MATEM MATEMA ATIKA TIK A 1– fiziˇcka cka hemij hemija a
NEODREDJENI INTEGRAL R, X R primitivna Kaˇzemo zemo da je funkcija F : X primitivna funkcija funkcija funkcije f : X f ( f (x)dx = dx = F F ((x) + C, C = const. const.
→
F (x) = f ( f (x), x ∈ X , X , i piˇsemo se mo 1. Osnovna Osnovna svojstva svojstva
(a) d (b) (c) (d)
⊂
→ R ako je
f ( f (x) dx = f = f ((x) dx
df ( f (x) = f ( f (x) + C
λf ( λf (x) dx = dx = λ λ
f ( f (x) dx, λ
(f ( f (x) + g (x)) dx = dx =
∈ R/ {0}
f ( f (x) dx +
g (x) dx
2. Tablica osnovnih integrala integrala (a) (b) (c) (d) (e) (f) (f )
xn dx = dx =
xn+1 n+1
ax ln a + C, ex dx = dx = e e x + C
ax dx = dx =
− | | − √ ± √ − √ √ − + C, n =
dx x = ln x + C dx = arctan x + 1+x 1+x2
1
(g )
(h)
C
dx x2 1
1 x−1 − = 2 ln x+1 + C
arcsin x + C arccos x + C
√ dx 2 = 1−x
√ dx2 = ln x + x2 x ±1
(i)
sin x dx = dx =
(j)
cos x dx = dx = sin x + C dx sin2 x
(k)
1 + C
a > 0 > 0,, a = 1
=
− cos x + C
− cot x + C
dx = cos2 x
(l)
tan x + C
Primeri: 1) 2)
( x + 1)(x 1)(x
x3
x + 1) dx = dx =
(6x (6x2 + 8x 8x + 3) dx = dx = 6
x2 dx + 8
1 dx = dx =
x dx + 3
x3/2 dx +
dx = dx = 52 x5/2 + x + C
dx = dx = 2x3 + 4x 4x2 + 3x 3x + C
3. Integracij Integracijaa prethodnim prethodnim svodjenjem na oblik diferencijala diferencijala
Ako je f ( f (x) dx = F ( F (x) + C, x X i x = ϕ(t), ϕ : Y bilna, tada je f ( f (ϕ(t)) ϕ (t) dt = dt = F F ((ϕ(t)) + C . Specijalno,
∈
·
Primeri: 1) 2) 3) 4) 5) 6)
dx x a = dx (x a)n
− −
± = =
1 a
− =
1 a
dx x2 a2
f ( f (ax + b) dx = dx = a1 F ( F (ax + b) + C .
| − a| + C = 1−1 n (x − a)1−n
dx x2 a2
dx a2 +x2
R, ϕ - neprekidna i diferencija-
ln x
√ dx = a2 −x2 √
→
d( x a)
2
−( ) x a
1
d( x a) 2
( xa )
x d( a )
1+(
±1
x 2 a
= arcsin xa + C
±
= ln x +
x a
= a1 arctan xa + C
) x d( a ) = 2 x ( a ) −1
1 x a 2a ln x+a
− + C
1
√ + C = ln x + x2 ± a2 0
+ C , C = C 0
− ln |a|
4. Parcijalna integracija
u, v - diferencijabilne funkcije:
u dv = uv
−
u = ln x du = x1 dx 2 dv = x dx v = x2
Primeri: 1)
xln x dx =
⇒
v du
x2 2
ln x
− 21
x dx =
x2 2
x
2
2 n
2
2 n
2
2 n
2
2 n+1
2
2 n
2
2
2 n
2
2 n+1
2
2
ln x
2
− x4
