![Page 1: Neodred¯eni integralpolj.uns.ac.rs/wp-content/uploads/files/matematika/10-11-vzb-integrali.… · Osnovni neodred¯eni integrali Z dx = x +C, α 6= −1 Z x α dx = xα+1 α +1 +C,](https://reader033.vdocuments.net/reader033/viewer/2022060607/605c9ee66024e8587d13a84f/html5/thumbnails/1.jpg)
Neodredeni integral
2010/2011
(Neodredeni integral) 2010/2011 1 / 1
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Neodredeni integral
Definicija 1
Funkcija F je primitivna funkcija za funkciju f : (a, b) → R na intervalu(a, b) ako vazi
F ′(x) = f (x), ∀x ∈ (a, b) .
Definicija 2
Neofredeni integral funkcije f : (a, b) → R na intervalu (a, b) je skupsvih primitivnih funkcija za funkciju f na intervalu (a, b), odnosno,∫
f (x) dx = F (x) + C ,
gde je C proizvoljna konstanta.
(Neodredeni integral) 2010/2011 2 / 1
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Neodredeni integral
Definicija 1
Funkcija F je primitivna funkcija za funkciju f : (a, b) → R na intervalu(a, b) ako vazi
F ′(x) = f (x), ∀x ∈ (a, b) .
Definicija 2
Neofredeni integral funkcije f : (a, b) → R na intervalu (a, b) je skupsvih primitivnih funkcija za funkciju f na intervalu (a, b), odnosno,∫
f (x) dx = F (x) + C ,
gde je C proizvoljna konstanta.
(Neodredeni integral) 2010/2011 2 / 1
![Page 4: Neodred¯eni integralpolj.uns.ac.rs/wp-content/uploads/files/matematika/10-11-vzb-integrali.… · Osnovni neodred¯eni integrali Z dx = x +C, α 6= −1 Z x α dx = xα+1 α +1 +C,](https://reader033.vdocuments.net/reader033/viewer/2022060607/605c9ee66024e8587d13a84f/html5/thumbnails/4.jpg)
Neodredeni integral
Definicija 1
Funkcija F je primitivna funkcija za funkciju f : (a, b) → R na intervalu(a, b) ako vazi
F ′(x) = f (x), ∀x ∈ (a, b) .
Definicija 2
Neofredeni integral funkcije f : (a, b) → R na intervalu (a, b) je skupsvih primitivnih funkcija za funkciju f na intervalu (a, b), odnosno,∫
f (x) dx = F (x) + C ,
gde je C proizvoljna konstanta.
(Neodredeni integral) 2010/2011 2 / 1
![Page 5: Neodred¯eni integralpolj.uns.ac.rs/wp-content/uploads/files/matematika/10-11-vzb-integrali.… · Osnovni neodred¯eni integrali Z dx = x +C, α 6= −1 Z x α dx = xα+1 α +1 +C,](https://reader033.vdocuments.net/reader033/viewer/2022060607/605c9ee66024e8587d13a84f/html5/thumbnails/5.jpg)
Neodredeni integral
Osobine neodredenog integrala∫A · f (x) dx = A
∫f (x) dx , gde je A konstanta∫
(f (x)± g(x)) dx =
∫f (x) dx ±
∫g(x) dx(∫
f (x) dx
)′= f (x)
(Neodredeni integral) 2010/2011 3 / 1
![Page 6: Neodred¯eni integralpolj.uns.ac.rs/wp-content/uploads/files/matematika/10-11-vzb-integrali.… · Osnovni neodred¯eni integrali Z dx = x +C, α 6= −1 Z x α dx = xα+1 α +1 +C,](https://reader033.vdocuments.net/reader033/viewer/2022060607/605c9ee66024e8587d13a84f/html5/thumbnails/6.jpg)
Neodredeni integral
Osobine neodredenog integrala
∫A · f (x) dx = A
∫f (x) dx , gde je A konstanta∫
(f (x)± g(x)) dx =
∫f (x) dx ±
∫g(x) dx(∫
f (x) dx
)′= f (x)
(Neodredeni integral) 2010/2011 3 / 1
