Neo Ne odred dred¯eni en i inte integral gralii funkciju F : I R, gde je I je I int interval, erval, kaˇ zemo zem o da je primitivna Definicija. Za Definicija. Za funkciju F funkcije f : I R ako je funkcija funkcije f
→ →
→ →
F (x) = f ( f (x) za svako x svako x
∈ I.
R primitivna funkcija za f : I R, tada je i Teorema 1. Ako je F : I funkcija F funkcija F ((x) + C, + C, C R, takod¯e primitivna primiti vna funkcija funkcije f. f .
∈ ∈
→ →
→ →
Definicija. Skup svih primitivnih funkcija funkcije f f naziva naziva se neod n eodred red¯eni ¯eni funkcije f i i oznaˇ oz naˇcava cava sa integral funkcije f
f ( f (x)dx.
Osnovne osobine neodred neod red¯enog integrala: integrala : 1o 2o 3o 4o
± f ( f (x)dx = f ( f (x) ,
F (x)dx = dx = F F ((x) + C + C ,
λf ( λf (x)dx = dx = λ λ f ( f (x)
f ( f (x)dx, λ
g(x) dx = dx =
∈ R \ {0} , f ( f (x)dx ± g (x)dx .
ˇ Cuvajmo drve´ ce. ce. Nemojte ˇ stampati stampati ovaj ovaj materijal, ukoliko to nije neophodno.
1
2 Tablica integrala xn+1 + C, + C, n = 1 n + 1 dx arctan x + C, + C, = arcctg x + C + C 1 + x + x2
xn dx = dx =
− √ − − − − − dx = 1 x2
−
arcsin x + C, + C, arccos x + C + C
ex dx = dx = e e x + C
sin x dx = dx =
cos x + C + C
dx = sin2 x
cot x + C + C
dx = ln x + x2 1 + C 2 x 1 ax x a dx = dx = + C + C,, a > 0 > 0,, a = 1 ln a cos x dx = dx = sin x + C + C dx = tan x + C + C cos2 x
sinh x dx = dx = cosh x + C + C dx = sinh2 x
dx = ln x + C x dx 1 x 1 = ln + C x2 1 2 x + 1
| | − √ − | ± | ± cosh x dx = dx = sinh x + C + C
coth x + C + C
dx = tanh x + C + C cosh2 x
Osnovne metode integracije:
METOD SMENE
Ako je f ( f (x)dx = dx = F F ((x) + C, tada C, tada je f ( f (u)du = du = F F ((u) + C. Uzmimo C. Uzmimo da je x = ϕ = ϕ((t), gde je ϕ je ϕ neprekidna funkcija zajedno sa svojim izvodom ϕ . Tada
f ϕ(t) ϕ (t)dt = dt = F F ϕ(t) + C.
METOD PARCIJALNE INTEGRACIJE Ako su u su u i i v v diferencijabilne diferencijabilne funkcije i funkcija uv funkcija uv ima primitivnu funkciju, tada je
udv = udv = uv uv
−
vdu.
2 Tablica integrala xn+1 + C, + C, n = 1 n + 1 dx arctan x + C, + C, = arcctg x + C + C 1 + x + x2
xn dx = dx =
− √ − − − − − dx = 1 x2
−
arcsin x + C, + C, arccos x + C + C
ex dx = dx = e e x + C
sin x dx = dx =
cos x + C + C
dx = sin2 x
cot x + C + C
dx = ln x + x2 1 + C 2 x 1 ax x a dx = dx = + C + C,, a > 0 > 0,, a = 1 ln a cos x dx = dx = sin x + C + C dx = tan x + C + C cos2 x
sinh x dx = dx = cosh x + C + C dx = sinh2 x
dx = ln x + C x dx 1 x 1 = ln + C x2 1 2 x + 1
| | − √ − | ± | ± cosh x dx = dx = sinh x + C + C
coth x + C + C
dx = tanh x + C + C cosh2 x
Osnovne metode integracije:
METOD SMENE
Ako je f ( f (x)dx = dx = F F ((x) + C, tada C, tada je f ( f (u)du = du = F F ((u) + C. Uzmimo C. Uzmimo da je x = ϕ = ϕ((t), gde je ϕ je ϕ neprekidna funkcija zajedno sa svojim izvodom ϕ . Tada
f ϕ(t) ϕ (t)dt = dt = F F ϕ(t) + C.
METOD PARCIJALNE INTEGRACIJE Ako su u su u i i v v diferencijabilne diferencijabilne funkcije i funkcija uv funkcija uv ima primitivnu funkciju, tada je
udv = udv = uv uv
−
vdu.
3 Prim Pr imer erii na jˇceˇ ceˇs´ce ce koriˇ ko riˇs´ceni ce nih h smen sm enaa: 1 dx = dx = d( d (ax + ax + b b)), a = 0, 0, smena: t = ax = ax + + b, b, dt = dt = a a dx, dx, a 1 xdx = xdx = d( d (x2 ), smena: t = x = x 2 , dt = dt = 2xdx, 2 1 xdx = xdx = d( d(ax2 + b) b), a = 0, 0 , smena: t = ax = ax2 + b, dt = dt = 2axdx, 2a 1 xn−1 dx = dx = d( d (xn ), n = 0, 0, smena: t = x = x n , dt = dt = nx nx n−1 dx, n dx = 2 d( x), smena: t = x, 2t dt = dt = dx, dx, x dx 2 2 = d( d ( ax + ax + b b)), a = 0, 0 , smena: t = ax + ax + b, b, t dt = dt = dx, dx, a a ax + ax + b b dx dx = d(ln d(ln x), smena: t = ln x, dt = dt = , x x dx dx = d ln ax , a = 0, 0, smena: t = ln ax, dt = dt = , x x ex dx = dx = d d ex , smena: t = e = e x , dt = dt = e e x dx,
√
√
√
√
1 ax ae
eax dx = dx = d d sin x dx = dx =
, a = 0,
−d(cos x),
cos x dx = dx = d d(sin (sin x), dx = d(tan d (tan x), cos2 x dx = d(cot x), sin2 x sinh x dx = dx = d d(cosh (cosh x),
−
−
smena: t = e = e ax , dt = dt = ae ae ax dx, smena: t = cos x, dt = dt =
− sin xdx,
smena: t = sin x, dt = dt = cos xdx,
dt = dx, 1 + t + t2 dt smena: t = cot x, = dx, 1 + t + t2 smena: t = cosh x, dt = dt = sinh xdx,
smena: t = sinh x, dt = dt = cosh xdx,
smena: t = tan x,
−
cosh x dx = dx = d d(sinh (sinh x), dx = d(tanh d (tanh x), cosh2 x dx = d(coth x), sinh2 x
√ √
smena: t = tanh x,
smena: t = coth x,
dt t2
− 1 = dx,
dt
1
− t2 = dx.
4 METOD REKURZIVNIH FORMULA Odred Odre d¯ivanje integrala integ rala I n = f n (x)dx, funkcij funkcijaa koje zavis zavisee od celobr celobroo jnog parametra n, parametra n, mogu´ mogu´ce ce je primenom parcijalne integracije ili nekog drugog metoda, meto da, svesti na izraˇcunavanja cunavanja integrala I m , 0 m < n, istog istog tipa. Takve relacije, oblika I n = Φ(I Φ(I n−1 , . . . , In −k ),
≤
zovemo rekurzivne formule . Da bi se na osnovu njih odredila vrednost integrala I n , neophodno je poznavati uzastopnih k uzastopnih k integrala I integrala I n−1 , . . . , In −k , kao i prvih k integrala I integrala I 0 , . . . , Ik −1 . INTEGRALI SA KVADRATNIM TRINOMOM 1. I =
mx + mx + n n dx ax2 + bx + bx + c c n dx n = 2 ax + bx + bx + c c a
dx n dx m = 0 : I = = 2 b c a x2 + a x + a x + 2ba + ca n dt 2 = t = x = x + + 2ba , h2 = ac 4ba2 = . a t2 h2 xdx dx m = 0 : I = m + n + n = mI 1 + nI + nI 2 ; ax2 + bx + bx + c c ax2 + bx + bx + c c
| − | ± −
I 1 =
= =
xdx 1 2axdx = ax2 + bx + bx + c c 2a ax2 + bx + bx + c c 2 1 2ax + ax + b b b 1 d(ax + bx + bx + c c)) dx = dx = 2a ax2 + bx + bx + c c 2a ax2 + bx + bx + c c 1 b ln ax2 + bx + bx + c c I 2 . 2a 2a
|
2 2
− 4ba
− 2ba I 2
|−
Izraˇcunavanje cun avanje integrala integ rala I 2 opisano je u prethodnom sluˇ caju caju (m = 0). 0). 2. I = 3. I = I =
mx + mx + n n dx, analogno dx, analogno tipu integrala pod 1. 1. ax2 + bx + bx + c c
√
4. I =
dx 1 √ , smenom mx smenom mx+ +n = svodi se na n a sluˇ s luˇcaj ca j 2. 2. t (mx + mx + n n)) ax2 + bx + bx + c c ax2 + bx + bx + c c dx, dx, reˇsava sava se dovod dovo d¯enjem enj em na potpu po tpun n kvadrat k vadrat izraza izra za
pod korenom 2
ax +bx+ bx+c = a
b 2 c b2 b c b2 2 2 2 x+ + = a t h , t = x + , h = . 2a a 4a2 2a a 4a2
−
±
−
5 INTEGRACIJA RACIONALNIH FUNKCIJA P ( P (x) , P ( P (x) i Q(x) su polinomi, jeste prava racionalna funkcija Q(x) ukoliko je stepen polinoma P polinoma P ((x) manji od stepena polinoma Q polinoma Q((x). Prava racionalna funkcija Funkcija
P ( P (x) = Q(x)
P ( P (x) n
m
·
(x
i=1
− ai)s
i
,
x2 + pi x + q + q i
i=1
ki
gde su nule kvadratnih trinoma x2 + pi x + q + q i , i = 1, . . . , m kompleksne, m kompleksne, ima razvoj P ( P (x) Q(x)
=
A11 A12 A1s1 + + + 2 x a1 (x a1 ) (x a1 )s1 A21 A22 A2s2 + + + + x a2 (x a2 )2 (x a2 )s2 + An1 An2 Ansn + + + + 2 x an (x an ) (x an )sn B11 x + C + C 11 B12 x + C + C 12 B x + C + C 1k1 11 12 + 2 + 2 + + 2 1k1 2 x + p1 x + q + q 1 (x + p1 x + q + q 1 ) (x + p1 x + q + q 1 )k1 + C 21 + C 22 + C 2k2 B21 x + C B22 x + C B2k2 x + C 21 22 + 2 + 2 + + x + p2 x + q + q 2 (x + p2 x + q + q 2 )2 (x2 + p2 x + q + q 2 )k2 + Bm1 x + C + C m1 Bm2 x + C + C m2 Bmk x + C + C mk mkm + 2 + 2 + + 2 m . 2 x + pm x + q + q m (x + pm x + q + q m ) (x + pm x + q + q m )km
−
···
−
− ··· −
−
−
···
−
−
···
−
··· ···
···
···
Konstante A Konstante A ij , Bij , C ij ¯enih koeficijenata. ij nalazimo metodom neodred INTEGRACIJA IRACIONALNIH FUNKCIJA: ax + ax + b b p1 /q1 ax + ax + b b p2 /q2 , , . . . dx, gde dx, gde je R je R neka racx + cx + d d cx + cx + d d cionalna funkcija i p i p i Z, q i N, uvod uvo d¯enjem ¯enje m smene sme ne
1. Integral Integral
R x,
∈
∈
ax + ax + b b = t n , n = NZS q 1 , q 2 , . . . , cx + cx + d d
{
svodi se na integral racionalne funkcije.
}
6 2.
P n (x)dx = Q n−1 (x) ax2 + bx + c + λ 2 ax + bx + c
√
dx , ax2 + bx + c
√
koeficijente polinoma Qn−1 (x) i λ odred¯ujemo metodom neodred¯enih koeficijenata nakon diferenciranja gornje jednakosti. 3. Integral pod 2. 4. Integral
1 dx smenom x + α = svodi se na sluˇcaj t ax2 + bx + c
√ (x + α)n
ax2 + bx + c)dx, gde je R neka racionalna funkcija,
R(x,
primenom Ojlerovih ili trigonometrijskih/hiperboliˇckih smena svodi se na integral racionalne funkcije. Ojlerove smene:
√ ax2 + bx + c = t ± √ a x ; √ √ II) Kada je c ≥ 0, smena glasi ax2 + bx + c = xt ± c ; III) U sluˇcaju ax2 + bx + c = a(x − x1 )(x − x2 ), x1 , x2 ∈ R, koristimo √ smenu ax2 + bx + c = t(x − x1 ). I) Ukoliko je a > 0, uvodimo smenu
Trigonometrijske i hiperboliˇcke smene:
√ 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; √ a a (c) ako se u integralu javi x2 − a2 , smena: x = ili x = ili (a) ako se u integralu javi
sin t
x = a cosh t.
cos t
5. Integracija binomnog diferencijala: I =
p
xr a + bxq dx, za a, b
∈ R, p,q,r ∈ Q
q 1 r1 Z i m N, , r = , gde su q 1 , r1 m m uvodimo smenu: x = tm , dx = mtm−1 dt. Polazni integral tada
a) U sluˇcaju p
∈
postaje I = m
Z, q =
∈
∈
p
tr1 +m−1 a + b tq1 dt.
r + 1 s , smena: a + bxq = t m ili x = t 1/q . Z, p = q m r + 1 s c) Za sluˇcaj + p Z, p = , smena: x = t 1/q ili ax −q + b = t m . q m
b) Kada je
∈
∈
7 INTEGRACIJA TRIGONOMETRIJSKIH FUNKCIJA Neka je R neka racionalna funkcija. Posmatramo integral oblika I =
R(sin x, cos x)dx.
