d(tanx)− In−2 =1

n− 1tann−1 x− In−2.

100 In =∫

cotn x dx =∫

cotn−2 x1− sin2 x

sin2 xdx =

∫cotn−2 x

dx

sin2 x− In−2

= −∫

cotn−2 x d(cotx)− In−2 = − 1n− 1

cotn−1 x− In−2.

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+1 du=−(n+ 1) a cosx−b sinx(a sinx+b cosx)n+2 dx

dv=(a sinx+b cosx)dx v=−(a cosx−b 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 cosx−b 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+2

dv = 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+2dx

=b sinx

(a+b cosx)n+1+aIn+1+(n+1)b2In+2−

∫(n+1)b2 cos2 x

(a+b cosx)n+2dx

=b sinx

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

∫ (a+b cosx−ab

)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 + 6

x3 − 5x2 + 6xdx 20

∫dx

(x− 1)2(x2 + 1)2

30

∫dx

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

∫x

x3 − 3x+ 2dx

50

∫dx

x3 + 160

∫x+ 1x3 − 1

dx 70

∫dx

x4 + 1

80

∫x2dx(

x2 + 2x+ 2)2 90

∫2x2 − 5

x4 − 5x2 + 6dx

100

∫x3 − 1x4 + x2

dx 110

∫x+ 2x3 + 1

dx 120

∫x+ 1

x(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+ 6

x3 − 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

=A1

x− 1+

A2

(x− 1)2+B1x+ C1

x2 + 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 jednakosti

1 = 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 +B1

0 =−A1 +A2 − 2B1 + C1

0 = 2A1 + 2B1 +B2 − 2C1

0 =−2A1 + 2A2 − 2B1 − 2B2 + 2C1 + C2

0 =A1 +B1 +B2 − 2C1 − 2C2

1 =−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

∣∣∣ x

x+ 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+ 1

2

∣∣

1.c)= log

∣∣∣ x

x+ 1

∣∣∣− 2√3

arctan2x+ 1√

3−∫

dt(t2 + 3

4

)2 −∫

dt(t2 + 3

4

)312.10

18.30= log

∣∣∣ x

x+1

∣∣∣− 143√

3arctan

2x+1√3− 2

32x+1

x2+x+1− 2x+1

6(x2+x+1

)2 +C.

51

40

∫x

x3 − 3x+ 2dx =

29

log∣∣∣1− x2 + x

∣∣∣+1

3(1− x)+ C.

50

∫dx

x3 + 1=

1√3

arctan2x− 1√

3+

16

log(x+ 1)2

x2 − x+ 1+ C

=1√3

arctan2x− 1√

3+

16

log∣∣∣(x+ 1)3

x3 + 1

∣∣∣+ C.

60

∫x+ 1x3 − 1

dx =13

log∣∣∣ (x− 1)2

x2 + x+ 1

∣∣∣+ C =13

log∣∣∣(x− 1)3

x3 − 1

∣∣∣+ C.

70

∫dx

x4+1=∫

dx

(x2+1)2−2x2=

12√

2

(arctan(1+

√2x)−arctan(1−

√2x)

)

+1

4√

2log∣∣∣x

2 +√

2x+ 1x2 −√2x+ 1

∣∣∣+ C.

80

∫x2dx(

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

1x2 + 2x+ 2

+ C.

90

∫2x2 − 5

x4 − 5x2 + 6dx =

12√

2log∣∣∣∣x−√2x+√

2

∣∣∣∣+1

2√

3log∣∣∣∣x−√3x+√

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)3

x3 + 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

∫dx√

x+ 2+ 3√x+ 2

20

∫1−√x+ 11+ 3√x+ 1

dx 30

∫dx

6√x− 1+ 3

√x− 1+

√x− 1

40

∫dx

(x+1)5√x2+2x

50

∫ √x+1x−1

dx 60

∫dx√

(x− 1)(2− x)

52

Resenje: 10

∫dx√

x+ 2 + 3√x+ 2

=∣∣∣∣t6 = x+ 2dx = 6t5dt

∣∣∣∣ = 6∫

t5dt

t3 + t2= 6

∫t3dt

t+ 1

= 6∫t3 + 1− 1t+ 1

dt = 6∫

(t2 − t+ 1)dt− 6∫

dt

t+ 1

= 2t3 − 3t2 + 6t− 6 log |t+ 1|+ C

= 2√x+ 2− 3 3

√x+ 2 + 6 6

√x+ 2− 6 log

∣∣ 6√x+ 2 + 1

∣∣+ C.