+ C ⇒ u = x ⇒ du = dx 2) xsin xdx = = −x cos x+ cos xdx = −x cos x+sin x+C dv = sin x dx ⇒ v = − cos x u = cos x ⇒ du = − sin x dx 3) I = ex cos x dx = = e x cos x + ex sin x dx = dv = e x dx ⇒ v = e x u = sin x ⇒ du = cos x dx = = e x cos x + ex sin x − ex cos x dx = e x (cos x +sin x) − I x x dv = e dx ⇒ v = e ⇒ 2I = ex(cos x + sin x) ⇒ I = e2 (cos x + sin x) + C −2nx dx 1 ⇒ u = (x +a du = dx ) x x (x +a ) 4) I n = (x +a ) = = (x +a +2n dx = ) (x +a ) dv = dx ⇒ v = x x dx = (x +a + 2n (x +a − 2na2 (x +adx) = (x +ax ) + 2nI n − 2na2I n+1 ) ) ⇒ I n+1 = 2na1 (x +ax ) + 2n2na−1 I n, n ≥ 1, I 1 = a1 arctan xa + C
=
2
2 n
2 n
2
2
2 n+1
5. Smena promenljive a) x = ϕ(t), t - nova promenljiva, ϕ - ima neprekidan izvod po t i ϕ (t) = 0 f (x) dx = f (ϕ(t))ϕ (t) dt
Primeri (trigonometrijske smene):
√ √ − √ − a2 x2
1) 2)
x2 dx smena: x = a cos t a a2 dx smena: x = cos t
a2 + x2 dx smena: x = a tan t
3)
b) u = ψ(x), f (x) dx = g(u) du g(u)du = F (u) + C f (x) dx = F (ψ(x)) + C
⇒
Primer:
dx √ 5x −2 = {smena : u = 5x − 2} =
1 5
√ du = u
2 5
√ u + C = 2 √ 5x − 2 + C 5
6. Integracija racionalnih funkcija P (x) r(x) R(x) = Q(x) = T (x) + Q(x) , P, Q, T , r - polinomi i deg r < deg Q deg Q = n Q - ima taˇcno n- nula (prostih ili viˇsestrukih, realnih ili kompleksnih) Q(x) = λ 0 (x a1 )k1 (x a2 )k2 . . . (x a p )kp (x2 + b1 x + c1 )l1 (x2 + b2 x + c2 )l2 . . . (x2 + bq x + cq )lq k1 + k2 + . . . + k p + 2(l1 + l2 + . . . + lq ) R(x) = (x−Aa)k R(x) = (x2Bx+C - proste racionalne funkcije +bx+c)l
⇒ −
+
−
∧
A = + (xA12 + . . . + (x 1ak1)k1 + a1 )2 1 B1l1 x+C 1l1 B11 x+C 11 + . . . + (x2 +b x+c )l1 + . . . + x2 +b1 x+c1 1 1
r(x) Q(x)
−
A11 x a1
−
−
−
Primeri: 1)
A p A A . . . + x pa1p + (x pa2p )2 + . . . + (x pk ap )kp Bql x+C ql Bq 1 x+C q 1 + . . . + (x2 +bq x+c )qlq x2 +bq x+cq q q
x3 +1 5x2 6x+1 dx = 1 + dx 3 2 x 5x +6x x3 5x2 +6x 5x2 6x+1 5x2 6x+1 B C = =A 3 2 x +x 2+x 3 x(x 2)(x 3) x 5x +6x
−
− −
− − −
− −
−
−
=
−
−
−
A(x2 5x+6)+B(x2 3x))+C (x2 2x) x(x 2)(x 3)
−
−
−
−
−
=
+
(A+B+C )x2 +( 5A 3B 2C )x+6A x3 5x2 +6x
− − − −
A + B + C = 5 5A + 3B + 2C = 6 6A =1 2)
⇒
−92 , C = 283
x3 +1 1 9 dx = 1 + 6x + 3(x28 3) dx = x+ 16 ln x 92 ln x 2 + 28 3 3 2(x 2) x 5x2 +6x x dx x3 3x+2 (B+C )x2 +(A+B 2C )x+(2A @B+C ) x x A B C = = + + = 3 2 2 x 1 x+2 x 3x+2 (x 1) (x+2) (x 1) x3 3x+2
− −
−
−
−
−
−
−
−
B + C = 0 A + B 2C = 1 A = 31 , B = 92 , C = 29 2A + 2B + C = 0 x3 +1 dx 2 dx dx = 31 (x− + 92 xdx −1 9 x+2 = x3 −5x2 +6x 1)2