1. U sluˇcaju
• R(− sin x, cos x) = −R(sin x, cos x), smena: t = cos x; • R(sin x, − cos x) = −R(sin x, cos x), smena: t = sin x; • R(− sin x, − cos x) = R(sin x, cos x), smena: t = tan x. x 2dt Inaˇce, smenom t = tan , dx = , imaju´ci u vidu transformacije 2 1 + t2 2t sin x = , 1 + t2
1 t2 cos x = , 1 + t2
−
polazni integral posta je integral racionalne funkcije. 2. Transformacije proizvoda u zbir 1 cos(a b)x cos(a + b)x , 2 1 sin ax cos bx = sin(a b)x + sin(a + b)x , 2 1 cos ax cos bx = cos(a b)x + cos(a + b)x , 2 sin ax sin bx =
svode integrale oblika
sin ax sin bxdx,
− − − −
sin ax cos bxdx,
cos ax cos bxdx,
na elementarne integrale. 3. Za I =
sinm x cosn xdx, gde su m, n
∈ Z razlikujemo slede´ce sluˇcajeve:
• n = 2k + 1 (m = 2k + 1 analogno): I =
sinm x(cos2 x)k cos xdx =
t = sin x dt = cos xdx
=
tm (1 t2 )k dt;
−
8
• m = 2k,
n = 2l : Sniˇzavamo stepen trigonometrijskih funkcija pod integralom primenom transformacija 1
sin2 x =
• m = −k,
n =
2
cos2 x =
1 + cos 2x ; 2
−l, k,l ∈ N :
I = =
= 4.
− cos2x ,
dx = sink x cosl x
1 sink x cosl+k−2 x k cos x
1 dx sink x cosl−2 x cos2 x dx cos2 x
tan−k x 1 + tan2 x
R tan x dx, smena: t = tan x, dx = R cot x dx, smena: t = cot x, dx =
l+k−2 2
d(tan x)
dt , takod¯e 1 + t2 dt − 1 + t . 2
5. Hiperboliˇcke funkcije analogno, uz napomenu o jednakostima koje vaˇze za ove funkcije: ex
− e−x ,
ex + e−x sinh x sinh x = cosh x = , tanh x = , 2 2 cosh x cosh2 x sinh2 x =1, sinh2x =2sinh x cosh x, cosh2x =cosh2 x+sinh2 x,
−
−
x cosh x 1 x sinh = sgn(x) , cosh = 2 2 2 tanh x sinh x = , cosh x = 1 tanh2 x 1
−
− √ − −
arc sinh x = ln x +
cosh x + 1 , 2 1 , tanh2 x
x2 + 1 , arccosh x = ln x +
√ x2 − 1 ,
− −
1 x + 1 1 x 1 arc tanh x = ln , arc coth x = ln , 2 x 1 2 x + 1 x 1 x + a x 1 x a arc tanh = ln , arc coth = ln , a 2 x a a 2 x + a
9
Zadaci 1. Dokazati za a > 0 i b = c : a) c) e) g) i) k) m) n)
dx = log x + b + C, x + b dx 1 x = arctan + C, a2 + x2 a a x2 x dx = x a arctan + C, a2 + x2 a dx x = arcsin + C, a a2 x2 x dx 1 = ln a2 x2 + C, 2 2 a x 2
√ − ± ±− ± dx
x a2
a2
x2
|
|
−
±
| ± |
b) d) f) h) j)
dx 1 x + b = ln +C, (x+b)(x+c) c b x + c dx 1 x + a = ln + C, a2 x2 2a x a x2 a x + a dx = x ln + C, 2 x2 a2 x a dx = ln x+ x2 a2 +C, x2 a2 x dx = a2 x2 + C, a2 x2 dx 1 a2 x2 a = log +C, 2a x a2 x2 a2 x2 +a
− √ − √ ± √ ±
1 x2 = ln +C, l) 2a2 a2 x2 x2 ) x a 2 x 2 2 2 x dx = a x + arcsin + C, 2 2 a 2 x a a2 dx = x2 a2 ln x + x2 a2 + C. 2 2
Reˇ senje: a) b)
± ± − ± ± ±
dx = x + b
d(x + b) = log x + b + C. x + b
| − − −
dx 1 = (x+b)(x+c) c b
1.a)
=
1
c b
− − − − ± ± ± √ √ ± − ±
|
(x+c) (x+b) 1 dx = (x+b)(x+c) c b
log
−
dx x+b
dx x+c
−
x+b +C. x+c
dx , x 2 1+ a x uvod¯enjem smene t = , dx = adt, polazni integral postaje tabliˇcni a c) Kako je I =
a2
dx 1 = 2 2 + x a
1 I = a
dt 1 = arctan t + C. 2 1 + t a
1 x Vra´ canjem smene, konaˇcno dobijamo I = arctan + C. a a
10 d) Integral I = a = a i b =
−
dx
jeste zapravo specijalan sluˇcaj integrala pod b) za a2 x2 a. Tako na osnovu rezulata pod b) zakljuˇcujemo da je
−
1 x + a I = ln + C. 2a x a
e) f)
− − − − − − − − − − √ √ − − − x2 dx = a2 +x2 x2
x2
dx = a2
a2 +x2 a2 dx = dx a2 2 2 a +x
dx 1.c) x = x a arctan +C. a2 +x2 a
x2 a2 +a2 dx = dx+a2 2 2 x a
dx 1.d) a x + a = ln + C. x x2 a2 2 x a
dx
g) I =
a2
x2
=
1 a
dx
1
= arcsin t + C = arcsin dx
x2
k) I =
1 = a a2
xdx 1 t = a2 x2 = = 2xdx = dt a2 x2 2
xdx 1 t = a2 x 2 = = 2xdx = dt 2 a2 x2 dx
x a2
x2 )
1 = 2 log x a
=
1
x 1 + C 1 = ln + a ln a.
=
dx = x a2 x2 dt
t2
1.d)
a2
1 a2
=
t = xa dx = a dt
=
dt 1 t2
=
x2
x2 a2
dt
t2
1
1 + C 1 = ln x +
x2
a2 + C,
dt 1 1 = ln t +C = ln a2 x2 +C. t 2 2 1
t
1 2
1 t2 dt = +C = 2 1/2
a2 x2 x2 1 dx = a2 x a2 x2
1 log a2 2 2a
1.i)
l) I =
x2 a2
=
t2
= ln t + gde je C = C 1
j) I =
x + C. a dx
t = xa dx = a dt
√ √ ± ± ± ± √ ± | ± | − ± ± ± | | ± ± ± ± √ ± ± − ± ± ± ± ± ± ∓ ∓ ± ± ± | |− ± ± √ √ √ ± ± − √ ± ± ± ± − √ √ ± − − ±
h) I =
i) I =
x2 a2
=
dx x
a2 x2 +C.
1 a2
x dx a2 x2
1 x2 x + C = 2 log 2 + C. 2a a x2 2
x dx = a2 x2
t = a2 x2 , x2 = dt = ax2 dxx2
1 t a 1 log + C = log 2a t + a 2a
a2 a2
t2
x2 a + C. x2 + a
a2
11
m) Oblast definisanosti podintegralne funkcije I =
− a2
√ a2 − x2 integrala
x2 dx
−
je segment Df = a, a . To opravdava uvod¯enje smene x = a sin t, pri tom je dx = a cos tdt, kada t na primer prolazi segment t [ π/2, π/2]. Za takav izbor vrednosti nove nezavisno promenljive t vaˇzi cos t 0. Tada je I = a
∈− ≥
− − a2 sin2 t cos t dt = a 2
a2
sin2 t cos t dt
1
a2 a2 cos t dt = (1 + cos 2t)dt = dt + cos2t dt 2 2 a 2 1 a2 1 = t + cos2t d(2t) = t + sin2t + C. 2 2 2 2 2
= a2
Primetimo da je
sin t =
x x , t = arcsin , a a
sin2t = 2 sin t cos t = 2 sin t Tada imamo a2 x x I = arcsin + 2 2 a a
n)
I =
x2
2x sin2 t = a
− a2
a2 dx =
x2
a2
=x
x2
a2
= x
Odatle je
1
x2
kao i
x + C = 2
− x a
1
− a2
x2
2
2x = 2 a
− a2
x2
a2
2I = x
a2 , du = x2dx 2 x a v = dx = x
u = x2 dv = dx,
x2 dx = x x2 a2 x2
I
x2
a2 dx
a2 ln x +
a2
x2
a2 ln x +
a2
x2
dx
a2
x2
x2
a2
x2
a2
a2 .
x2
x2 .
a 2 x + arcsin + C. 2 a
± √ ± √ ± ± − √ ± − √ ± ∓ ± − ± ± ∓ √ ± ± − ± | ± | ± ± ± | ± |
=x
1.h)
−
a2 + C 1 ,
a2
a2
dx
12 odnosno
x I = 2
± x2
a2
±
a 2 ln x + 2
|
± x2
a2 + C.
|
Joˇs jedan pogodan naˇcin izraˇcunavanja integrala I dat je u nastavku. I =
± x2
x2 a 2 dx = (c0 x + c1 ) x2 x2 a2
√ ±
a2 dx =
±
±
a2 + λ
√
dx
x2
± a2
Diferenciranjem ove jednakosti nalazimo x2 a2 = c 0 x2 a2
√ ±±
Posle mnoˇzenja sa
± x2
a2 + (c0 x + c1 )
√ x2x± a2 + λ √ x2dx± a2 .
√ x2 ± a2 i sred¯ivanja izraza, dobijamo x2
± a2 = 2c0x2 + c1x ± c0a2 + λ.
Izjednaˇcavanjem koeficijenata uz iste stepene promenljive x dolazimo do sistema jednaˇcina 1 = 2c0 ,
±a2 = ±c0a2 + λ,
0 = c 1 ,
ˇcije je reˇsenje c0 = 1/2,
c1 = 0,
λ =
±a2/2.
Time traˇzeni integral postaje x I = 2
± x2
a2
±
a 2 2
√
dx
x2
1.h)
± a2
=
x 2
± x2
a2
±
a 2 ln x + 2
|
± x2
a2 + C.
|
2. Odrediti slede´ce integrale: 10 40
√
Reˇ senje:
3
x(x + 2) dx
dx x2 + 4x + 4
10
√
20 50
3
x(x + 2) dx =
2
20
x (1 + x) dx
30
dx x2 + x
60
−2
x dx (x + 9) 10
− 1
2x2 + x4 dx
x1/2 x3 + 6x2 + 12x + 8 dx
13
√
x7/2 + 6x5/2 + 12x3/2 + 8x1/2 dx
=
2 12 24 16 x x4 + x3 + x2 + x + C. 9 7 5 3
=
0
2
0
3
20
50
60
− = (t
2 20
1) t dt = (t2
− 2t + 1)t20dt
t23 t22 t 21 = t 2t + t dt = 2 + + C 23 22 21 (1 + x)23 (1 + x)22 (1 + x)21 = + + C. 23 11 21
−
22
t
8
20
−
− − =
(t 9)t−10 dt =
9
dx x2 + x 1
2
=
dx (x + 2)(x
2x2 + x4 dx =
(1
1.b)
1)
=
− ≥ ∈− ∈ −∞ − ∈ 1
t
9
9t
10
dt
1 x 1 ln + C. 3 x + 2
2x2
+ x4 dx =
− − −
x2 )2 dx =
za 1 x2 0, tj. x [ 1, 1]. Za x ( , 1) ili x (1, + ) vaˇzi
−
− − −
d(x + 2) 1 = + C. 2 (x + 2) x + 2
− − − − − x2
−
1 1 − − 9 + C = + C. −8 −9 (x + 9)9 8(x + 9) 8
dx = + 4x + 4
t
21
−
x dx t = x + 9 = 10 dt = dx (x + 9)
=
40
t =1 + x x (1 + x) dx= dt = dx 2
1 x2 dx = x
∞
− − 1
x2 dx =
3
− x − x3
+ C 1 .
