MATEMATIKA 1– fiziˇcka hemija
NEODREDJENI INTEGRAL
Kaˇzemo da je funkcija F : X → R, X ⊂ R primitivna funkcija funkcije f : X → R ako je
F′(x) =f(x), x∈X, i piˇsemo R
f(x)dx=F(x) +C, C =const.
1. Osnovna svojstva
(a) d R
f(x)dx
=f(x)dx
(b) R
df(x) =f(x) +C
(c) R
λf(x)dx=λR
f(x)dx, λ∈R/{0} (d) R
(f(x) +g(x)) dx=R
f(x)dx+R
g(x)dx
2. Tablica osnovnih integrala
(a) R
xndx= xnn+1+1 +C, n6=−1 (g) R
axdx= lnaxa+C, a >0, a6= 1 (b) R dx
x = ln|x|+C (h)
R
exdx=ex+C
(c) R dx
1+x2 = arctanx+C (i)
R
sinx dx=−cosx+C
(d) R dx
x2−1 = 12ln
x−1
x+1
+C (j)
R
cosx dx= sinx+C
(e) R dx √
1−x2 =
arcsinx+C
−arccosx+C (k)
R dx
sin2x =−cotx+C
(f) R dx √
x2±1 = ln
x+
√ x2±1
+C (l) R dx
cos2x = tanx+C
Primeri:
1) R
(√x+ 1)(x−√x+ 1)dx=R√
x3−1 dx=R
x3/2dx+R
dx= 25x5/2+x+C
2) R
(6x2+ 8x+ 3)dx= 6R
x2dx+ 8R
x dx+ 3R
dx= 2x3+ 4x2+ 3x+C
3. Integracija prethodnim svodjenjem na oblik diferencijala
Ako je R
f(x)dx = F(x) +C, x ∈ X i x = ϕ(t), ϕ : Y → R, ϕ - neprekidna i diferencija-bilna, tada je R
f(ϕ(t))·ϕ′(t)dt=F(ϕ(t)) +C. Specijalno, R
f(ax+b)dx= a1F(ax+b) +C .
Primeri: 1) R dx
x−a = ln|x−a|+C
2) R dx
(x−a)n = 1−1n(x−a)1−n
3) R dx √
a2−x2 =
R d(xa)
q
1−(x a)
2 = arcsin
x a +C
4) R dx √
x2±a2 =
R d(xa)
q
(x a)
2
±1 = ln x+
q
x a
±1
+C0 = ln x+
√
x2±a2
+C, C =C0−ln|a|
5) R dx
a2+x2 = 1a
R d(xa) 1+(x
a)
2 = a1arctanxa+C
6) R dx
x2−a2 = 1a
R d(xa) (x
a)
2
−1 = 1 2aln
x−a x+a
4. Parcijalna integracija
u, v - diferencijabilne funkcije: R
u dv =uv−R
v du
Primeri:
1) R
xlnx dx=
u= lnx ⇒ du= 1xdx dv=x dx ⇒ v= x22
= x22 lnx−12R
x dx= x22 lnx−x42 +C
2) R
xsinxdx=
u=x ⇒ du=dx dv = sinx dx ⇒ v=−cosx
=−xcosx+R
cosxdx=−xcosx+sinx+C
3) I =R
excosx dx=
u= cosx ⇒ du=−sinx dx dv=exdx ⇒ v=ex
=excosx+R
exsinx dx=
=
u= sinx ⇒ du= cosx dx dv=exdx ⇒ v=ex
=excosx+exsinx−R
excosx dx=ex(cosx+ sinx)−I ⇒ 2I =ex(cosx+ sinx) ⇒ I = e2x(cosx+ sinx) +C
4) In=R dx (x2+a2)n =
(
u= (x2+1a2)n ⇒ du= −2nx
(x2+a2)n+1 dx
dv =dx ⇒ v=x
)
= (x2+xa2)n+2n
R x2
(x2+a2)n+1dx=
= (x2+xa2)n + 2n
R dx
(x2+a2)n −2na2
R dx
(x2+a2)n+1 = (x2+xa2)n + 2nIn−2na2In+1
⇒ In+1= 2na12(x2+xa2)n +2n−1
2na2In, n≥1, I1 = 1aarctanxa +C
5. Smena promenljive
a) x=ϕ(t), t- nova promenljiva, ϕ- ima neprekidan izvod potiϕ′(t)6= 0 R
f(x)dx=R
f(ϕ(t))ϕ′(t)dt
Primeri (trigonometrijske smene): 1) R√
a2−x2 dx smena: x=acost 2) R√
x2−a2 dx smena: x= a
cost
3) R√
a2+x2 dx smena: x=atant
b) u=ψ(x), f(x)dx=g(u)du
R
g(u)du=F(u) +C ⇒R
f(x)dx=F(ψ(x)) +C
Primer: R dx
√
5x−2 ={smena: u= 5x−2}= R 15 du
√
u =
2 5
√
u+C = 25√5x−2 +C
6. Integracija racionalnih funkcija
R(x) = PQ((xx)) =T(x) +Qr((xx)), P, Q, T, r - polinomi i degr <degQ
degQ=n ⇒ Q- ima taˇcnon- nula (prostih ili viˇsestrukih, realnih ili kompleksnih)
Q(x) =λ0(x−a1)k1(x−a2)k2. . .(x−ap)kp(x2+b1x+c1)l1(x2+b2x+c2)l2. . .(x2+bqx+cq)lq k1+k2+. . .+kp+ 2(l1+l2+. . .+lq)