−
3)
A = 61 , B =
⇒
| |−
| − |
− −
−
ln x
| − 3|+C
−
−
−3(x1−1) + 92 ln |x − 1| − 92 ln |x + 2| + C
dx x3 +1 (A+B)x2 +( A+B+C )x+(A+C ) 1 A Bx+C = + = A = 31 , B = 13 , C = 32 3 2 x+1 x +1 x x+1 x3 +1 x 12 dx 1 dx 1 x 2 1 dx 1 = dx = dx + 21 x2 dxx+1 = 3 x+1 3 3 x+1 3 x3 +1 x2 x+1 x2 x+1 d(x2 x+1) 1 dx 1 dx = 31 x+1 +2 = 31 ln x + 1 16 ln x2 x + 1 + 13 arctan 2x 31 +C 1 2 3 6 x2 x+1 (x ) +
−
− − ⇒ − − | |− − ∈ ≥ − − − −
− −
− −
− −
−2
4
sinm x cosn x dx, m, n
7. Integrali oblika I mn =
sin10 x(1
2) m = 2k, n = 2l Transformacije: sin2 x = 21 (1
√
√ −
N
sin2k x cosn x d(cos x) =
1) m = 2k + 1, k 0 : I mn = Analogno za n = 2k + 1 Primer: sin10 x cos3 x dx =
−
cos2 x)k cosn x d(cos x)
(1
11
13
− sin2 x)d(sin x) = sin11 x − sin13 x + C
− cos2x), cos2 x = 21 (1 + cos 2x), sin x cos x = 21 sin2x
Primer: sin4 x cos2 x dx = (cos x sin x)2 sin2 x dx = 41 sin 2 2x 21 (1 cos2x) dx = 1−cos4x = 81 (sin2 2x sin2 2x cos2x) dx = 81 sin2 2x cos2x = 2 1 1 1 1 1 1 = 16 dx 16 cos 4x dx 16 sin2 2x d(sin 2x) = 16 x 64 sin4x 48 sin 3 2x + C
−
−
−
−
· − −
−
8. Integrali oblika R(sin x, cos x) dx, R = R(u, v) - racionalna funkcija, u = sin x, v = cos x 1) R( u, v) =
−
Primer: dx sin x =
−R(u, v)
- smena: t = cos x
sin x dx = 1 cos2 x
−
−dt = ln t−1 + C = ln t+1 1−t2
cos x 1 cos x+1
− + C
2) R(u, v) = R(u, v) - smena: t = sin x Primer: cos x dx = dt = 13 t−3 + C = 3sin1 3 x + C t4 sin4 x
−
−
−
−
∈ − π2 , π2
3) R( u, v) = R(u, v) - smena: t = tan x, x
− −
Primer: dx sin4 x cos2 x
=
−
1 3tan3 x
=
−
1 dt 1+t2 4 1 t (1+t2 )2 1+t2
2 tan x +
·
=
tan x + C
(1+t2 )2 t4
dt =
, sin x =
√ t 2 , cos x = √ 1 2 1+t 1+t
t−4 + 2t−2 + 1 dt =
−13 t−3 − 2t−1 + t + C =
4) opˇsta smena: t = tan x2 , x ( π, π), Primer: dx 1+sin x+cos x
=
2 dt 1+t2 2 1+ 2t 2 + 1−t2 1+t 1+t
dt t+1
=
∈ Z - smena:
Primer: x(1 +
√ √
2t , 1+t2
sin x =
cos x =
1 t2 1+t2
−
= ln t + 1 + C = ln tan x2 + 1 + C
|
|
xm (a + bxn ) p dx, m,n,p
9. Integrali binomnih diferencijala I mnp = 1) p
2 dt , 1+t2
dx =
∈ −
∈ Q
x = t λ , λ - najmanji zajedniˇcki sadrˇzalac imenilaca brojeva m, n
√ x)−1 dx = x 1 + x −1 dx m = 21 , n = 31 , p = −1 ∈ Z, λ = N ZS (2, 3) = 6 ⇒ smena: x = t 6 √ t 1 x(1 + x)−1 dx = t3 (1 + t2 )−1 · 6t5 dt = 6 1+t dt = 6 t6 − t4 + t2 + 1 + 1+t √ = 76 t7 − 56 t5 + 2t3 + 6t + 6 arctan t + C = 76 x − 56 x + 2x + 6x + 6 arctan x + C 2)