x3 +C, 3
14 3. Metodom smene odrediti: x3 dx 20 8 x 2 x9 dx 50 x5 + 1 xdx 80 4 1 + x
− √
10 40 70
Reˇ senje: 10
20
−
3
x8 dx = (x3 1)3
−
0
4
x9 1 dx = x5 + 1 5 1 = 5
30
2
+1
dx x x2 + 1
x8 dx (x3 1)3 1 x7 dx x 1 + x7 1 x + x2 + x + 1 dx 2
60 90
√ − √ √ −
d(x4 ) x4 1.d) 1 = ln (x4 )2 ( 2)2 8 2 x4 +
− √
2 + C. 2
x4 1 t 1 t = 1 + x4 3 x dx = = dt 3 dt = 4x dx (1 + x4 )2 4 t2 1 t 1 t 2 dt = ln t + 1/t + C 4 1 ln 1 + x4 + + C. 1 + x4
||
(x3 )2 1 (t + 1) 2 t = x 3 1 2 x dx = = dt dt = 3x2 dx (x3 1)3 3 t3 1 2 1 1 2 1 + 2 + 3 dt = ln t + C t t t 3 t 2t2 2 1 ln x3 1 + C. 3 3 x 1 2(x 1)2
− | − |−
1 3 1 = 3 =
x10
− − −
1 4 1 = 4 =
0
x
x3 dx 1 = 8 x 2 4
x7 dx = (1 + x4 )2
x7 dx (1 + x4 )2 dx
−− √
− | | − − −
x5 d x5 1 = x5 + 1 5
−
−
− − − d x
5
x5 + 1 1 d x5 5 x +1
d x5 + 1 x5 + 1
x5 = 5
1 log 1 + x5 + C. 5
15
50
60
dx
x9 dx
− 2
=
2
1 x7 dx = x 1+x7
− =
1 10
1 x7 1 t = x 7 6 x dx = = 6 dt = 7x dx 7 x7 1+ x7
− − − xdx 1 = 2 1 + x4
70
√
80
√
dt t(t + 1)2
d(x2 )
1 = ln x2 + 2 1 + (x2 )2
dx = x x2 + 1
1 x + 2
dt
−
1.d) 1
−
+ C.
1 + x4 + C.
√ t = x2 + 1, x2 = t 2 − 1 dt =
√ x 2dx
x +1
−
t 1 1 = ln + C 2 = ln 1 2 t + 1 2
x2 + x + 1 dx =
2
1 t dt t(1+ t)
x dx √ = x2 x2 + 1
t2
√ 1 = 2
| |
1 + t 2t 1 x7 dt = log 7 t(1 + t) x7 + 1
=
9
=
t = x 10 dt = 10x9 dx
x x10 + 1 x10 x10 + 1 1 t + 1 t 1 dt 1 d(t + 1) = dt = 10 t(t + 1) 2 10 t(t + 1) 10 (t + 1)2 t 1 1 x10 1 1b) 1 = log + +C = log 10 + +C. 10 t+1 10(t+1) 10 x +1 10 x10 +1
1 = 7
0
√ √
x2 + 1 1 + C 2 . x2 + 1 + 1
t = x 2 + x + 1 dt = (2x + 1)dx = 2 x +
1 t3/2 1 t dt = + C = 2 3/2 3
−
(x2 + x + 1) 3 + C.
1 2
dx
16 4. Metodom smene odrediti: dx 5x 2
√
10
40
20
60
−
90
Reˇ senje:
0
2
0
3
10
√
−
− −
x
√ − 4
1
30
dx
d( 5x
x
x2 + 1
2
3
x2 + 1
2 x dx 2 + x
2) =
√ − 2 + C.
2 5x 5
√ − 1−t 1 4 = (t2 − 1)dt 8 − −1 √ 1 − 4x (1 + 2x) + C. + C = 2
12
√ −
t = 4 1 4x, t4 = 1 4x 4x dx= = 4 x = 1−4t , dx = t3 dt
4x dx
x3 dx
80
t = 1 4x, x = dx = 12 t dt
−
t5 t dt = + C 5
−
−
√ − 4x 4x − 1 + C.
= 41
40
1
x + 3
3/2 3
4
dx
x dx
1 + x2 2 (2 x)2
√ −
√ √ − −√
− − √ −
dx 2 = 5 5x 2
√
dx = 4x 1 t 3 = t 8 3 1
x
− 1 − 4x √ x + 1 dx √ 50 + x−1
x5 dx 70 2 1 x
√
√
4
−
5
√
− √ − − − − √ −
√
− √ −
dx x+1 x 1 1 = dx = x+1 x 1 dx x + 1 (x 1) 2 x+1 + x 1 1 1 = x + 1 d(x + 1) x 1 d(x 1) 2 2 1 (x+1) 3/2 1 (x 1)3/2 1 = +C = (x+1) 3/2 (x 1)3/2 +C. 2 3/2 2 3/2 3
√
√ − √
−
−
−
− −
17
0
5
dx = x+3 x 2
√
− √ −
=
60
70
80
90
√
x5
√
√ − − −
x+3 + x 2 1 dx = (x+3) (x 2) 5
2 (x + 3)3/2 + (x 15
√
− x2 dx = 1 = − (1 − t2 )2 dt = 1
−
x3 dx
x2 + 1
(2
3
x2 + 1
4
t
=
t
2
−
−
3
x)2
6
=
dt =
+ C.
t = 1 x2 , x2 = 1 dx dt = √ x1− x2
−
x dx
√ − − −
√ x − 2
√ − t2 − xdx = − x2 1 2 1 x2 (1 − x2 )2 − (1 − x2 ) + 1 5 3
1 t = √ 1+x 2 = = x dx 3/2 dt = (1+x 2 )3/2 1+ x2
x + 3 +
− 2)3/2
− − −
x4
√
dt = t+C =
−
dt =
t−3 3
√
−5
−
t − − −5 +C = 5
=
1
(x2 +1) 5
3 3 = t2 + C = 4 4
3
3
2 x 2+x
2
2 + x 2 x
−
t2
+ C.
− 1 dt
t6
1
3 (x2 +1) 3
−− −
t = 3 2+x 2 x 2 x dx = 2 + x dt = (24/3x)2
dx
− √ 1+1 x2 +C.
√ t = x2 + 1, x2 = t 2 − 1 x dx x2 +1
=
dx 2
+ C.
3 2
t dt
+C.
18 5. Metodom smene izraˇcunati: 10 40 70
Reˇ senje: 10
0
2
√
dx x + ln x 20 dx 30 x ln x x dx ln(1 + x) ln x 50 dx 60 x(1 + x) x ln x 2 x ln(1 + x) + ln(1 x) dx x2 1
√
−
dx = x ln x
√
x + ln x dx= x
√
−
dx = x ln x
40
50
ln x d(ln x)
dx i ln4x =ln2x+ln2, uvodimo smenu t =ln4x : x
|
−
| √
dt = t t
− ln2 ln |t| + C
(ln x)−1/2 d(ln x) = 2 ln x + C.
− − − − − ln dx = x(1+ x) dt = 1+11
ln(1+x) ln x dx= x(1+ x)
−
xx (1 + log x)dx =
t = ln 1+ x1
1+ x1
x
=
60
dx x1/2 ln x = + x 1/2
ln 2x t ln 2 dx = dt = dt ln 2 x ln 4x t = ln 4x ln2 ln ln 4x + C.
√
dx +
1 =2 x + ln 2 x + C. 2
−
xx (1 + log x)dx
| |
1/2
30 S obzirom da je d(ln4x) =
−
ln 2x dx x ln 4x
d(ln x) = ln ln x + C. ln x
− x
−
t dt =
t2 2
+ C =
t = xx , log t = x log x dt t
= (1 + log x)dx
dx = x2
dx x(1+x)
1 2 1 ln 1 + + C. 2 x
= dt = t + C = xx + C.
19
70
x ln(1+x)+ln(1 x) x2 1
−
−
1 = 2
2
− − dx = ln 2 1 x2
1 1 t2 dt = t 3 + C = ln 3 1 6 6
6. Metodom smene odrediti: 10 40
dx ex + 1
20
x2 1
x dx 50
Reˇ senje:
10
dx = ex +1
−
2x dx 1 x2
√ − − − − − − − − −
dx 1 + e2x 2 −1/2 x e (1+x ) (1 + x2 )3
dx ex 1+ e
x dx = x2 1 dt =
x2 + C.
√ − − − − − e
t = ln 1 x2
− −−
x
ex = ln + C = x 1 + ex
x + 1 dx x 1 + x ex
dx 60
d(1+ e 1+e
=
ex 1 dx 1 + 3e x
30
x)
x
=
ln 1+ e
x
+C
ln 1 + ex + C.
Isti rezultat moˇzemo da dobijemo i na slede´ci naˇcin. dx 1 + ex ex = dx = dx ex + 1 ex + 1 = x ln ex + 1 + C.
20
−
dx = 1 + e2x
√
−
√ −| ex
dx = e 2x + 1
d ex + 1 ex + 1
ex dx = x ex + 1
t = e −x dt = e−x dx
− √ − − −
dt t2 + 1
=
− ln |t + t2 + 1 + C = − ln(e x + e 2x + 1) + C ex √ = ln + C = x − ln 1 + 1 + e2x + C. 1 + 1 + e2x =
30
√ − − ex
1 + 3e
1
dx = x
ex
ex dx
− 1 √
ex + 3
ex
−1 =
√ t = ex − 1, ex = t 2 + 1 dt =
1 ex dx 2 ex 1
√ −
√ −
t2 t ex 1 1.e) =2 2 dt = 2t 4 arctan +C = 2 ex 1 4 arctan +C. t +4 2 2
−
√ − −
20
40
50
60
t = x2 1 1 = x dx = dt = 2x dx
− −
− − − e
x2
xe
1
(1+x2 )− 2
− 12 e−x −1 + C. 2
1
dt =
1 2 − e −(1+x ) 2 x dx 3 (1+x2 ) 2
1 2 )− 2
= dt = t+C = e−(1+x
(x + 1)ex dx = xex 1 + x ex
x + 1 dx = x 1 + x ex
et dt =
(1+x2 )− 2
t=e
dx =
(1+ x2 )3
1 2
− − −
t = xex dt = (1 + x)ex dx
+C.
dt t xex 1.b) = ln + C = ln + C. t(t + 1) t + 1 xex + 1
=
7. Metodom smene odrediti: 10 40
Reˇ senje: 10 20
tan x dx
20
cot x dx
50
tan x dx =
dx = cos x
dx cos x dx sin x
sin x dx = cos x
cos x dx = cos2 x
30 60
−
dx 1 + cos x dx 1 + sin x
d(cos x) = cos x
− ln | cos x| + C.
d(sin x) 1.b) 1 1 + sin x = ln + C. 2 1 sin x 1 sin2 x
−
−
1 (1 + sin x)2 1 + sin x = ln + C = ln + C 2 2 cos x 1 sin x cos x2 + sin x2 1 + tan x2 = log +C = log +C. cos x2 sin x2 1 tan x2 30 40
dx = 1 + cos x
−
−
dx 2cos2
x 2
=
cot x dx = log sin x + C.
|
|
−
d x2 x = tan + C. cos2 x2 2
21 50
0
6
dx x = log tan + C. sin x 2
dx dt dt = x = t + π/2, dx = dt = = 1 + sin x 1 + sin(t + π/2) 1+cos t π x π x 2 tan 2 tan 4 7.30 = tan t/2+ C =tan +C = +C 2 1+tan x2 tan π4 tan x2 1 cos x 2 = + C = + C = + C 1 . tan x2 + 1 1 + sin x 1 + cot x2
−
−
−
−
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
Reˇ senje:
20
10
cos x dx =
C.
1 + cos 2x 1 dx = 2 2 x sin2x = + + C. 2 4
cos2 x dx =
3
2
cos x cos x dx =
1 dx + 4
−
cos2x d(2x)
1 sin2 x d(sin x) = sin x
3
− sin3 x +
Drugi oblik primitivne funkcije moˇze se dobiti sniˇzavanjem stepena trigonometrijske funkcije pod integralom. cos3 x dx =
1 2 1 = 4 =
cos2 x cos x dx =
1 2
1 + cos 2x cos x dx
1 cos x+ (cos3x+cos x) dx 2 3 1 3cos x+cos3x dx = sin x+ sin3x+C. 4 12 cos x+cos2x cos x dx =
1 2
22
2
1+2 cos 2x+cos2 2x cos x dx= cos x dx = dx = dx 4 1 1 1 = dx + cos2x d(2x) + (1 + cos 4x)dx 4 4 8 3 1 1 = x + sin2x + sin 4x + C. 8 4 32
0
3
4
2
x 2
− 14 sin2x + C.
sin2 x dx =
40
50 60
1+cos2x 2
2
3 1 cos x + cos 3x + C = 4 12
sin3 x dx =
−
sin4 x dx =
3 x 8
−
cos 3 x cos x + + C. 3
− 14 sin 2x + 321 sin4x + C.
