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Ako se pravilno ne potkrijepe pouzdanim izvorima, sporne rečenice i navodi mogli bi biti izbrisani. Pomozite Wikipediji tako što ćete navesti validne izvore putem referenci te nakon toga možete ukloniti ovaj šablon.
Tablični integrali predstavljaju grupu osnovnih integrala koji se rješavaju šablonski. Ovdje su dati neki od takvih integrala. Ovaj članak prikazuje spisak nekih najčešćih antiderivacija; kompletniju listu možete pronaći na članku spisak integrala .
Osnovna pravila integriranja [ uredi | uredi izvor ]
∫
a
f
(
x
)
d
x
=
a
∫
f
(
x
)
d
x
(
a
constant)
{\displaystyle \int af(x)\,dx=a\int f(x)\,dx\qquad {\mbox{(}}a{\mbox{ constant)}}\,\!}
∫
[
f
(
x
)
+
g
(
x
)
]
d
x
=
∫
f
(
x
)
d
x
+
∫
g
(
x
)
d
x
{\displaystyle \int [f(x)+g(x)]\,dx=\int f(x)\,dx+\int g(x)\,dx}
∫
f
(
x
)
g
(
x
)
d
x
=
f
(
x
)
∫
g
(
x
)
d
x
−
∫
[
f
′
(
x
)
(
∫
g
(
x
)
d
x
)
]
d
x
{\displaystyle \int f(x)g(x)\,dx=f(x)\int g(x)\,dx-\int \left[f'(x)\left(\int g(x)\,dx\right)\right]\,dx}
∫
[
f
(
x
)
]
n
f
′
(
x
)
d
x
=
[
f
(
x
)
]
n
+
1
n
+
1
+
C
(for
n
≠
−
1
)
{\displaystyle \int [f(x)]^{n}f'(x)\,dx={[f(x)]^{n+1} \over n+1}+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!}
∫
f
′
(
x
)
f
(
x
)
d
x
=
ln
|
f
(
x
)
|
+
C
{\displaystyle \int {f'(x) \over f(x)}\,dx=\ln {\left|f(x)\right|}+C}
∫
f
′
(
x
)
f
(
x
)
d
x
=
1
2
[
f
(
x
)
]
2
+
C
{\displaystyle \int {f'(x)f(x)}\,dx={1 \over 2}[f(x)]^{2}+C}
Više na: Spisak integrala racionalnih funkcija
∫
d
x
=
x
+
C
{\displaystyle \int \,{\rm {d}}x=x+C}
∫
x
n
d
x
=
x
n
+
1
n
+
1
+
C
if
n
≠
−
1
{\displaystyle \int x^{n}\,{\rm {d}}x={\frac {x^{n+1}}{n+1}}+C\qquad {\mbox{ if }}n\neq -1}
∫
d
x
x
=
ln
|
x
|
+
C
{\displaystyle \int {dx \over x}=\ln {\left|x\right|}+C}
∫
d
x
a
2
+
x
2
=
1
a
arctan
x
a
+
C
{\displaystyle \int {dx \over {a^{2}+x^{2}}}={1 \over a}\arctan {x \over a}+C}
Više na: Spisak integrala iracionalnih funkcija
∫
d
x
a
2
−
x
2
=
sin
−
1
x
a
+
C
{\displaystyle \int {dx \over {\sqrt {a^{2}-x^{2}}}}=\sin ^{-1}{x \over a}+C}
