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Neodredeni integral

2010/2011

(Neodredeni integral) 2010/2011 1 / 1

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

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

Osobine neodredenog integrala

∫A · f (x) dx = A


(f (x)± g(x)) dx =

∫f (x) dx ±

∫g(x) dx(∫

f (x) dx

)′= f (x)


Neodredeni integral


∫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



(f (x)± g(x)) dx =

∫f (x) dx ±

∫g(x) dx

(∫f (x) dx

)′= f (x)


Neodredeni integral



(f (x)± g(x)) dx =

∫f (x) dx ±

∫g(x) dx(∫

f (x) dx

)′= f (x)


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



∫dx = x + C , α 6= −1

∫xα dx =

xα+1

α + 1+ C , α 6= −1∫

1

xdx = ln |x |+ C , x 6= 0∫


ax

ln a+ C , a > 0, a 6= 1∫




∫dx = x + C , α 6= −1∫xα dx =

xα+1

α + 1+ C , α 6= −1

∫1

xdx = ln |x |+ C , x 6= 0∫


ax

ln a+ C , a > 0, a 6= 1∫




∫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∫




∫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∫




∫dx = x + C , α 6= −1∫xα dx =

xα+1

α + 1+ C , α 6= −1∫

1

xdx = ln |x |+ C , x 6= 0∫


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



∫dx = x + C , α 6= −1∫xα dx =

xα+1

α + 1+ C , α 6= −1∫

1

xdx = ln |x |+ C , x 6= 0∫


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



∫dx = x + C , α 6= −1∫xα dx =

xα+1

α + 1+ C , α 6= −1∫

1

xdx = ln |x |+ C , x 6= 0∫


ax

ln a+ C , a > 0, a 6= 1∫




∫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



∫1


{(2k + 1)

π

2|k ∈ Z

}

∫1


1


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



∫1


{(2k + 1)

π

2|k ∈ Z

}∫

1

sin2 xdx = − ctg x + C , x ∈ R\ {kπ|k ∈ Z}

∫1


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



∫1


{(2k + 1)

π

2|k ∈ Z

}∫

1


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



∫1


{(2k + 1)

π

2|k ∈ Z

}∫

1


1


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



∫1


{(2k + 1)

π

2|k ∈ Z

}∫

1


1


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



∫1


{(2k + 1)

π

2|k ∈ Z

}∫

1


1


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


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