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Neodredeni integral
Osobine neodredenog integrala∫A · f (x) dx = A
∫f (x) dx , gde je A konstanta
∫(f (x)± g(x)) dx =
∫f (x) dx ±
∫g(x) dx(∫
f (x) dx
)′= f (x)
(Neodredeni integral) 2010/2011 3 / 1
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Neodredeni integral
Osobine neodredenog integrala∫A · f (x) dx = A
∫f (x) dx , gde je A konstanta∫
(f (x)± g(x)) dx =
∫f (x) dx ±
∫g(x) dx
(∫f (x) dx
)′= f (x)
(Neodredeni integral) 2010/2011 3 / 1
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Neodredeni integral
Osobine neodredenog integrala∫A · f (x) dx = A
∫f (x) dx , gde je A konstanta∫
(f (x)± g(x)) dx =
∫f (x) dx ±
∫g(x) dx(∫
f (x) dx
)′= f (x)
(Neodredeni integral) 2010/2011 3 / 1
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Osnovni neodredeni integrali
∫dx = x + C , α 6= −1∫xα dx =
xα+1
α + 1+ C , α 6= −1∫
1
xdx = ln |x |+ C , x 6= 0∫
ex dx = ex + C∫ax dx =
ax
ln a+ C , a > 0, a 6= 1∫
sin x dx = − cos x + C , x 6= 0∫cos x dx = sin x + C , x 6= 0
(Neodredeni integral) 2010/2011 4 / 1
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Osnovni neodredeni integrali
∫dx = x + C , α 6= −1
∫xα dx =
xα+1
α + 1+ C , α 6= −1∫
1
xdx = ln |x |+ C , x 6= 0∫
ex dx = ex + C∫ax dx =
ax
ln a+ C , a > 0, a 6= 1∫
sin x dx = − cos x + C , x 6= 0∫cos x dx = sin x + C , x 6= 0
(Neodredeni integral) 2010/2011 4 / 1
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Osnovni neodredeni integrali
∫dx = x + C , α 6= −1∫xα dx =
xα+1
α + 1+ C , α 6= −1
∫1
xdx = ln |x |+ C , x 6= 0∫
ex dx = ex + C∫ax dx =
ax
ln a+ C , a > 0, a 6= 1∫
sin x dx = − cos x + C , x 6= 0∫cos x dx = sin x + C , x 6= 0
(Neodredeni integral) 2010/2011 4 / 1
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Osnovni neodredeni integrali
∫dx = x + C , α 6= −1∫xα dx =
xα+1
α + 1+ C , α 6= −1∫
1
xdx = ln |x |+ C , x 6= 0
∫ex dx = ex + C∫ax dx =
ax
ln a+ C , a > 0, a 6= 1∫
sin x dx = − cos x + C , x 6= 0∫cos x dx = sin x + C , x 6= 0
(Neodredeni integral) 2010/2011 4 / 1
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Osnovni neodredeni integrali
∫dx = x + C , α 6= −1∫xα dx =
xα+1
α + 1+ C , α 6= −1∫
1
xdx = ln |x |+ C , x 6= 0∫
ex dx = ex + C
∫ax dx =
ax
ln a+ C , a > 0, a 6= 1∫
sin x dx = − cos x + C , x 6= 0∫cos x dx = sin x + C , x 6= 0
(Neodredeni integral) 2010/2011 4 / 1
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Osnovni neodredeni integrali
∫dx = x + C , α 6= −1∫xα dx =
xα+1
α + 1+ C , α 6= −1∫
1
xdx = ln |x |+ C , x 6= 0∫
ex dx = ex + C∫ax dx =
ax
ln a+ C , a > 0, a 6= 1
∫sin x dx = − cos x + C , x 6= 0∫cos x dx = sin x + C , x 6= 0
(Neodredeni integral) 2010/2011 4 / 1
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Osnovni neodredeni integrali
∫dx = x + C , α 6= −1∫xα dx =
xα+1
α + 1+ C , α 6= −1∫
1
xdx = ln |x |+ C , x 6= 0∫
ex dx = ex + C∫ax dx =
ax
ln a+ C , a > 0, a 6= 1∫
sin x dx = − cos x + C , x 6= 0
∫cos x dx = sin x + C , x 6= 0
(Neodredeni integral) 2010/2011 4 / 1