R(x) = (x−Aa)k ∧ R(x) =
Bx+C
(x2+bx+c)l - proste racionalne funkcije
r(x)
Q(x) =
A11
x−a1 +
A12
(x−a1)2 +. . .+
A1k1
(x−a1)k1
+. . .+ Ap1
x−ap +
Ap2
(x−ap)2 +. . .+
Apkp
(x−ap)kp
+ +B11x+C11
x2+b1x+c1 +. . .+
B1l1x+C1l1
(x2+b1x+c1)l1
+. . .+Bq1x+Cq1
x2+bqx+cq +. . .+
Bqlqx+Cqlq
(x2+b
qx+cq)lq
Primeri:
1) R x3+1
x3−5x2+6xdx=
R
1 + 5x2−6x+1
x3−5x2+6x
dx
5x2−6x+1
x3−5x2+6x = 5x 2−6x+1
x(x−2)(x−3) = Ax+xB−2+xC−3 =
A(x2−5x+6)+B(x2−3x))+C(x2−2x)
x(x−2)(x−3) =
(A+B+C)x2+(−5A−3B−2C)x+6A
A+ B+ C= 5
7. Integrali oblika Imn=R
sinmx cosnx dx, m, n∈N
8. Integrali oblika R
4) opˇsta smena: t= tanx2, x∈(−π, π), dx= 1+2dtt2, sinx= 1+2tt2, cosx= 1−t 2
1+t2
Primer: R dx
1+sinx+cosx =
R
2 1+t2 dt
1+ 2t
1+t2+
1−t2
1+t2
=R dt
t+1 = ln|t+ 1|+C = ln tanx
2 + 1 +C
9. Integrali binomnih diferencijala Imnp=Rxm(a+bxn)pdx, m, n, p∈Q
1) p∈Z - smena: x=tλ, λ- najmanji zajedniˇcki sadrˇzalac imenilaca brojevam, n
Primer: R √
x(1 +√3x)−1dx=R
x12
1 +x13
−1
dx
m= 12, n= 13, p=−1∈Z, λ=N ZS(2,3) = 6 ⇒ smena: x=t6
R √
x(1 +√3x)−1dx=R
t3(1 +t2)−1·6t5dt= 6R t8
1+t2 dt= 6
R
t6−t4+t2+ 1 +1+1t2
dt= = 67t7−6
5t5+ 2t3+ 6t+ 6 arctant+C= 67x
7 6 −6
5x
5
6 + 2x12 + 6x16 + 6 arctan√6x+C
2) mn+1 ∈Z - smena: x= tν−a b
1n
, ν- imenilac broja p
Primer: R x
√
1+√3x2 dx=
R
x1 +x23
−12
dx
m= 1, n= 23, p=−12, mn+1 = 3∈Z, ν= 2 ⇒ smena: x= (t2−1)32
R x
√
1+√3x2 dx=
R (t2−1)
3 2
t ·3t(t2−1)
1
2 dt= 3R(t2−1)2dt= 3
5t5−2t3+ 3t+C= = 35(x23 + 1)52 −2(x23 + 1)32 + 3(x23 + 1)12 +C
3) mn+1 +p∈Z - smena: x=tνa−b
n1
, ν- imenilac brojap
Primer: R √3
3x−x3dx=R
x13(3−x2)13
m= 13, n= 2, p= 13, mn+1 = 1∈Z, ν = 3 ⇒ smena: x=t33+1
12
R √3
3x−x3dx=R 3
t3+1
16
3−t33+1
13
·−32√3 t
2
(t3+1)32 dt=−
9 2
R t3
(t3+1)2 dt=. . .
10. Integrali oblika R
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: R dx
x+√x2+x+1 =
n
smena: √x2+x+ 1 =t−xo=R −
2(1−t+t2)
(1−2t)2 t dt=
=R
−2t −(2t−1)3 2 +2t3−1
dt=−2 ln|t|+2(2t3−1) +32ln|2t−1|+C= =−2 ln(x+√x2+x+ 1) + 3
2(2x+2√x2+x+1−1) +
3
2ln(2x+ 2
√
x2+x+ 1−1) +C
2) II Ojlerova smena: c >0, smena: √ax2+bx+c=xt±√c
Primer: R x dx
1+√1+x−x2 =
n
smena: √1 +x−x2 =tx−1o=R
21−t−t2
(1+t2)21tdt=
= 2R dt
t −
R dt 1+t2 dt−
R 1+4t−t2 (1+t2)2 dt−
R 2t
1+t2 dt= 2 lnt−arctant−1+t−t22 −ln(1 +t2) +C=
= 1−√1 +x−x2−arctan1+√1+x−x2
x + ln(3−x+ 2 √
3) III Ojlerova smena: ax2+bx+c=a(x−λ)(x−µ), λ6=µ, λ, µ∈R, smena: √ax2+bx+c=t(x−λ) ∨ t(x−µ)
Primer: R √
2x−x2dx=nsmena: √2x−x2 =txo=R 2t
t2−1(t2−4−1)t3 dt=−8
R t2
(t2−1)4 dt=
=−8R dt (t2−1)3 −8
R dt
(t2−1)4 =−8 (I3+I4)
11. Integrali oblika R
R
x, αxγx++βδp1/n1, . . . ,αxγx++βδpk/nk
dx
Smena: x= αδt−nγt−βn, n=N ZS(n1, n2, . . . , nk)
Primer: R dx
√
2x−1−√42x−1
α= 2, β=−1, γ = 0, δ = 1, p1 =p2 = 1, n1= 2, n2 = 4 ⇒ smena: x= t
4+1
2 R dx
√
2x−1−√42x−1 = R 2t3
t2−tdt= 2
R t2
t−1 dt= 2 R
t+ 1 +t−11 dt=t2+ 2t+ 2 ln|t−1|+C= =√2x−1 + 2√4
2x−1 + ln4
√
