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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.
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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.
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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.
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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
∣∣∣.
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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.
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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.
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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;
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• 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
∣∣∣∣,
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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.
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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.
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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.