∫
−
d
x
a
2
−
x
2
=
cos
−
1
x
a
+
C
{\displaystyle \int {-dx \over {\sqrt {a^{2}-x^{2}}}}=\cos ^{-1}{x \over a}+C}
∫
d
x
x
x
2
−
a
2
=
1
a
sec
−
1
|
x
|
a
+
C
{\displaystyle \int {dx \over x{\sqrt {x^{2}-a^{2}}}}={1 \over a}\sec ^{-1}{|x| \over a}+C}
Više na: Spisak integrala logaritamskih funkcija
∫
ln
x
d
x
=
x
ln
x
−
x
+
C
{\displaystyle \int \ln {x}\,dx=x\ln {x}-x+C}
∫
log
b
x
d
x
=
x
log
b
x
−
x
log
b
e
+
C
{\displaystyle \int \log _{b}{x}\,dx=x\log _{b}{x}-x\log _{b}{e}+C}
Više na: Spisak integrala eksponencijalnih funkcija
∫
e
x
d
x
=
e
x
+
C
{\displaystyle \int e^{x}\,dx=e^{x}+C}
∫
a
x
d
x
=
a
x
ln
a
+
C
{\displaystyle \int a^{x}\,dx={\frac {a^{x}}{\ln {a}}}+C}
Više na: Spisak integrala trigonometrijskih funkcija i Spisak integrala arkusnih funkcija
∫
sin
x
d
x
=
−
cos
x
+
C
{\displaystyle \int \sin {x}\,dx=-\cos {x}+C}
∫
cos
x
d
x
=
sin
x
+
C
{\displaystyle \int \cos {x}\,dx=\sin {x}+C}
∫
tan
x
d
x
=
−
ln
|
cos
x
|
+
C
{\displaystyle \int \tan {x}\,dx=-\ln {\left|\cos {x}\right|}+C}
∫
cot
x
d
x
=
ln
|
sin
x
|
+
C
{\displaystyle \int \cot {x}\,dx=\ln {\left|\sin {x}\right|}+C}
∫
sec
x
d
x
=
ln
|
sec
x
+
tan
x
|
+
C
{\displaystyle \int \sec {x}\,dx=\ln {\left|\sec {x}+\tan {x}\right|}+C}
∫
csc
x
d
x
=
ln
|
csc
x
−
cot
x
|
+
C
{\displaystyle \int \csc {x}\,dx=\ln {\left|\csc {x}-\cot {x}\right|}+C}
∫
sec
2
x
d
x
=
tan
x
+
C
{\displaystyle \int \sec ^{2}x\,dx=\tan x+C}
∫
csc
2
x
d
x
=
−
cot
x
+
C
{\displaystyle \int \csc ^{2}x\,dx=-\cot x+C}
∫
sec
x
tan
x
d
x
=
sec
x
+
C
{\displaystyle \int \sec {x}\,\tan {x}\,dx=\sec {x}+C}
∫
csc
x
cot
x
d
x
=
−
csc
x
+
C
{\displaystyle \int \csc {x}\,\cot {x}\,dx=-\csc {x}+C}
∫
sin
2
x
d
x
=
1
2
(
x
−
sin
x
cos
x
)
+
C
{\displaystyle \int \sin ^{2}x\,dx={\frac {1}{2}}(x-\sin x\cos x)+C}
∫
cos
2
x
d
x
=
1
2
(
x
+
sin
x
cos
x
)
+
C
{\displaystyle \int \cos ^{2}x\,dx={\frac {1}{2}}(x+\sin x\cos x)+C}
∫
sin
n
x
d
x
=
−
sin
n
−
1
x
cos
x
n
+
n
−
1
n
∫
sin
n
−
2
x
d
x
{\displaystyle \int \sin ^{n}x\,dx=-{\frac {\sin ^{n-1}{x}\cos {x}}{n}}+{\frac {n-1}{n}}\int \sin ^{n-2}{x}\,dx}