9. Metodom smene odrediti: dx cos3 x dx sin3 x
10 40 70
Reˇ senje: 10 Za x dx = cos3 x 1.n)
= =
20 50
tan2 x dx 80
100
dx cos4 x dx sin4 x
t 2
60
tan3 x dx 90
∈ − π2 + 2kπ, π2 + 2kπ
tan4 x dx
110
, k
1 1+ t2 + ln t+ 2
1 1+t2
1+ t2
tan x 1 1 + ln tan x + 2cos x 2 cos x
sin x cos3 x dx 1 + cos2 x
∈ Z, imamo cos x > 0. Tada je
t =tan x, dt= cosdx2 x
√
1 dx = cos x cos2 x cos x =
30
sin2x dx 4cos2 x + 9 sin2 x
dx sin x cos2 x dx sin2 x cos x
+C
=
1+ t2 dt
+ C
cos x2 + sin x2 sin x 1 sin x + 1 sin x 1 = + ln + C = + ln + C. 2cos2 x 2 cos x 2cos2 x 2 cos x2 sin x2
−
23 Integral, takod¯e moˇzemo odrediti na slede´ci naˇcin:
dx = cos3 x =
sin2 x + cos 2 x dx = cos3 x
u = sin x x) dv = d(cos cos3 x
−
sin x 1 = + 2cos2 x 2
0
2
dx = cos4 x
30
40 50 60 70
0
8
dx 1 1 = 8 cos2 sin3 x dx = sin4 x
−
t = tan x, dt =
dx cos2 x
1 1+tan2 x
cos2 x =
=
1 1+t2
dx cos x
t 3 1 + C = tan x + tan 3 x + C. 3 3
sin x dx + cos2 x
dx sin x
d(cos x) x 1 x + ln tan = + ln tan + C. cos2 x 2 cos x 2
1 − x sin2 x 2
2
1 x + log tan + C. 2 2
− cot x − 13 cot3 x + C.
tan x dx=
cos x2 + sin x2 dx = log cos x2 sin x2 sin2 x cos x 2
dx cos x
−
sin2 x + cos 2 x dx = sin x cos2 x
−
7.50
=
sin x 1 dx = + 2cos2 x 2 cos x 1 + tan x2 dx 7.20 sin x 1 = + log + C. cos x 2cos2 x 2 1 tan x2
1 dx = cos2 x cos2 x
dx = sin x cos2 x
sin2 x dx + cos3 x
du = cos x dx v = 2cos1 2 x
(1 + t2 )dt = t +
=
tan x dx= =
−
sin2 x dx = cos2 x
3
−
1 + C. sin x
1 cos2 x dx = cos2 x
−
2
−
tan x tan x dx = tan x d(tan x)
tan x
1
dx cos2 x
−
dx =tan x x+C.
−
− cos2 x dx cos2 x 7.10
tan x dx =
1 tan 2 x + ln cos x + C. 2
|
|
24
90
tan4 x dx =
1 cos2 x tan2 x dx = cos2 x
−
tan3 x = 3
9.70
100
tan2 x d(tan x) dx
tan2 x dx
− tan x + x + C.
t =4cos2 x+9sin2 x 1 sin2x dx = = 5 4cos2 x+9sin2 x dt =5sin2x dx
−
dt 1 = ln t + C t 5
||
1 = ln 4cos2 x+9sin2 x + C. 5
110
sin x cos3 x t = 1 + cos 2 x dx= dt = 2cos x sin x dx 1 + cos2 x
−
− =
1 2
1
t
t
dt
1 dt 1 1 = dt = ln t t + C 2 t 2 2 1 1 + cos2 x 2 = ln 1 + cos x + C. 2 2
− −
| |−
10. Metodom smene odrediti: 10 40 70 100
− √ x
arctan 2x dx 1 + 4x2 1 dx cos x x2 dx sin x + 2 cos x + 3 tan x dx (1 + cos x)2
20 50 80 110
sin2x dx sin x + 1
√ √
sin x cos2 x dx
dx 2cos x + 3 dx (3 + cos x)sin x
30 60 90 120
√ − − 1
sin2x dx
cos2 x sin2x dx dx 4 3cos x sin3x + cos 3x dx sin3x cos3x
25 Reˇ senje: 10
x
− √ arctan 2x dx =
− √ √ −
1 + 4x2
x dx 1 + 4x2
arctan 2x dx 1 + 4x2
1 d(1 + 4x2 ) 1 = arctan 2x d(arctan 2x) 8 1 + 4x2 2 1 1 arctan3/2 2x = ln(1 + 4x2 ) + C 8 2 3/2 1 1 = ln(1 + 4x2 ) arctan3/2 2x + C. 8 3
− −
20
sin2x dx=2 sin x+1
√
=4 =
30
√ − 1
t2
1 dt =
√
50 za x
60
cos
−
− 2) + C.
2sin x cos x dx
(sin x + cos x)2 dx = (sin x + cos x)dx cos x + C,
1 1 cos d = x x
za sin x + cos x
≥ 0.
1 sin + C. x
(sin x)3/2 sin x cos x dx = sin x d(sin x) = +C, 3/2 (2kπ, 2kπ + π/2), k Z, gde je sin x, cos x 0. sin x cos2 x dx =
∈
− 4t + C = 43 t(t2 − 3) + C
sin2 x + cos2 x
sin2x dx=
1 dx = x x2
√
− − − − √ √ = sin x
40
4 3 t 3
4 sin x + 1 (sin x 3
=
√
t = sin x+1, sin x = t2 1 sin x cos x dx = cos x dt = 12 √ sin dx sin x+1 x+1
√ −
cos2 x sin2x dx=2 =
∈
≥
−
cos3 x sin x dx =
− 12 cos4 x + C.
2
cos3 x d(cos x)
26
t = tan x2 dx = 2t sin x+2cos x+3 sin x = 1+t 2
70
=2
− √
2dt t =tan x2 , dx = 1+t dx 2 = = 2 − 1 t 2cos x+3 cos x = 1+t2
0
9
4
−
dx = 3cos x 1.c)
=
100
0
dt t2 +2t+5
=2
tan x2 + 1 d(t + 1) t + 1 = arctan + C = arctan + C. (t + 1) 2 + 4 2 2
80
11
2dt dx = 1+t 2 1−t2 cos x = 1+t2
t = tan x2 , dx = 1 t2 cos x = 1+t 2
√ 27 arctan
tan x2 2dt 1.c) 2 = arctan +C. t2 +5 5 5
2dt 1+t2
7tan
=
−
√
2dt 1 + 7t2
x + C. 2
tan x sin x dx dx = = 2 (1 + cos x) cos x(1 + cos x)2 dt 1 + t t = = dt = t(1 + t)2 t(1 + t)2 1 + cos x 1 1.b) = log + + C. cos x 1 + cos x
√
−
−
−
t = cos x dt = sin x dx dt d(1 + t) t(1 + t) (1 + t)2
−
−
dx d(cos x) t = cos x = = 2 dt = sin x dx (3+cos x)sin x (3+cos x)(1 cos x) dt 1 t + 1 (t 1) = = dt (3 + t)(t2 1) 2 (3 + t)(t 1)(t + 1) 1 dt 1 dt = 2 (3 + t)(t 1) 2 (3 + t)(t + 1) t 1 1 t+3 1 (1 cos x)(3+cos x) 1.b) 1 = log + log +C = log +C 8 t+3 4 t+1 8 (1+cos x)2
−
− − − −
−
− − −
−
−
27
120
sin3x + cos 3x dx = sin3x cos3x
1 = 3
dt 1 + cos t 3
1 = log 3
dt sin t
1 + tan 3x 2 1
dx + cos3x =
− tan 3x2
dx 1 = sinh x 2
Reˇ senje: 10
2
sinh x dx = cosh2x
√
1.h)
=
0
3
40
sinh
x x 2 cosh 2
=
sinh x dx cosh2x dx
√
√ 3
cosh2 x tanh2 x
1 2
dx
tanh
2 x x 2 cosh 2
sinh x dx 2cosh2 x
1 ln cosh x + 2
1
=
1 2
cosh2 x
−
cosh 2x+1 2 2
√ √ +
cosh2 x
− 21
1 + C. 2
sinh 2x dx
(sinh2 x)2 + (cosh2 x)2
sinh2x dx
cosh2x 1 2 2
d(cosh x)
=
sinh 2x dx
2(cosh2 2x + 1)
1 dt 1 t =cosh2x = = = ln t+ dt =2sinh2x dx 2 2 t2 +1 2 2 1 = ln cosh2x + cosh2 2x + 1 + C. 2 2
√
+C
d tanh x2 x = ln tanh + C. x tanh 2 2
1 dx = 2 sinh4 x + cosh 4 x
1 = 2
20
dx
sinh x cosh x
t = 3x dx = dt 3
√ − √ − =
0
−
+ C.
11. Metodom smene odrediti: dx 10 sinh x sinh x cosh x 30 dx sinh4 x + cosh 4 x
1+tan 2t 1 t log +log tan t 3 2 1 tan 2
7.20 ,7.50
tan 3x 2
dx = sin3x
√
t2 +1 +C
28
40
dx
√ 3
cosh2 x tanh2 x
=
√
tanh−2/3 x d(tanh x) = 3 tanh x + C. 3
12. Primenom metoda parcijalne integracije odrediti: dx (a2 + x2 )2 x3 dx a2 + x2
√
10 30
10
Reˇ senje:
40 dx
√ − −
50
x2 dx (x2 a2 )2 x3 dx a2 x2
20
x2
√ −
−
5
a2
a2 + x2 x2 dx (a2 + x2 )2
dx 1 = 2 2 2 2 (a + x ) a
1 dx 1 x2 x 1 1.c) 1 = 2 dx = arctan a a2 +x2 a2 (a2 +x2 )2 a3 a a2 u = x du = dx d(a2 + x2 ) 1 = dx 1 dv = (a2x+x v = = 2 )2 2 2 2 2 2 (a + x ) 2(a + x2 ) 1 x 1 x 1 dx = 3 arctan + a a a2 2(a2 + x2 ) 2 a2 + x2 1 x x = 3 arctan + 2 2 + C. 2a a 2a (a + x2 )
0
2
−
−
−
x2 dx
(x2 a2 )2
−
= 1.d)
=
−
u=x
−
dv = (x2x−dx a2 )2
du = dx v =
2(a2
x
− x2)
+
− − −
− 2(x 1 a ) 2
2
=
x 2(x2
1 x a log + C. 4a x + a
−
x2 dx (a2 +x2 )2
1 + a2 ) 2
dx x2 a2
−
29
30
a2 + x2
dx =
= x2 a2 +x2 =x
2
u = x 2
√ − −
x3
√
a2 +x2
dv =
√ − −
du = 2x dx 1. j)
x dx a2 +x2
a2 + x2
v =
2 x a2 +x2 dx = x2 a2 +x2 2 3
x 3 a2 +x2 +C =
2
2a2 3
a2 +x2 d a2 +x2
a2 +x2 +C.
Drugi naˇcin odred¯ivanja ovog integrala podrazumeva koriˇs´cenje rezultata P n (x)dx = Q n−1 (x) ax2 + bx + c + λ ax2 + bx + c
√
dx . ax2 + bx + c
√
x3 Tada je I = dx = c0 x2 + c 1 x + c 2 a2 + x2 + λ 2 2 a + x gde c 0 , c1 , c2 i λ odred¯ujemo metodom neodred¯enih koeficijenata. Diferenciranjem poslednje jednakosti po x dobijamo
√
3
dx , a2 + x2
√
√ a2x+ x2 = (2c0x + c1) a2 + x2 + c0x2 + c1x + c2 √ a2x+ x2 + √ a2λ+ x2 . √ Nakon mnoˇzenja sa a2 + x2 i sred¯ivanja, izraz postaje
x3 = 3c0 x3 + 2c1 x2 + (2a2 c0 + c2 )x + a2 c1 + λ. Jednaˇcenjem koeficijenata uz iste stepene od x na levoj i desnoj strani jednakosti, dobijamo 1 c0 = , c1 = 0, c2 = 3 ˇsto ponovo daje vrednost integrala I =
40
u = x 2
√ − − − − − − − − −
√
x3
a2
x2
dx =
= x2 a2 2
= x
a2
dv =
x dx a2 x2
x2
−
2a2 , λ = 0, 3
− 2a2 3
a2 + x2 + C.
− √ − − − − − − − −
du = 2x dx 1. j)
v =
a2
x2
x2 +2 x a2 x2 dx = x2 a2 x2 x2
2 3
a2
3 x2 +C =
x2 +2a2 3
a2 x2 d a2 x2
a2 x2 +C.