20

∫1−√x+ 11 + 3√x+ 1

dx =∣∣∣∣t6 = x+ 1dx = 6t5dt

∣∣∣∣ = 6∫

(1− t3)t5

1 + t2dt

= 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(

6√x+ 1

)7 +65(

6√x+ 1

)5 +32(

3√x+ 1

)2 − 2√x+ 1

−3 3√x+ 1 + 6 6

√x+ 1 + 3 log( 3

√x+ 1 + 1)− 6 arctan 6

√x+ 1 + C.

30

∫dx

6√x− 1 + 3

√x− 1 +

√x− 1

=∣∣∣∣t6 = x− 1dx = 6t5dt

∣∣∣∣ = 6∫

t5

t+ t2 + t3dt

= 6∫

(t2 − t)dt+ 6∫

t dt

t2 + t+ 1= 2t3 − 3t2 + 3

∫2t+ 1− 1t2 + t+ 1

dt

= 2t3 − 3t2 + 3∫d(t2 + t+ 1)t2 + t+ 1

dt− 3∫

dt

t2 + t+ 1

= 2t3 − 3t2 + 3 log |t2 + t+ 1| − 3∫

dt

(t+ 12)2 + 3

4

1.c)= 2t3 − 3t2 + 3 log

∣∣∣ t3 − 1t− 1

∣∣∣− 3√3/2

arctant+ 1

2√3/2

+ C

= 2√x−1−3 3

√x−1+3 log

∣∣∣∣√x−1−1

6√x−1−1

∣∣∣∣−2√

3 arctan2 6√x−1+1√

3+C.

53

40 I =∫

dx

(x+1)5√x2+2x

=

∣∣∣∣∣x+1= 1

t , x, t > 0

dx=−dtt2

∣∣∣∣∣=−∫

t4dt√1−t2

=(a0t

3 + a1t2 + a2t+ a3

)√1− t2 + λ

∫dt√

1− t2 .

Diferenciranjem po t i mnozenjem sa√

1− t2 dobijamo

−t4 = −4a0t4 − 3a1t

3 + (3a0 − 2a2)t2 + (2a1 − a3)t+ a2 + λ.

Izjednacavanjem koeficijenata uz stepene od t formiramo sistem jednacina

−1 = −4a0 0 = 2a1 − a3

0 = −3a1 0 = a2 + λ

0 = 3a0 − 2a2

cije je resenje

a0 = 1/4, a1 = 0, a2 = 3/8, a3 = 0, λ = −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 je√t2 = −t.

Ovaj integral takode se moze odrediti primenom Ojlerove smene.

I=∫

dx

(x+ 1)5√x2 + 2x

=

∣∣∣∣∣

√x2 + 2x = t x, x = 2

t2−1

dx = − 4t dt(t2−1)2

∣∣∣∣∣

=−4∫

(t2 − 1)4

(t2 + 1)5dt = −4

∫(t2 + 1− 2)4

(t2 + 1)5dt

=−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

arctan√x2 + 2xx

+ C.

50

∫ √x+ 1x− 1

dx =

∣∣∣∣∣t =

√x+1x−1

dx = −4t dt(t2−1)2

∣∣∣∣∣ = −4∫

t2

(t2 − 1)2dt

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, π/2)dx = 2 sin t cos t dt,√

(x− 1)(2− x) =√

sin2 t(1− sin2 t)

∣∣∣∣∣∣

=2∫dt = 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 = t2+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− x230

∫x−√x2 + 3x+ 2x+√x2 + 3x+ 2

dx

40

∫ √x3 + x4 dx 50

∫ √x(

1 + 3√x)2 dx 60

∫3√

3x− 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 uzimamo√x2 + 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

=A1

t+

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 = 2∫dt

t−3∫

dt

2t+1− 3

2

∫d(2t+1)(2t+1)2

=2 log |t|− 32

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

+C1

= 2 log∣∣x+

√x2+x+1

∣∣− 32

log∣∣2x+2

√x2+x+1+1

∣∣−x+√x2+x+1+C,

gde je C = C1 − 1/2.

56

20 I =∫

dx

1+√

1−2x−x2=

∣∣∣∣∣

√1−2x−x2 =xt−1

dx=2−t2+2t+1

(t2+1)2 dt

∣∣∣∣∣=∫ −t2+2t+1t(t−1)(t2+1)

dt.

−t2 + 2t+ 1t(t− 1)(t2 + 1)

=A1

t+

A2

t− 1+Bt+ C

t2 + 1A1 = −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− x2

1 +√

1− 2x− x2

∣∣∣− 2 arctan1 +√

1− 2x− x2

x+ C.