∫
cos
n
x
d
x
=
cos
n
−
1
x
sin
x
n
+
n
−
1
n
∫
cos
n
−
2
x
d
x
{\displaystyle \int \cos ^{n}x\,dx={\frac {\cos ^{n-1}{x}\sin {x}}{n}}+{\frac {n-1}{n}}\int \cos ^{n-2}{x}\,dx}
∫
tan
−
1
x
d
x
=
x
arctan
x
−
1
2
ln
|
1
+
x
2
|
+
C
{\displaystyle \int \tan ^{-1}{x}\,dx=x\,\arctan {x}-{\frac {1}{2}}\ln {\left|1+x^{2}\right|}+C}
Više na: Spisak integrala hiperboličkih funkcija
∫
sinh
x
d
x
=
cosh
x
+
C
{\displaystyle \int \sinh x\,dx=\cosh x+C}
∫
cosh
x
d
x
=
sinh
x
+
C
{\displaystyle \int \cosh x\,dx=\sinh x+C}
∫
tanh
x
d
x
=
ln
|
cosh
x
|
+
C
{\displaystyle \int \tanh x\,dx=\ln |\cosh x|+C}
∫
csch
x
d
x
=
ln
|
tanh
x
2
|
+
C
{\displaystyle \int {\mbox{csch}}\,x\,dx=\ln \left|\tanh {x \over 2}\right|+C}
∫
sech
x
d
x
=
arctan
(
sinh
x
)
+
C
{\displaystyle \int {\mbox{sech}}\,x\,dx=\arctan(\sinh x)+C}
∫
coth
x
d
x
=
ln
|
sinh
x
|
+
C
{\displaystyle \int \coth x\,dx=\ln |\sinh x|+C}
Inverzne hiperboličke funkcije [ uredi | uredi izvor ]
∫
sinh
−
1
x
d
x
=
x
sinh
−
1
x
−
x
2
+
1
+
C
{\displaystyle \int \sinh ^{-1}x\,dx=x\sinh ^{-1}x-{\sqrt {x^{2}+1}}+C}
∫
cosh
−
1
x
d
x
=
x
cosh
−
1
x
+
x
2
−
1
+
C
{\displaystyle \int \cosh ^{-1}x\,dx=x\cosh ^{-1}x+{\sqrt {x^{2}-1}}+C}
∫
tanh
−
1
x
d
x
=
x
tanh
−
1
x
+
1
2
log
(
1
−
x
2
)
+
C
{\displaystyle \int \tanh ^{-1}x\,dx=x\tanh ^{-1}x+{\frac {1}{2}}\log {(1-x^{2})}+C}
∫
csch
−
1
x
d
x
=
x
csch
−
1
x
+
log
[
x
(
1
+
1
x
2
+
1
)
]
+
C
{\displaystyle \int {\mbox{csch}}^{-1}\,x\,dx=x{\mbox{csch}}^{-1}\ x+\log {\left[x\left({\sqrt {1+{\frac {1}{x^{2}}}}}+1\right)\right]}+C}
∫
sech
−
1
x
d
x
=
x
sech
−
1
x
−
tan
−
1
(
x
x
−
1
1
−
x
1
+
x
)
+
C
{\displaystyle \int {\mbox{sech}}^{-1}\,x\,dx=x{\mbox{sech}}^{-1}\ x-\tan ^{-1}{\left({\frac {x}{x-1}}{\sqrt {\frac {1-x}{1+x}}}\right)}+C}
∫
coth
−
1
x
d
x
=
x
coth
−
1
x
+
1
2
log
(
x
2
−
1
)
+
C
{\displaystyle \int \coth ^{-1}x\,dx=x\coth ^{-1}x+{\frac {1}{2}}\log {(x^{2}-1)}+C}
Postoje funkcije čiji se integrali ne mogu predstaviti u zatvorenom intervalu (integral [a,b]).