30 Drugi naˇcin odred¯ivanja : I =
√
x3
a2
− x2
2
dx = c0 x + c1 x + c2
− a2
x2 + λ
√
dx
a2
− x2 .
Diferenciranjem po x i sred¯ivanjem koeficijenata uz iste stepene od x dobijamo c0 = tj., I =
0
5
−
x2 + 2a2 3
dx
−
− a2
1 a2
1 , c1 = 0, c2 = 3
−
2a2 , λ = 0, 3
x2 + C.
x2 (x2 a2 )
1 x2 1 dx dx 5 a2 (x2 a2 ) 2 a2 (x2 a2 ) 32
√ − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − √ − − − − − − − √ − − − √ − − √ − − √ − − − √ − − √ √ − − x2 a2
5 =
u = x
dv =
=
du = dx v =
1 a2 3(x2 1 = 3a2 (x2 1 = 3a2 (x2 1 = 3a2 (x2 =
u = x
=
5
(x2 a2 ) 2
1 2
x dx (x2 a2 )5/2
d(x2 a2 ) (x2 a2 )5/2
1 3(x2 a2 )3/2
=
x 1 dx 1 dx + a2 (x2 a2 )3/2 a2 )3/2 3 (x2 a2 )3/2 x 2 dx (x2 a2 )3/2 a2 )3/2 3a2 x 2 x2 (x2 a2 ) dx (x2 a2 )3/2 a2 )3/2 3a4 x 2 x2 dx dx a2 )3/2 3a4 (x2 a2 )3/2 x2 a2 dv = (x2 x adx 2 )3/2
du = dx v =
1 2
d(x2 a2 ) (x2 a2 )3/2
1 x 2 3a2 (x2 a2 )3/2 3a4 2 x 1 = 4 3a x2 a2 3a2 =
dx=
1 x2 a2
=
x
+
x 2 a2 x
x2
a2
3
dx dx x2 a2 x2 a2 x 2x2 3a2 + C = 4 + C. 3 3a x2 a2
31 13. Primenom metoda parcijalne integracije odrediti: 10 20 30
log x dx
40
log2 x dx
50
x2 log2 x dx 60
Reˇ senje:
10
log x dx =
= x log x
0
2
2
log x dx = 13.10
0
3
2
2
x log x dx =
=
x log
70 80
2 + x dx 90 2 x
−
√ √ −
a2 + x2 dx
log x +
x2 log x +
a2 + x2 dx
x log(x + 1 + x2 ) dx 1 + x2
u = log x du = dx x dv = dx v = x
= x log x
dx
− x + C = x(log x − 1) + C = x log xe + C.
u =log2 x du =2log x dx x dv = dx v =x
= x log2 x
log x dx x2 log2 x dx x2
du = 2logxx dx
u =log x
3
dv = x2 dx v = x3 du =
dv = x 2 dx v =
dx x
x3 3
2
−
= x log x 2
− 2x(log x − 1) + C = x 2
u = log x
log2
log x dx
x + 1 + C. e
x3 2 = log2 x x 2 log x dx 3 3
x3 = log2 x 3
−
−
2 x3 log x 3 3
−
1 3
x2 dx
1 3 2 3 2 3 x3 2 = x log x x log x+ x +C = 9log2 x 6log x+2 +C 3 9 27 27 3 x = (3log x 1)2 + 1 + C. 27
− −
0
4
u = log x du =
dx x
−
log x 1 dx dx= = log x+ = 2 dx 1 x2 x x dv = x2 v = x 1 1 = (log x + 1) + C = log ex + C. x x
−
−
−
−
− x1 log x − x1 +C
32
0
5
2
log x dx = x2 13.40
=
0
6
u =log2 x du = 2logxx dx
v=
− x1
1 = log 2 x+2 x
−
− x1 log2 x − x2 log ex + C = − x1
u =log 2+x 2−x dv = x dx
2 +x x log dx = 2 x du = 44−dx x2
dv = dx x2
−
2
v = x2
log x dx x2
log2 ex + 1 + C.
x2 2+x = log +2 2 2 x
−
x2
x2 4
−
dx
x2 2+ x 2+ x x2 4 2+ x = log +2x 2log +C = log +2x+C. 2 2 x 2 x 2 2 x
1.f)
−
0
7
−
log x+
a2 + x2 dx =
u = log x+ a2 + x2 dx a2 +x2
du =
dv = dx v = x
x dx 1. j) = x log x+ a2 +x2 2 2 a +x
√ √ − √ − − x2 log x +
a2 + x2 dx =
x3 log x + a2 + x2 3 3 12.30 x = log x + a2 + x2 3 =
90
−
√ √ − √ − = x log x+ a2 +x2
80
−
−
a2 + x2
dx a2 +x2 x3
du =
1 3 x 2
9
dv = x 2 dx v =
x3 3
dx a2 + x2 2a2 a2 + x2 + C.
√ √ √ √ − −
√ x log(x + 1 + x2 ) √ 1 + x2 dx =
u = log x +
a2 +x2 +C.
= 1+ x2 log x+ 1+ x2
u = log x + du =
dx 1+x2
dx =
1 + x2
dv = v =
x dx 1+x2
1 + x2
1+ x2 log x+ 1+ x2
x+C.
33
14. Primenom metoda parcijalne integracije odrediti: 10 40 70
x3 sin x dx
20
x sin2 x dx
30
dx dx cos5 x
50
tan7 x dx
60
(2x
80
sin3 x cos3 x dx 90
− 1)sin3x dx
Reˇ senje: 10
x2 cos2x dx
x sin
−
2
u = x du = 2x dx x2 cos x dx = − dv = cos x dx v = sin x u = x = −x3 cos x + 3 x2 sin x 2 x sin x dx = dv = sin x dx = −x3 cos x + 3x2 sin x 6 x cos x + cos x dx = −x3 cos x + 3x2 sin x + 6x cos x − 6sin x + C. = x3 cos x + 3
0
2
2
x sin x dx =
−
−
3
−
1 cos2x 2
−
x2 x x 1 sin2x sin2x dx 2 4 2 4 x2 x x2 1 x2 = sin2x cos 2x+C = 2 4 4 8 4 =
0
u = x dv = sin2 x dx = du = dx v = x2 41 sin2x
− −
2
− −
x cos2x dx =
dx
du = dx v = cos x
−
− x4 sin 2x − 18 cos 2x+C.
u = x 2 du = 2x dx dv = cos 2x dx v = 21 sin2x
√ x dx
dx (sin x + cos x)4
u = x 3 du = 3x2 dx dv = sin x dx v = cos x
− − −
x3 sin x dx =
x2 u = x du = dx = sin2x x sin2x dx = dv = sin 2x dx v = 12 cos 2x 2 x2 x 1 = sin2x + cos 2x cos2x dx 2 2 2 x2 x 1 2x2 1 x = sin2x+ cos2x sin2x+C = sin2x+ cos 2x+C. 2 2 4 4 2
−
−
−
−
−
34
0
4
dx = cos5 x
sin2 x + cos2 x dx = cos5 x
u =sin x = x dx dv = sin cos5 x
sin2 x dx + cos5 x
dx cos3 x
sin x 3 du =cos x dx = + v = 4cos1 4 x 4cos4 x 4
dx cos3 x
cos x2 + sin x2 sin x 3 sin x 3 = + + log + C. 4cos4 x 8 cos2 x 8 cos x2 sin x2
9.10
50
sin7 x dx = cos7 x
tan7 x dx =
−
u = sin6 x x dx dv = sin cos7 x
du = 6 sin5 x cos x dx v = 6cos1 6 x
−
1 sin5 x u =sin4 x du =4sin3 x cos x dx = tan 6 x dx = x dx 1 dv = sin v = 4cos 6 cos5 x 4x cos5 x 1 1 = tan 6 x tan 4 x + tan3 x dx 6 4 1 1 9.80 1 = tan 6 x tan 4 x + tan 2 x + log cos x + C. 6 4 2 Integral se moˇze takod¯e izraˇcunati i primenom metoda smene.
− −
7
tan x dx = =
5
60
5
tan x d(tan x)
√ x sin x dx = 14.10
= 2
2
tan x tan x dx =
1 = tan 6 x 6 9.80 1 = tan 6 x 6
|
− −
−
−
tan5 x
1
|
− cos2 x dx
cos2 x 1 tan5 x dx = tan 6 x 6
1 cos2 x tan x dx cos2 x
−
3
−
1 tan 4 x + tan3 x dx 4 1 1 tan 4 x + tan 2 x + log cos x + C. 4 2
|
√
t = x, x = t 2 dx = 2t dt
− =2
t3 cos t + 3t2 sin t + 6t cos t
|
t3 sin t dt
6sin t + C √ 3 √ √ √ √ √ = −2 x cos x + 6x sin x + 12 x cos x − 12sin x + C.
35
0
7
0
8
−
− −
1 sin x cos x dx= 8
3
3
−
− −
u =sin2 2x du =4sin2x cos2x dx sin 2x dx= dv =sin2x dx v = 12 cos2x
−
3
−
− 12 sin2 2x cos2x + 2 sin2x cos2 2x dx 1 = − sin 2 2x cos2x cos2 2x d(cos2x) 2 1 1 = − sin 2 2x cos2x − cos 3 2x + C. 2 3 =
90
u = 2x 1 dv = sin 3x dx (2x 1)sin3x dx = du = 2dx v = 13 cos 3x 2x 1 2 1 2x 2 = cos3x+ cos3x dx = cos3x+ sin 3x+C. 3 3 3 9
dx = (sin x + cos x)4 = =
u =
1
d(sin x + cos x) (cos x sin x)(sin x + cos x)4
cos x sin x
dv =
−
d(sin x+cos x) (sin x+cos x)4
−1
−
du = v =
3(sin x+cos x)3 (cos x
−1
=
x − (cossinxx+cos −sin x) dx 2
1 3(sin x+cos x)3
−
− sin x)
1 3
dx (cos x sin x)2 (sin x+cos x)2 1 dx 3 (cos2 x sin2 x)2
− −
3(sin x + cos x)2 (cos2 x sin2 x) 1 1 dx = 3(1 + sin 2x)cos2x 3 cos2 2x 1 1 = tan 2x + C. 3(1 + sin 2x)cos2x 6
−
− −
−
−
−
−
36 15. Primenom metoda parcijalne integracije odrediti: 10 40 70
√ −
arcsin x dx
20
arcsin2 x dx
50
x arcsin x dx 80 2 1 x
Reˇ senje: 10
x arccos dx 2
x arcsin2 x dx 60
u =arccos x2 x arccos dx = du = √ 4dx 2 −x2
1. j)
0
3
x 2
− 1
− − 4
=
arcsin2 x dx =
= x arcsin2 x
3
3
−
arccos x
− x
2
+2 9
x dx 4 x2
√
1
x3 1 = arccos x+ 3 3
−
x3 dx 1 x2
√
−
− x2 + C.
√ − √ − − − − − − u =arcsin2 x du =2arcsin x √ 1dx −x2 2 arcsin x
−
x2 + C.
u =arccos x x arccos x dx = x 3 du = √ 1dx v = 2 3 −x 2
x dx 1 x2
− √
x2 + C.
dv = x2 dx
12.40 x
40
dv = dx x = x arccos + v=x 2
−
= x arccos
arcsin x dx x2
arcsin x arccos x dx
= x arcsin x +
2
x2 arccos x dx
u =arcsin x dv = dx arcsin x dx= = x arcsin x du = √ 1dx −x2 v = x 1. j)
0
30
dv = dx v=x
u =arcsin x du = √ 1dx −x2
x dx = 1 x2 dv =
= x arcsin2 x + 2
1
x2 arcsin x
2
= x arcsin2 x + 2
1
x2 arcsin x
2x + C.
dx
x dx 1 x2
1. j)
v=
−√ 1 − x2
37
50
x arcsin2 x dx== =
x2 arcsin2 x 2
u = arcsin 2 x du = 2 arcsin x √ 1dx −x2
−
2
√ x1 −dxx2
arcsin x
√ − √ √ − − − − − − √ − √ − √ − − − − − − − u = x arcsin x
=
dv = x dx 2 v = x2
x dx 1 x2
dv =
du = arcsin x +
x 1 x2
1. j)
dx v =
1
x2
x2 = arcsin2 x+x 1 x2 arcsin x 1 x2 arcsin x dx 2 u = arcsin x du = 1dxx2 = 1.m) dv = 1 x2 dx v = x2 1 x2 + 21 arcsin x
x dx
1 x x 2 1 arcsin2 x + 1 x2 arcsin x + x dx 2 2 2 2 1 + arcsin x d(arcsin x) 2 2x2 1 x x 2 = arcsin2 x + 1 x2 arcsin x + C. 4 2 4 =
60
x2
−
arcsin x dx = x2
1.l)
=
0
7
du =
√ dx 2 v = 1−x
1 1 arcsin x + log x 2
−
√ − − −
−
=
1 1
dx u =arcsin x dv = √ x1− x2
x arcsin x dx = 1 x2 du =
√
dx x2 1 x
− √ √ − −
u = arcsin x dv =
= x
dx 1 x2
1
x2 arcsin x + C.
dx x 1 x2
√
−
x2 1 + C. x2 + 1
√ v = − 1 − x2 1. j)
−
arcsin x + x
−
− − =
1 x2 arcsin x+ dx
38
80
arcsin x arccos x dx=
− √ − − − − u =arccos x
dv =arcsin x dx 15.10
dx 1 x2
du =
= x arcsin x arccos x+ 1 x2 arccos x+ arcsin x 15.70
= x arcsin x arccos x +
1
√ −
v = x arcsin x+ 1 x2
x2 arccos x
√ x1 −dxx2 + dx
arcsin x + 2x + C.