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Osnovni neodredeni integrali
∫dx = x + C , α 6= −1∫xα dx =
xα+1
α + 1+ C , α 6= −1∫
1
xdx = ln |x |+ C , x 6= 0∫
ex dx = ex + C∫ax dx =
ax
ln a+ C , a > 0, a 6= 1∫
sin x dx = − cos x + C , x 6= 0∫cos x dx = sin x + C , x 6= 0
(Neodredeni integral) 2010/2011 4 / 1
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Osnovni neodredeni integrali
∫1
cos2 xdx = tg x + C , x ∈ R\
{(2k + 1)
π
2|k ∈ Z
}∫
1
sin2 xdx = − ctg x + C , x ∈ R\ {kπ|k ∈ Z}∫
1
1 + x2dx = arctg x + C∫
1√1− x2
dx = arcsin x + C , |x | < 1∫1√
x2 + 1dx = ln
∣∣∣x +√
x2 + 1∣∣∣ + C∫
1√x2 − 1
dx = ln∣∣∣x +
√x2 − 1
∣∣∣ + C , |x | > 1
(Neodredeni integral) 2010/2011 5 / 1
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Osnovni neodredeni integrali
∫1
cos2 xdx = tg x + C , x ∈ R\
{(2k + 1)
π
2|k ∈ Z
}
∫1
sin2 xdx = − ctg x + C , x ∈ R\ {kπ|k ∈ Z}∫
1
1 + x2dx = arctg x + C∫
1√1− x2
dx = arcsin x + C , |x | < 1∫1√
x2 + 1dx = ln
∣∣∣x +√
x2 + 1∣∣∣ + C∫
1√x2 − 1
dx = ln∣∣∣x +
√x2 − 1
∣∣∣ + C , |x | > 1
(Neodredeni integral) 2010/2011 5 / 1
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Osnovni neodredeni integrali
∫1
cos2 xdx = tg x + C , x ∈ R\
{(2k + 1)
π
2|k ∈ Z
}∫
1
sin2 xdx = − ctg x + C , x ∈ R\ {kπ|k ∈ Z}
∫1
1 + x2dx = arctg x + C∫
1√1− x2
dx = arcsin x + C , |x | < 1∫1√
x2 + 1dx = ln
∣∣∣x +√
x2 + 1∣∣∣ + C∫
1√x2 − 1
dx = ln∣∣∣x +
√x2 − 1
∣∣∣ + C , |x | > 1
(Neodredeni integral) 2010/2011 5 / 1
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Osnovni neodredeni integrali
∫1
cos2 xdx = tg x + C , x ∈ R\
{(2k + 1)
π
2|k ∈ Z
}∫
1
sin2 xdx = − ctg x + C , x ∈ R\ {kπ|k ∈ Z}∫
1
1 + x2dx = arctg x + C
∫1√
1− x2dx = arcsin x + C , |x | < 1∫
1√x2 + 1
dx = ln∣∣∣x +
√x2 + 1
∣∣∣ + C∫1√
x2 − 1dx = ln
∣∣∣x +√
x2 − 1∣∣∣ + C , |x | > 1
(Neodredeni integral) 2010/2011 5 / 1
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Osnovni neodredeni integrali
∫1
cos2 xdx = tg x + C , x ∈ R\
{(2k + 1)
π
2|k ∈ Z
}∫
1
sin2 xdx = − ctg x + C , x ∈ R\ {kπ|k ∈ Z}∫
1
1 + x2dx = arctg x + C∫
1√1− x2
dx = arcsin x + C , |x | < 1
∫1√
x2 + 1dx = ln
∣∣∣x +√
x2 + 1∣∣∣ + C∫
1√x2 − 1
dx = ln∣∣∣x +
√x2 − 1
∣∣∣ + C , |x | > 1
(Neodredeni integral) 2010/2011 5 / 1
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Osnovni neodredeni integrali
∫1
cos2 xdx = tg x + C , x ∈ R\
{(2k + 1)
π
2|k ∈ Z
}∫
1
sin2 xdx = − ctg x + C , x ∈ R\ {kπ|k ∈ Z}∫
1
1 + x2dx = arctg x + C∫
1√1− x2
dx = arcsin x + C , |x | < 1∫1√
x2 + 1dx = ln
∣∣∣x +√
x2 + 1∣∣∣ + C
∫1√
x2 − 1dx = ln
∣∣∣x +√
x2 − 1∣∣∣ + C , |x | > 1
(Neodredeni integral) 2010/2011 5 / 1
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Osnovni neodredeni integrali
∫1
cos2 xdx = tg x + C , x ∈ R\
{(2k + 1)
π
2|k ∈ Z
}∫
1
sin2 xdx = − ctg x + C , x ∈ R\ {kπ|k ∈ Z}∫
1
1 + x2dx = arctg x + C∫
1√1− x2
dx = arcsin x + C , |x | < 1∫1√
x2 + 1dx = ln
∣∣∣x +√
x2 + 1∣∣∣ + C∫
1√x2 − 1
dx = ln∣∣∣x +
√x2 − 1
∣∣∣ + C , |x | > 1
(Neodredeni integral) 2010/2011 5 / 1