∫
0
∞
x
e
−
x
d
x
=
1
2
π
{\displaystyle \int _{0}^{\infty }{{\sqrt {x}}\,e^{-x}\,dx}={\frac {1}{2}}{\sqrt {\pi }}}
(također pogledajte Gama funkcija )
∫
0
∞
e
−
x
2
d
x
=
1
2
π
{\displaystyle \int _{0}^{\infty }{e^{-x^{2}}\,dx}={\frac {1}{2}}{\sqrt {\pi }}}
(Gausov integral )
∫
0
∞
x
e
x
−
1
d
x
=
π
2
6
{\displaystyle \int _{0}^{\infty }{{\frac {x}{e^{x}-1}}\,dx}={\frac {\pi ^{2}}{6}}}
(također pogledajte Bernulijev broj )
∫
0
∞
x
3
e
x
−
1
d
x
=
π
4
15
{\displaystyle \int _{0}^{\infty }{{\frac {x^{3}}{e^{x}-1}}\,dx}={\frac {\pi ^{4}}{15}}}
∫
0
∞
sin
(
x
)
x
d
x
=
π
2
{\displaystyle \int _{0}^{\infty }{\frac {\sin(x)}{x}}\,dx={\frac {\pi }{2}}}
∫
0
π
2
sin
n
x
d
x
=
∫
0
π
2
cos
n
x
d
x
=
1
⋅
3
⋅
5
⋅
⋯
⋅
(
n
−
1
)
2
⋅
4
⋅
6
⋅
⋯
⋅
n
π
2
{\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{n}{x}\,dx=\int _{0}^{\frac {\pi }{2}}\cos ^{n}{x}\,dx={\frac {1\cdot 3\cdot 5\cdot \cdots \cdot (n-1)}{2\cdot 4\cdot 6\cdot \cdots \cdot n}}{\frac {\pi }{2}}}
(if n is an even integer and
n
≥
2
{\displaystyle \scriptstyle {n\geq 2}}
)
∫
0
π
2
sin
n
x
d
x
=
∫
0
π
2
cos
n
x
d
x
=
2
⋅
4
⋅
6
⋅
⋯
⋅
(
n
−
1
)
3
⋅
5
⋅
7
⋅
⋯
⋅
n
{\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{n}{x}\,dx=\int _{0}^{\frac {\pi }{2}}\cos ^{n}{x}\,dx={\frac {2\cdot 4\cdot 6\cdot \cdots \cdot (n-1)}{3\cdot 5\cdot 7\cdot \cdots \cdot n}}}
(if
n
{\displaystyle \scriptstyle {n}}
is an odd integer and
n
≥
3
{\displaystyle \scriptstyle {n\geq 3}}
)
∫
0
∞
x
z
−
1
e
−
x
d
x
=
Γ
(
z
)
{\displaystyle \int _{0}^{\infty }x^{z-1}\,e^{-x}\,dx=\Gamma (z)}
(gdje je
Γ
(
z
)
{\displaystyle \Gamma (z)}
gama funkcija )
∫
−
∞
∞
e
−
(
a
x
2
+
b
x
+
c
)
d
x
=
π
a
exp
[
b
2
−
4
a
c
4
a
]
{\displaystyle \int _{-\infty }^{\infty }e^{-(ax^{2}+bx+c)}\,dx={\sqrt {\frac {\pi }{a}}}\exp \left[{\frac {b^{2}-4ac}{4a}}\right]}
(gdje je
exp
[
u
]
{\displaystyle \exp[u]}
eksponencijalna funkcija
e
u
{\displaystyle e^{u}}
.)
∫
0
2
π
e
x
cos
θ
d
θ
=
2
π
I
0
(
x
)
{\displaystyle \int _{0}^{2\pi }e^{x\cos \theta }d\theta =2\pi I_{0}(x)}
(gdje je
I
0
(
x
)
{\displaystyle I_{0}(x)}
modificirana Beselova funkcija prve vrste)
∫
0
2
π
e
x
cos
θ
+
y
sin
θ
d
θ
=
2
π
I
0
x
2
+
y
2
{\displaystyle \int _{0}^{2\pi }e^{x\cos \theta +y\sin \theta }d\theta =2\pi I_{0}{\sqrt {x^{2}+y^{2}}}}
∑
n
=
1
∞
n
−
n
=
∫
0
1
x
−
x
d
x
(
=
1.291285997
…
)
{\displaystyle \sum _{n=1}^{\infty }n^{-n}=\int _{0}^{1}x^{-x}\,dx\quad \quad (=1.291285997\dots )}
∑
n
=
1
∞
−
(
−
1
)
n
n
−
n
=
∫
0
1
x
x
d
x
(
=
0.783430510712
…
)
{\displaystyle \sum _{n=1}^{\infty }-(-1)^{n}n^{-n}=\int _{0}^{1}x^{x}\,dx\quad \quad (=0.783430510712\dots )}
(Pogledajte Johann Bernoulli i sofomorov san ).