Bez koriˇs´cenja rezultata 15.10 i 15.70 , integral se moˇze izraˇcunati na slede´ci naˇcin.
−
u =arcsin x arccos x
arcsin x arccos x dx=
arcsin x du = arccos√ x1− dx v = x −x2
= x arcsin x arccos x
=
(arccos x
v=
2
= x arcsin x arccos x +
−√ 1 − x2 1 − x2 (arccos x − arcsin x) + 2 dx 1 − x2 (arccos x − arcsin x) + 2x + C.
− √ 2dx 1−x
= x arcsin x arccos x +
− arcsin x) √ 1x −dxx2
dx u =arccos x − arcsin x dv = √ x1− x2
du =
dv = dx
16. Primenom metoda parcijalne integracije odrediti: 10 40 70
x arctan x dx
20
√ arctan x dx
50
x earctan x dx (1 + x2 )3/2
Reˇ senje: 10
x arctan x dx =
x2 = arctan x 2
−
1 2
x2 arctan x dx
30
x2 arctan x dx 60 1 + x2
1 x arctan dx x2 3 2x arctan dx 1 x2
u = arctan x du = dv = x dx
v =
dx 1+x2
x2 2
−
2 x2 1.e) x + 1 dx = arctan x 1 + x2 2
− x2 + C.
39
0
2
2
x arctan x dx =
dx u =arctan x du = 1+x 2 3
v = x3
dv = x2 dx
x3 1 = arctan x 3 3
x3 dx 1+ x2
− − − | |
x3 1 t 1 x3 1 t =1+ x2 = = arctan x dt = arctan x t log t +C dt =2x dx 3 6 t 3 6 x3 1 = arctan x 1 + x2 log(1 + x2 ) + C. 3 6
0
3
− −
−
u =arctan x3
1 x arctan dx = 3 dx x2 3 du = 9+x 2
−
− −
dv = dx x2 1 x
v=
− x1 arctan x3 + 16 log
1.k)
=
0
4
√ arctan x dx =
16.10
=
50
√
t = x, t2 = x 2t dt = dx
t2 + 1 arctan t
x2 arctan x 1 + x2
dx =
= x arctan x
6
2x arctan dx = 1 x2 2x = x arctan 1 x2
−
−
x2 + C. x2 + 9
=2
3 dx x 9+ x2
t arctan t dt
− t + C = (x + 1) arctan √ x − √ x + C.
u = arctan x du = dv =
1.e)
x2 dx 1+x2
2
dx 1+x2
v = x
− arctan x
x dx + arctan x d(arctan x) 1 + x2 1 log 1 + x2 + C. 2
−
− arctan x 1 1.i) = x arctan x − arctan2 x − 2 0
1 x = arctan + x 3
2dx u = arctan 1−2xx2 du = 1+x 2 dv = dx v = x x dx 1.i) 2x 2 = x arctan log 1+ x2 +C. 2 2 1+ x 1 x
−
− −
40
70
I =
= =
x earctan x dx = (1 + x2 )3/2
xe arctan x
√ 1+ x2
xe arctan x 1 + x2
−
du =
dv =
dx (1+x2 )3/2
−
earctan x
earctan x dx 1+x2
v = e arctan x
1 u = √ 1+x 2
du =
− (1+xx dx)
2 3/2
dx = arctan x (1+ x2 )3/2 dv = e 1+x2 dx v = e arctan x
arctan x
− e√ 1 + x2
√
x u = √ 1+x 2
x earctan x (x dx = (1 + x2 )3/2
−√ 1) earctan x − I. 1 + x2
earctan x (x 1) Odatle je I = + C. 2 1 + x2
√ −
17. Primenom metoda parcijalne integracije odrediti: 10 40 60
x 3 2x = e 2 =
x 3 e2x dx 20 e ax cos bxdx
− −
3 x2 2x e 2 2
2 2x
x e dx =
3x2 2x 3x 2x e + e 4 4
70
x
e
dx
e ax sin bxdx
e 2x cos2 x dx
u = x 2 du = 2x dx dv = e 2x dx v = 21 e2x
x3 2x xe dx = e 2 2x
−
√
u = x 3 du = 3x2 dx dv = e 2x dx v = 21 e2x
x3 e2x dx =
3 2
−
50
e 2x sin2 x dx
u = x du = dx 2x dv = e dx v = 21 e2x
x 3 = e2x 2
x2 e x dx 30 (x + 2) 2
−
Reˇ senje: 10 I = x 3 2x = e 2
x3 = e2x 2
−
−
3x2 2x 3 e + 4 2
3x2 2x 3 x 2x e + e 4 2 2
xe2x dx
−
1 2
e2x dx
3 2x e2x e +C = 4x3 6x2 +6x 3 +C. 8 8
−
−
41
20
x 2 ex (x + 2)2
dx =
u = x 2 ex dv =
du = x(x + 2)ex dx
dx (x+2)2
x2 x = e + x + 2
−
x2 x = e +xex x + 2 x 2 x = e + C. x + 2
−
−
0
3
√ √ √ √ I =
− 2e
=
5
−
u = x du = dx x dv = e dx v = e x
xex dx = x
e dx =
−
√ u= x
x2 x e + xex x + 2
+ C = 2e
√
x
− ex +C
−1
+ C.
−
−
b2 I . a2
−
u = sin bx du = b cos bxdx dv = e ax dx v = a1 eax e
ax
cos bxdx
a2 + b2 eax I = a cos bx + b sin bx + C 1 , a2 a2 eax I = 2 a cos bx + b sin bx + C. a + b2
−
b eax sin bxdx = a b 1 ax b e sin bx a a a
e
u = cos bx du = b sin bxdx dv = e ax dx v = a1 eax
e ax a cos bx + b sin bx a2
Odatle je
√
eax cos bxdx =
1 = eax cos bx + a 1 = eax cos bx + a
0
1 − x+2
√ x x dx √ √ √ dv = e e dx x x √ √ − e x dx = x = =2 xe √ x v =2e x x x du = 2dx √ √ √ x x x √ x dx
= 2 xe
40
v =
odnosno
u = sin bx du = b cos bxdx e sin bxdx = dv = e ax dx v = a1 eax ax 1 b ax 17.40 e = eax sin bx e cos bxdx = 2 2 a sin bx b cos bx +C. a a a +b ax
−
42
60
0
7
e
u = sin2 x du = 2 sin x cos x dx = sin 2x dx dv = e 2x dx v = 21 e2x
e2x sin2 x dx =
2x
1 = e2x sin2 x 2
2
cos x dx = =
0
= − e2xsin2x dx 17.5
e2x 2 cos2x sin2x +C. 8
−
− e
17.60
2x
1
2
sin x dx =
e
−
2x
dx
e2x 2 + cos 2x + sin 2x + C. 8
−
e2x sin2 x dx
18. Odrediti rekurentnu formulu za izraˇcunavanje integrala I n za n i a, b = 0.
10 I n = 30 I n = 50 I n = 70 I n = 90 I n =
√ ±
x log n x dx
dx (x2 + a2 )n x2n
110 I n =
Reˇ senje:
10
20 I n =
x2
40 I n = 60 I n =
a2 dx
− ± 1
x2
n/2
∈ N, n ≥ 2
dx
dx (ax2 + bx + c)n x2n+1
x2
a2 dx
dx cosn x
sinn x dx
80 I n =
tann x dx
100 I n =
cotn x dx
120 I n =
dx (a + b cos x)n
dx (a sin x + b cos x)n
I n =
√
n
x log x dx =
2 = x3/2 logn x 3
−
2n 3
√
u = log n x du = n logn−1 x dx x dv = x dx v = 32 x x
√
x logn−1 x dx =
√
2 3/2 x logn x 3
− 2n3 I n−1.
43 Iskoristimo izvedenu relaciju za izraˇ cunavanje, na primer, integrala I 3 . Kako je
√
I 1 =
x log x dx =
2 = x3/2 log x 3
dv =
dx x
−
u = log x du =
√ x
2 = x3/2 log x 3
v = 32 x3/2
− 49 x3/2 + C = 29 x3/2 3log x
−
2 3
√
x dx
2 + C,
na osnovu izvedene rekurentne relacije imamo
0
2
2 I 2 = x3/2 log2 x 3
− 43 I 1 = 272 x3/2 9log2 x − 12 log x + 8
2 I 3 = x3/2 log3 x 3
− 2I 2 = 272 x3/2 9log3 x − 18log2 x + 24 log x − 16
I n =
1
x2
n/2
dx =
x 2 1 x2
2
= I n
2+
x 1 x2 n
n + 1 I n = I n n
1
x2 1
Kako je
x2
(n 2)/2
+ C,
n−2 2
x2
I 2 =
I 4 =
1
−
dv = x 1 dx dx = u = x du = dx v = n1 (1 x2 )n/2 1 1 x n/2 1 x2 dx = I n +I n 2 + 1 x2 n n n
n/2
x n/2 1 x2 , pa je n n x n/2 I n = I n 2 + 1 x2 . n + 1 n + 1
3 x I 1 + 1 4 4
+ C.
dx
n−2 2
n/2
2 +
1.m) x Znamo da je I 1 = 1 x2 dx = 2 izvedene rekurentne relacije dobijamo
I 3 =
− − − − − − − − − − − − − − − − − − − − − √ − − − − − − − −
= I n
Dakle,
x2
3/2
1 x2 + arcsin x + C. Na osnovu 2
3 x = arcsin x + 8 8
x2 dx = x
4 x I 2 + 1 5 5
1
x2
2
1
x2 5
2x2 + C.
x 3 + C, moˇzemo odrediti 3
+ C = x
2 3 1 5 x + x + C. 3 5
.
44
0
3
I n = = =
dx 1 = (x2 +a2 )n a2
u = x
−
x dx (x2 +a2 )n 1 (2 2n)(x2 +a2 )n−1
dv =
du = dx v =
x2 +a2 x2 1 dx = I n−1 (x2 +a2 )n a2
−
1 a2
−
x2 dx (x2 +a2 )n
2n 3 1 x I n−1 + . 2 2 2 (2n 2)a (2n 2)a (x + a2 )n−1
− −
−
Primetimo da su ovaj integral i dobijena rekurentna relacija samo specijalan sluˇ caj integrala i relacije iz narednog primera (a = 1, b = 0, c = a 2 ).
0
4
I n = = = =
dx ax2 + bx + c = dx (ax2 + bx + c)n (ax2 + bx + c)n+1 d(ax2 + bx + c) = (2ax + b)dx
ax2 + bx + c =
1 4a
(2ax + b)2 + 4ac
(2ax+b)2 +4ac b2 1 dx = 4a(ax2 +bx+c)n+1 4a
−
u = 2ax + b d(ax2 +bx+c) dv = (ax 2 +bx+c)n+1
⇒
− b2
(2ax+b)2 4ac b2 dx + I n+1 (ax2 +bx+c)n+1 4a
du = 2a dx 1 v = n(ax2 +bx+c) n
−
−
1 (2ax + b) 2a dx 4ac b2 = + + I n+1 4a n(ax2 + bx + c)n n (ax2 + bx + c)n 4a 2ax + b 1 4ac b2 = + I n + I n+1 . 4an(ax2 + bx + c)n 2n 4a 4ac b2 1 2ax + b I n+1 = I n I n + n 4a 2n 4an ax2 + bx + c 2a 2n 1 2ax + b I n+1 = I + n n. 4ac b2 n n 4ac b2 ax2 + bx + c
−
−
−
⇒
−
−
− −
−
−
45
0
5
I n = =
⇒ ⇒
60
7
0
x2n
x2n
x2
2
a2 dx = 3 2
x2 a2
x2n
dx a2 I n
2
x2
1=
a2
a2
x2
a2
x2
3 2
a2 dx dv = x2n 2 dx
−
u= 2n−1 du = 3 x2 a2 xdx v = x2n−1
3 x2n 1 2 3 = x a2 2 I n a2 I n 1 2n 1 2n 1 2n 1 3 2n + 2 x I n = x2 a2 2 a2 I n 1 2n 1 2n 1 2n 1 3 x 2n 1 2 I n = x2 a2 2 a I n 1 . 2n + 2 2n + 2
I n =
⇒ ⇒
± − ± ∓ ± − ± ∓ √ ± − ± − ± − − ∓ − − − ∓ − ± − − − ± ∓ − − ± √ ± ± − − ± ± − ± − ± ∓ − ± ∓ −
du = 2nx2n−1 dx
x2n+1
x2
a2 dx =
u = x 2n dv =
x2
a2 xdx v =
3 3 1 2n = x2n x2 a2 2 x2n 1 x2 a2 2 dx 3 3 3 1 2n = x2n x2 a2 2 I n a2 I n 1 3 3 3 2n + 3 1 2n 2 2n 2 I n = x x a2 2 a I n 1 3 3 3 3 1 2na2 2n 2 2 2 I n = x x a I n 1 . 2n + 3 2n + 3
n
x2
± a2
3 2
I n = sin x dx = sin − x(1 − cos2 x)dx = I n−2 − cos2 x sinn−2 x dx u = cos x dv = sin n−2 x d(sin x) = 1 n−1 x du = − sin x dx v = n− 1 sin 1 1 = I n−2 − sin n−1 x cos x − I n n
−1
n 2
1 3
n
−1
⇒ n −n 1 I n = I n−2 − n −1 1 sinn−1 x cos x ⇒ I n = n −n 1 I n−2 − n1 sinn−1 x cos x.