30 I=∫x−√x2+3x+2x+√x2+3x+2

dx=

∣∣∣∣∣

√x2+3x+2=

√(x+2)(x+1)= t(x+1)

x = 2−t2t2−1

, dx = − 2tdt(t2−1)2

∣∣∣∣∣

=−2∫

t2 + 2t(t− 2)(t− 1)(t+ 1)3

dt.

−2t2 + 2t

(t− 2)(t− 1)(t+ 1)3= − 16

27(t− 2)+

34(t− 1)

− 17108(t+ 1)

+5

18(t+ 1)2+

13(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 + x−1)1/2 dx =

∣∣∣∣∣1 + x−1 = t2

dx = −2t dt(t2−1)2

∣∣∣∣∣

=−2∫

t2dt

(t2 − 1)4= −2

∫t2 − 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)n−1− 2n− 3

2(n− 1)In−1

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 = − t

2(t2 − 1)− 1

4log∣∣∣ t− 1t+ 1

∣∣∣+ C2.

I3 =3t3 − 5t

8(t2 − 1)2+

316

log∣∣∣ t− 1t+ 1

∣∣∣+ C3.

I4 = −15t5 − 40t3 + 33t48(t2 − 1)3

− 532

log∣∣∣ t− 1t+ 1

∣∣∣+ C4.

I = −3t5 − 8t3 − 3t24(t2 − 1)3

− 116

log∣∣∣ t− 1t+ 1

∣∣∣+ C

=124

√x2 + x(8x2 + 2x− 3) +

18

log∣∣∣√x+

√1 + x

∣∣∣+ C.

58

50

∫ √x(

1+ 3√x)2 dx=

∣∣∣∣x= t6

dx=6t5dt

∣∣∣∣=6∫

t8 dt

(t2+1)2

= 6∫

(t4 − 2t2 + 3)dt− 6∫

4t2 + 3(t2 + 1)2

dt

=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

=6√x

5(1 + 3√x )(6x− 14 3

√x2 + 70 3

√x+ 105

)− 21 arctan 6√x+ C.

60

∫3√

3x−x3 dx=∫x(−1+3x−2)1/3dx=

∣∣∣∣∣∣−1+3x−2 = t3, x=

√3

t3+1

dx=− 3√

3t2dt2(t3+1)3/2

∣∣∣∣∣∣

= −92

∫t3dt

(t3+1)2=

∣∣∣∣∣u= t dv= t2dt

(t3+1)2

du=dt v=− 13(t3+1)

∣∣∣∣∣=3t

2(t3+1)− 3

2

∫dt

t3+1

19.50

=3t

2(t3 + 1)−√

32

arctan2t− 1√

3− 1

4log∣∣∣(t+ 1)3

t3 + 1

∣∣∣+ C

=118

log

(3√

3x− x3 + x)3

3x+x2

9+

19√

3arctan

2 3√

3x− x3 − x√3x

+ C.

22. Svesti sledece integrale na integral racionalne funkcije.

10

∫3 tanx+ 13− tanx

dx 20

∫dx

4 + 3 tanx

30

∫1− 2 cosx+ 3 sinx2 + 3 cosx− 4 sinx

dx 40

∫dx

(sinx+ 2 cosx)3

50

∫dx

sin2 x cos4 x60

∫dx(

sinx+2

cosx

)2

70

∫ √2x− 13− x dx 80

∫3

√x+ 22x− 1

dx

59

Resenje: 10

∫3 tanx+ 13− tanx

dx =∣∣∣∣t = tanxdx = dt

1+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 x

2 dx = 2dt1+t2

sinx = 2t1+t2

cosx = 1−t21+t2

∣∣∣∣∣

= 2∫

3t2 + 6t− 1(5− 8t− t2)(1 + t2)

dt .

40

∫dx

(sinx+2 cosx)3=

∣∣∣∣∣t=tan x

2 dx= 2dt1+t2

sinx= 2t1+t2

cosx= 1−t21+t2

∣∣∣∣∣=12

∫(1+t2)dt

(1+t−t2)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+2

cosx

)2 =

∣∣∣∣∣t=tan x

2 dx= 2dt1+t2

sinx= 2t1+t2

cosx= 1−t21+t2

∣∣∣∣∣=12

∫(1+t2)(1−t2)2dt(t(1−t2)+(1+t2)2

)2 .

70

∫ √2x− 13− x dx =

∣∣∣∣∣∣t =

√2x−13−x

dx = 10t dt(2+t2)2

∣∣∣∣∣∣= 10

∫t2dt

(2 + t2)2.

80

∫3

√x+ 22x− 1

dx =

∣∣∣∣∣∣t = 3

√x+22x−1

dx = −15t2dt(2t3−1)2

∣∣∣∣∣∣= −15

∫t3dt

(2t3 − 1)2.