m+1 n
1 2
3
3
tν a b
−
− ·
∈ Z - smena: x =
Primer:
√ √
x
1+
√ 2 dx = x 1 + x
3
x
1 3
2 3
1 n
1+ x 2 5 3 3 + 1) 2 (x 5
= 3)
√ 2 dx = 3
− 12
m+1 n
+ p
1 m+1 2, n = 3 2 (t 1) 2 3t(t2 t
−
− 2(x
m = 3
−
1 3,
3x
−
x3
dx =
5 6
1 2
2
1 6
6
dt =
3
dx 3 2
∈ Z, ν = 2 ⇒ smena: x = (t2 − 1) − 1) dt = 3 (t2 − 1)2 dt = 53 t5 − 2t3 + 3t + C = 3
1 2
2
1
a
1
tν b
−
1 n
, ν - imenilac broja p
1
x2 ) 3
− ∈ − · x 3 (3
n = 2, p =
2
+ 1) 2 + 3(x 3 + 1) 2 + C
∈ Z - smena: x =
Primer: 3 3x x3 dx =
√ √
2 3
8
, ν - imenilac broja p
m = 1, n = 32 , p = x
7 6
1 m+1 3, n
=1
1 6
3 t3 +1
Z, ν = 3 1 3
3 t3 +1
3
⇒ smena:
√ − 3 3 2
t2 3 3 (t +1) 2
− 3 t3 +1
x =
dt =
9 2
1 2
t3 (t3 +1)2
dt = . . .
√
10. Integrali oblika R(x, ax2 + bx + c) dx – Ojlerove smene R = R(u, v), ax2 + bx + c = 0 - nema dvostruko reˇsenje 1) I Ojlerova smena: a > 0, smena:
√ ax2 + bx + c = t ± √ ax
Primer:
− −
− 2(1−t+t2)
√
−
(1−2t)2 dx 2 + x + 1 = t = smena : x x = dt = t x+ x2 +x+1 2 3 = + 2t3 1 dt = 2 ln t + 2(2t3 1) + 23 ln 2t 1 + C = t (2t 1)2 = 2ln(x + x2 + x + 1) + 2(2x+2 x32 +x+1 1) + 23 ln(2x + 2 x2 + x + 1
√
−
√ −
−
− | | √
2) II Ojlerova smena: c > 0, smena: Primer:
−
−
| √ − |
− 1) + C
√ ax2 + bx + c = xt ± √ c
√ 1 + x − x2 = tx − 1 = 2 1−t−t 1 dt = √ smena : (1+t ) t − −t dt − 2t dt = 2 ln t − arctan t − t−2 − ln(1 + t2 ) + C = − dt − 1+4t (1+t ) √ 1+t 1+t √ √ −x + ln(3 − x + 2 1 + x − x2 ) + C = 1 − 1 + x − x2 − arctan 1+ 1+x x
x dx = 1+ 1+x x2 dt = 2 dtt 1+t2
2 2 2
2
2
2 2 2
2
3) III Ojlerova smena: ax2 + bx + c = a(x λ)(x smena: ax2 + bx + c = t(x λ) t(x µ)
√
−
Primer: 2x x2 dx = smena :
√ =
∨
−
− µ), λ = µ, λ, µ ∈ R,
−
√ − − − − − 8
dt (t2 1)3
11. Integrali oblika Smena: x =
8
−
R x,
δtn β α γt n ,
−
−
dt (t2 1)4
−
αx+β γx+δ
2x
=
x2 = tx =
8 (I 3 + I 4 )
p1 /n1
αx+β γx+δ
, ... ,
2t
−4t dt = −8 t2 −1 (t2 −1)3
pk /nk
t2 (t2 1)4
−
dt =
dx
n = N ZS (n1 , n2 , . . . , nk )
Primer:
√ 2x−1dx 4 2x−1 − √
4
−1, γ = 0, δ = 1, p1 = p2 = 1, n1 = 2, n2 = 4 ⇒ smena: x = t 2+1 √ 2x−1dx √ 2x−1 = t2t−t dt = 2 t−t 1 dt = 2 t + 1 + t−1 1 dt = t 2 + 2t + 2ln |t − 1| + C = − √ √ √ = 2x − 1 + 2 2x − 1 + ln 2x − 1 − 1 + C α = 2, β = 4
4
3
2
4
2