46
0
8
I n = = = =
0
9
I n = =
10
0
sin2 x + cos 2 x sin2 x dx dx = dx + cosn x cosn x cosn−2 x u = sin x du = cos x dx sin2 x dx + I = d(cos x) n − 2 1 dv = cosn x v = (n−1)cos cosn x n−1 x
(n
−
(n
−
I n = =
dx = cosn x
−
sin x 1 I n−2 + I n−2 1)cosn−1 x (n 1) sin x n 2 + I n−2 . − n 1 1)cos x n 1
− − − −
n
tan x dx =
1 − cos2 x − tan x dx = 2
n 2
tann−2 x d(tan x)
−
n
cot x dx =
−
tann−2 x
cos x 1 I n−2 = tan n−1 x n 1
−
1 − sin2 x cot − x dx = 2
n 2
cotn−2 x d(cot x)
sin x
dx cos2 x
− I n−2
− I n−2.
cotn−2 x
dx sin2 x
− I n−2 = − n −1 1 cotn−1 x − I n−2.
− I n−2
47
11
0
I n =
dx = (a sin x + b cos x)n
u = (a sin x+b1cos x)n+1
a sin x + b cos x dx (a sin x + b cos x)n+1 a cos x−b sin x du = (n + 1) (a sin dx x+b cos x)n+2
− = dv =(a sin x+b cos x)dx v = −(a cos x − b sin x) a cos x − b sin x (a cos x − b sin x)2 =− − (n + 1)
⇒
12
0
−
dx (a sin x + b cos x)n+1 (a sin x + b cos x)n+2 a cos x b sin x a2 +b2 (a sin x+b cos x)2 = (n + 1) dx (a sin x+b cos x)n+1 (a sin x+b cos x)n+2 a cos x b sin x = (n + 1) a2 + b2 I n+2 I n n+1 (a sin x + b cos x) n 1 a cos x b sin x I n+2 = I . n (n + 1) a2 + b2 (n + 1) a2 + b2 (a sin x + b cos x)n+1
−
−
−
−
−
−
−
−
−
dx a + b cos x I n = = dx (a + b cos x)n (a + b cos x)n+1 dx cos x dx cos x dx = a +b = aI +b n +1 (a+b cos x)n+1 (a+b cos x)n+1 (a+b cos x)n+1 =
1 u = (a+b cos x)n+1 dv = cos x dx
du = (n + 1)b (a+bsincosx dx x)n+2 v = sin x
sin x sin2 x dx = aI n+1 +b (n+1)b (a+b cos x)n+1 (a+b cos x)n+2 b sin x 1 cos2 x 2 = aI n+1 + + (n + 1)b dx (a + b cos x)n+1 (a + b cos x)n+2 b sin x (n+1)b2 cos2 x 2 = +aI +(n+1)b I dx n+1 n+2 (a+b cos x)n+1 (a+b cos x)n+2
−
−
−
− a+b cos x a b
2
dx b sin x 2 2 = +aI n+1 (n+1)b I n+2 (n+1)b (a+b cos x)n+1 (a + b cos x)n+2 b sin x = (n+1)I n +a(2n+3)I n+1 (n+1) a2 +b2 I n+2 . (a+b cos x)n+1
−
−
−
−
48
⇒ ⇒
b sin x (a + b cos x)n+1 b sin x I n+2 = (n + 1) a2 + b2 (a + b cos x)n+1 (2n + 3)a + I n+1 . (n + 1) a2 + b2 (n + 1) a2 + b2 I n+2 =
− (n + 2)I n + a(2n + 3)I n+1 − (n + 1)n +a22 + b2
I n
19. Razlaganjem racionalne funkcije, izraˇcunati slede´ce integrale 10 30 50 80 100
x3 + 6 dx 20 x3 5x2 + 6x dx 40 x(x + 1)(x2 + x + 1)3 dx x + 1 0 6 dx x3 + 1 x3 1 x2 dx 90 2 2 x + 2x + 2 x3 1 x + 2 0 dx 11 dx x4 + x2 x3 + 1
− −
Reˇ senje:
10
I =
x3 prave racionalne funkcije.
−
dx (x
−
1)2 (x2
+ 1)2
x dx x3 3x + 2 dx 70 x4 + 1 2x2 5 dx x4 5x2 + 6 x + 1 120 dx x(x2 + 2x + 2)
− −
−
x3 + 6 dx; integral najpre svedemo na integral 5x2 + 6x
−
x3 + 6 x3 = 5x2 + 6x
− 5x2 + 6x + 5x2 − 6x + 6 = 1 + 5x2 − 6x + 6 . x3 − x3 − 5x2 + 6x x3 − 5x2 + 6x Kako je x3 − 5x2 + 6x = x(x2 − 5x + 6) = x(x − 2)(x − 3), to je 5x2 − 6x + 6 I = dx + dx. x(x − 2)(x − 3)
Nova racionalna podintegralna funkcija dozvoljava razvoj: 5x2 6x + 6 A B C = + + , x(x 2)(x 3) x x 2 x 3
− − −
gde je koeficijente A,B, C potrebno odrediti.
−
−
(1)
49 Metod neodred¯enih koeficijenata podrazumeva slede´ce: pomnoˇzimo jednakost (1) sa x(x 2)(x 3). Ona postaje jednakost dva polinoma 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.
Izjednaˇcavanjem koeficijenata uz iste stepene promenljive x dolazimo do sistema jednaˇcina 5 = A + B + C,
−6 = −5A − 3B − 2C, 6 = 6A. Reˇsavanjem ovog sistema nalazimo A = 1, B = −7 i C = 11.
Time polazni integral svodimo na zbir ˇcetiri elementarna integrala: I =
dx +
dx x
− 7
dx
dx
−2 x−3 = x + log |x| − 7log |x − 2| + 11 log |x − 3| + C.
20
I =
(x
(x
−
−
x
+ 11
dx , podintegralna funkcija dozvoljava razvoj: 1)2 (x2 + 1)2
1 A1 A2 B 1 x + C 1 B 2 x + C 2 = + + + . 2 1)2 (x2 + 1)2 x 1 (x 1)2 x2 + 1 x2 + 1
−
−
(2)
Odredimo A j , B j , C j metodom neodred¯enih koeficijenata. Nakon mnoˇzenja (2) sa (x 1)2 (x2 + 1)2 dolazimo do jednakosti
−
− 1) x2 + 1 2 + A2 x2 + 1 2 + (B1x + C 1)(x − 1)2 x2 + 1 +(B2 x + C 2 )(x − 1)2 = x5 (A1 + B1 ) + x4 (−A1 + A2 − 2B1 + C 1 ) + x3 (2A1 + 2B1 + B2 − 2C 1 ) +x2 (−2A1 + 2A2 − 2B1 − 2B2 + 2C 1 + C 2 ) +x(A1 + B1 + B2 − 2C 1 − 2C 2 ) − A1 + A2 + C 1 + C 2 .
1 = A1 (x
Izjednaˇcavanjem koeficijenata dva polinoma uz odgovaraju´ce stepene od x, dolazimo do sistema jednaˇcina 0 = A1 + B1 0 = A1 + A2 2B1 + C 1 0 = 2A1 + 2B1 + B2 2C 1 0 = 2A1 + 2A2 2B1 2B2 + 2C 1 + C 2 0 = A1 + B1 + B2 2C 1 2C 2 1 = A1 + A2 + C 1 + C 2 ,
− − −
−
−
− − − −
50 ˇcijim reˇsavanjem nalazimo vrednosti A1 =
−1/2, A2 = 1/4, B1 = 1/2, C 1 = 1/4, B2 = 1/2, C 2 = 0.
Polazni integral tada postaje 1 dx 1 dx 1 2x + 1 1 x dx I = + + dx + 2 x 1 4 (x 1)2 4 x2 + 1 2 x2 + 1 1 1 1 d x2 + 1 1 dx 1 = log x 1 + + + 2 4(x 1) 4 x2 + 1 4 x2 + 1 4 1 x2 + 1 x2 + x = log + arctan x + C. 4 (x 1)2 (x 1) x2 + 1
− −
30 I =
− | − |−
− −
− − − 2
d x2 + 1 x2 + 1
dx ; primetimo slede´ci razvoj podintegralne funkcije x(x + 1)(x2 + x + 1) 3
1 x2 + x + 1 x(x + 1) = x(x + 1)(x2 + x + 1)3 x(x + 1)(x2 + x + 1)3 1 1 = 2 2 2 x(x + 1)(x + x + 1) (x + x + 1) 3 x2 + x + 1 x(x + 1) 1 = x(x + 1)(x2 + x + 1)2 (x2 + x + 1) 3 1 1 1 = x(x + 1)(x2 + x + 1) (x2 + x + 1)2 (x2 + x + 1)3 1 1 1 1 = . x(x + 1) x2 + x + 1 (x2 + x + 1)2 (x2 + x + 1)3
−
−
−
−
−
−
−
−
−
Time polazni integral postaje I =
1.b)
= log
dx x(x + 1)
dx x2 + x + 1 d(x + 21 )
= log
12.10 18.30
= log
dx (x2 + x + 1) 2 d(x + 21 )
− − − − − − √ √ − − √ √ −
x x + 1
x +
1 2 2
+
d(x + 21 )
x +
1.c)
2
x x + 1 x x+1
1 2 2
+
3 4
3 4
3
x +
=
2 2x + 1 arctan 3 3
14 2x+1 arctan 3 3 3
t = x +
1 2 2 1 2
dt t2 +
2 2x+1 3 x2 +x+1
dx (x2 + x + 1) 3
− − − +
3 4
2
dt
3 2 4
t2 +
3 3 4
2x+1
6 x2 +x+1
2
+C.
51
40
0
5
x3
−
x 2 1 x 1 dx = log + + C. 3x + 2 9 2 + x 3(1 x)
−
dx 1 2x 1 1 (x + 1)2 = arctan + log 2 + C x3 + 1 6 x x + 1 3 3 1 2x 1 1 (x + 1) 3 = arctan + log + C. 6 x3 + 1 3 3
√ − √ −
√ √
60
−
−
√
− − − √ x2 + 2x + 2
= arctan(x + 1) + 2
110 120
− −
√
−
−
√
2 x)
+ C.
1 + C. x2 + 2x + 2
− √ √
2x2 5 1 x dx = log x4 5x2 + 6 2 2 x +
100
√
√ 2 x + 1 − √ 2 x + 1
x2 + + log 4 2 x2 1
x2 dx
90
−
dx 1 = arctan(1+ 2 x) arctan(1 (x2 +1) 2 2x2 2 2
√
80
−
dx = x4 +1
7
x + 1 1 (x 1)2 1 (x 1)3 dx = log + C = log + C. x3 1 3 x2 + x + 1 3 x3 1
0
−
− √ √
2 1 x + log 2 2 3 x +
√
3 + C. 3
x3 1 1 1 dx = + log(1 + x2 ) + arctan x + C. 4 2 x + x x 2 x + 2 dx = x3 + 1
2x 3 arctan
√ −3 1 + 16 log
(x + 1) 3 + C. x3 + 1
x + 1 1 1 x2 dx = arctan(x + 1) + log 2 + C. x(x2 + 2x + 2) 2 4 x + 2x + 2
20. Odrediti slede´ce integrale iracionalnih funkcija 10 40
dx x + 2+ 3 x + 2
20
dx (x+1) 5 x2 +2x
50
√
√
√
−√ √
1 x + 1 dx 30 3 1+ x + 1 x+1 dx x 1
−
60
dx √ √ 1+ x − 1+ x − 1
√ − − 6
x
(x
3
dx 1)(2
− x)
52
t6 = x + 2 dx = 6t5 dt
√ −
dx = x + 2 + 3 x + 2
√
Reˇ senje: 10
t3 + 1 1 dt = 6 t + 1
=6
(t2
t5 dt =6 t3 + t2
t3 dt t + 1
− − =6
t + 1)dt
dt t + 1
6
= 2t3
− 3t2 + 6t − 6log |t + 1| + C √ √ √ √ = 2 x + 2 − 3 x + 2 + 6 x + 2 − 6log x + 2 + 1 3
0
2
− √ √ − 1 1+
x + 1 dx = 3 x + 1
6
+ C.
(1 t3 )t5 =6 dt 1 + t2 t 1 t + 1)dt + 6 dt t2 + 1 2t dt 3t2 + 6t + 3 6 t2 + 1
t6 = x + 1 dx = 6t5 dt
( t6 + t4 + t3
=6
6
− t2
− − − −
6 7 6 5 3 4 dt = t + t + t 2t3 2 7 5 2 t +1 6 7 6 5 3 4 = t + t + t 2t3 3t2 + 6t + 3 log(t2 + 1) 6 arctan t + C 7 5 2 6 6 6 6 3 3 7 5 2 = x + 1 + x + 1 + x + 1 2 x + 1 7 5 2 3 3 x + 1 + 6 6 x + 1 + 3 log( 3 x + 1 + 1) 6 arctan 6 x + 1 + C.
− − − √ − √
− − − √
0
3
√ 6
x
−
=6
= 2t
√
dx 1+ 3 x 1+
√ −
(t2
3
2
+3
√
√ − x
1
=
t2
= 2t3
− − √
√
−
t6 = x 1 dx = 6t5 dt
t5 dt t + t2 + t3 2t + 1 1 dt t2 + t + 1
− − |− −
t dt = 2t3 + t + 1 d(t2 + t + 1) dt 3 t2 + t + 1
− t)dt + 6
− 3t
√
−
=6
3t2 + 3
−
dt t2 + t + 1 dt (t + 21 )2 + 43
− 3t2 + 3 log |t2 + t + 1 3 t + 21 t3 − 1 3 1.c) 3 2 − √ 3/2 arctan √ 3/2 + C = 2t − 3t + 3 log t−1 √ x − 1 −1 √ √ √ √ 2 x − 1+ 1 − 2 3 arctan √ 3 +C. = 2 x − 1 − 3 x − 1+3log √ x−1−1 3
6
6
53
40
x+1 = 1t , x, t > 0
dx = (x+1) 5 x2 +2x dx =
I =
√
− √ − √
t4 dt 1 t2
=
− dtt √ dt = a0 t3 + a1 t2 + a2 t + a3 1 − t2 + λ . 1 − t2 √ Diferenciranjem po t i mnoˇzenjem sa 1 − t2 dobijamo 2
−t4 = −4a0t4 − 3a1t3 + (3a0 − 2a2)t2 + (2a1 − a3)t + a2 + λ. Izjednaˇcavanjem koeficijenata uz stepene od t formiramo sistem jednaˇcina
−1 = −4a0 0 = 2a1 − a3 0 = −3a1 0 = a 2 + λ 0 = 3a0 − 2a2 ˇcije je reˇsenje a0 = 1/4, a1 = 0, a2 = 3/8, a3 = 0, λ =
−3/8.
Nakon sred¯ivanja nalazimo da je polazni integral I =
1 3 + 2x + 4(x + 1)4 8(x + 1) 2
x2
−
3 1 arcsin + C. 8 x + 1
Za x < 2 integral se izraˇcunava analognim postupkom, vode´ci raˇcuna da je t2 = t. Ovaj integral takod¯e se moˇze odrediti primenom Ojlerove smene.
√
− −
I =
√ −
dx = (x + 1) 5 x2 + 2x
− − −
= 4 = 4 = 4
√ (t2 − 1)4 dt =
x2
+ 2x = t x, x = dx = (t24t−dt1)2
−
2
t2 1
−
(t2 + 1 2)4 4 dt (t2 + 1)5 (t2 + 1)5 (t2 + 1)4 8(t2 + 1)3 + 24(t2 + 1)2 32(t2 + 1) + 16 dt (t2 + 1)5 dt dt dt dt + 32 96 + 128 t2 + 1 (t2 + 1)2 (t2 + 1)3 (t2 + 1)4 dt 64 . 2 (t + 1)5
−
−
−
−
−
54 Koriste´ci izvedenu rekurentnu relaciju iz zadatka 18.30 nalazimo I =
−
3 5t7 3t5 + 3t3 arctan t + 2 2(t2 + 1)4
−
− 5t + C.
Vra´ canjem smene konaˇcno dobijamo I =
5
3x2 + 6x + 5 2 x + 2x 4(x + 1)4
x + 1 dx = x 1
0
−
=
60
(x
√ x2 + 2x
dx 1)(2
−
x
− −− − − − − − t =
dx = 2t
x+1 x 1 4t dt (t2 1)2
=
t2
4
(t2
t + 1 + C t2 1 t 1 = (x + 1)(x 1) + log x +
12.20
−
−
3 arctan 2
1)2
dt
+ log
− 1)
∈ (0, π/2)
(x + 1)(x
x (1, 2) x = 1 + sin 2 t, t = dx = 2 sin t cos tdt, x) (x 1)(2 x) = sin2 t(1
∈ −
=2
+ C.
⇒
− sin2 t) √ dt = 2t + C = 2 arcsin x − 1 + C. −
+ C.
Integral moˇzemo reˇsiti i primenom tre´ce Ojlerove smene. I = =
− (x
dx 1)(2
− (x
− x)
=
x =
−2 arctan t + C = −2 arctan
1)(2
t2 +2 t2 +1
2 x
− x) = t(x − 1)
dx =
2t dt (t2 +1)2
−
− x + C. −1
− =
2
dt t2 + 1
21. Svod¯enjem na integral racionalne funkcije, izraˇcunati slede´ce integrale 10 40
dx x2 + x + 1
√ x +
x3 + x4 dx
20 50
dx 1 + 1 2x x2 x dx 2 1+ 3 x
√ − − √ √
30 60
− √ √ − x x + 3
3x
x2 + 3x + 2 dx x2 + 3x + 2 x3 dx
55
Reˇ senje: 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 sred¯ivanju podintegralne funkcije. Kako je t =
∓x +
x2 + x + 1,
posle pored¯enja sa podintegralnom funkcijom uzimamo Tada je x2 + x + 1 = x 2
− 2xt + t2 ⇒
√ x2 + x + 1 = −x + t.
x + 1 = t 2
− 2xt,
t2 1 t2 + t + 1 tj. x = , pa je dx = 2 dt. Integral onda glasi 1 + 2t (2t + 1) 2
−
I = 2
t2 + t + 1 dt. t(2t + 1) 2
Rastavimo racionalnu funkciju 2
t2 + t + 1 A1 A2 A3 = + + . 2 t(2t + 1) t 2t + 1 (2t + 1) 2
Metodom neodred¯enih koeficijenata nalazimo da je A1 = 2,
A2 =
−3,
A3 =
−3.
Integral se transformiˇse u elementarne integrale I = 2
dt t
dt 2t+1
− 3
= 2 log x+ gde je C = C 1
3 2
x2 +x+1
− 1/2.
d(2t+1) 3 3 =2 log t log 2t+1 + +C 1 (2t+1) 2 2 2(2t+1) 3 log 2x+2 x2 +x+1+1 x+ x2 +x+1+C, 2
− | |− | | − −
56
20
I =
dx √ = 1+ 1 − 2x − x2
√ − − −
x2 = xt
− − −
t2 +2t+1 dt. t(t 1)(t2 +1)
1 2x 1 = t2 +2t+1 dx =2 (t2 +1)2 dt
−t2 + 2t + 1 = A1 + A2 + B t + C t(t − 1)(t2 + 1) t t−1 t2 + 1 A1 = −1, A2 = 1, B = 0, C = −2 dt dt dt t−1 − − − 2 arctan t + C I = 2 = log 2 t−1 t t +1 t √ √ 1 − x + 1 − 2x − x2 1 + 1 − 2x − x2 √ = log + C. − 2 arctan x 1 + 1 − 2x − x2
30
I =
−√ √ −
x2 +3x+2
√
x2 +3x+2 =
x dx = x = x+ x2 +3x+2
= 2
(t
−
2 t2 , t2 1
t2 + 2t dt. 2)(t 1)(t + 1) 3
−
− −
(x+2)(x+1)= t(x+1)
dx =
− (t 2tdt −1) 2
2
2
2t 16 3 17 5 1 = + + + −2 (t − 2)(tt −+1)(t + − − 1)3 27(t − 2) 4(t − 1) 108(t + 1) 18(t + 1)2 3(t + 1) 3
I = =
3 17 5 1 log |t − 2| + log |t − 1| − log |t + 1 | − + C − 16 − 27 4 108 18(t + 1) 6(t + 1)2
log − 16 27
−
18
x + 2 x + 1
− −
− −
3 2 + log 4
5 x + 2 + 1 x + 1
x + 2 x + 1
1
1
6
x + 2 + 1 x + 1
2
17 log 108
+ C.
x + 2 + 1 x + 1
57
40
I =
− −
= 2 = 2
gde je I n =
x3 + x4 dx =
1 + x−1 = t 2 − 2 1 1/2 x (1 + x ) dx = dt dx = (t−22t −1)2
t2 dt t2 1 + 1 = 2 dt (t2 1)4 (t2 1)4 dt dt 2 = 2I 3 2 3 2 (t 1) (t 1)4
− −
− −
− −
− − 2I 4,
−
dt
(t2
− 1)n . Rekurentnu relaciju za izraˇcunavanje I n I n =
2(n
−
t 1)(t2
−
1)n 1
−
2n − 3 − 2(n − 1) I n−1
moˇzemo dobiti, na primer, iz 18.40 za a = 1, b = 0 i c = da je I 1 =
dt t2
−
1 t 1 = log + C 1 . 1 2 t + 1
−
Tada je t
−1, imaju´ci u vidu
1 I 2 = log 2(t2 1) 4 3t3 5t 3 + log I 3 = 8(t2 1)2 16 15t5 40t3 + 33t I 4 = 48(t2 1)3
t 1 + C 2 . − − t + 1 − t 1 + C 3 . t + 1 − − − − − 325 log tt +− 11 + C 4. 3t5 − 8t3 − 3t 1 t−1 I = − log + C − 24(t2 − 1)3 16 t + 1 √ √ 1 1 = x2 + x(8x2 + 2x − 3) + log x + 1 + x 24 8
−
− −
+ C.
58
0
5
√ √ −
t8 dt x = t6 dx = =6 2 dx =6t5 dt (t2 +1) 2 1+ 3 x x
− − − −
4t2 + 3 = 6 (t 2t + 3)dt 6 dt (t2 + 1)2 6 4(t2 + 1) 1 = t5 4t3 + 18t 6 dt 5 (t2 + 1)2 6 dt dt = t5 4t3 + 18t 24 + 6 5 t2 + 1 (t2 + 1)2 t 12.10 6 5 = t 4t3 + 18t 21 arctan t + 3 2 + C 5 t +1 6 x 3 = 6x 14 x2 + 70 3 x + 105 21 arctan 6 x + C. 3 5(1 + x ) 4
2
− −
− √ √
60
−
−
√
√
− − − − − √ √ − − − √ − 3
=
19.50
=
3x x3 dx = 9 2
2 1/3
)
dx = 2
− − − √ − 1+ 3x
dx =
2 = t3 ,
3
3 2t 1 arctan 2 3
3x
x3 + x
3x
3
x=
3 2(t3 +1)3/2
3 2
√ − − x √ + C.
x2 1 2 3 3x x3 + + arctan 9 9 3 3x
30 50 70
3tan x + 1 dx 3 tan x 1 2cos x + 3 sin x dx 2 + 3 cos x 4sin x dx sin2 x cos4 x
−− −
−
2x 1 dx 3 x
−
dt t3 +1
1 (t + 1) 3 log 3 + C 4 t +1
√
22. Svesti slede´ce integrale na integral racionalne funkcije. 10
3 t3 +1
3t2 dt
dt u=t dv = (t3t+1) t3 dt 3t 2 = = 1 (t3 +1) 2 du = dt v = 3(t3 +1) 2(t3 +1)
3t 2(t3 + 1)
1 = log 18
x( 1+ 3x
√
−
20 40 60 80
dx 4 + 3 tanx dx (sin x + 2 cos x)3 dx 2 2 sin x + cos x 3 x + 2 dx 2x 1
−
