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Lyn 111:
[[bs:Tabela integrala]]
[[bs:Tabela integrala]]
[[de:Tabelle von Ableitungs- und Stammfunktionen]]
[[de:Tabelle von Ableitungs- und Stammfunktionen]]
[[fr:Table de primitives]]
[[id:Tabel integral]]
[[it:Tavola degli integrali più comuni]]
[[ko:적분표]]
[[nl:Lijst van integralen]]
[[pt:Tábua de integrais]]
[[ro:Tabel de integrale]]
[[ru:Список интегралов элементарных функций]]
[[ru:Список интегралов элементарных функций]]
[[sl:Tabela integralov]]
[[sr:Таблични интеграли]]
[[sr:Таблични интеграли]]
[[tr:İntegral tablosu]]
[[uk:Таблиця інтегралів]]
Wysiging soos op 19:25, 8 Maart 2013
Integrasie is een van die hoofbewerkings van calculus . Vir differensiasie kan die eenvoudiger dele van 'n funksie maklik gedifferensieer word, wat differensiasie dan vergemaklik, maar dit kan egter nie met integrasie gedoen word nie. Vir gevalle waar daar met komplekse funksies gewerk word, is dit makliker om 'n lys van integrale byderhand te hou.
Vir die doeleindes van hierdie lys word K as arbitrêre-integrasiekonstante gebruik.
Reëls by die integreer van algemene funksies
∫
a
f
(
x
)
d
x
=
a
∫
f
(
x
)
d
x
(
a
konstant)
{\displaystyle \int af(x)\,dx=a\int f(x)\,dx\qquad {\mbox{(}}a{\mbox{ konstant)}}\,\!}
∫
[
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
+
K
(vir
n
≠
−
1
)
{\displaystyle \int [f(x)]^{n}f'(x)\,dx={[f(x)]^{n+1} \over n+1}+K\qquad {\mbox{(vir }}n\neq -1{\mbox{)}}\,\!}
∫
f
′
(
x
)
f
(
x
)
d
x
=
ln
|
f
(
x
)
|
+
K
{\displaystyle \int {f'(x) \over f(x)}\,dx=\ln {\left|f(x)\right|}+K}
∫
f
′
(
x
)
f
(
x
)
d
x
=
1
2
[
f
(
x
)
]
2
+
K
{\displaystyle \int {f'(x)f(x)}\,dx={1 \over 2}[f(x)]^{2}+K}
Integrale van eenvoudige funksies
Rasionale funksies
∫
d
x
=
x
+
K
{\displaystyle \int \,{\rm {d}}x=x+K}
∫
x
n
d
x
=
x
n
+
1
n
+
1
+
K
mits
n
≠
−
1
{\displaystyle \int x^{n}\,{\rm {d}}x={\frac {x^{n+1}}{n+1}}+K\qquad {\mbox{ mits }}n\neq -1}
∫
d
x
x
=
ln
|
x
|
+
K
{\displaystyle \int {dx \over x}=\ln {\left|x\right|}+K}
∫
d
x
a
2
+
x
2
=
1
a
(bgtan)
x
a
+
K
{\displaystyle \int {dx \over {a^{2}+x^{2}}}={1 \over a}{\mbox{(bgtan)}}{x \over a}+K}
Irrasionale funksies
∫
d
x
a
2
−
x
2
=
sin
−
1
x
a
+
K
{\displaystyle \int {dx \over {\sqrt {a^{2}-x^{2}}}}=\sin ^{-1}{x \over a}+K}
∫
−
d
x
a
2
−
x
2
=
cos
−
1
x
a
+
K
{\displaystyle \int {-dx \over {\sqrt {a^{2}-x^{2}}}}=\cos ^{-1}{x \over a}+K}
∫
d
x
x
x
2
−
a
2
=
1
a
sec
−
1
|
x
|
a
+
K
{\displaystyle \int {dx \over x{\sqrt {x^{2}-a^{2}}}}={1 \over a}\sec ^{-1}{|x| \over a}+K}
Logaritmes
∫
ln
x
d
x
=
x
ln
x
−
x
+
K
{\displaystyle \int \ln {x}\,dx=x\ln {x}-x+K}
∫
log
b
x
d
x
=
x
log
b
x
−
x
log
b
e
+
K
{\displaystyle \int \log _{b}{x}\,dx=x\log _{b}{x}-x\log _{b}{e}+K}
Eksponensiaalfunksies
∫
e
x
d
x
=
e
x
+
K
{\displaystyle \int e^{x}\,dx=e^{x}+K}
∫
a
x
d
x
=
a
x
ln
a
+
K
{\displaystyle \int a^{x}\,dx={\frac {a^{x}}{\ln {a}}}+K}
Trigonometriese funksies
∫
sin
x
d
x
=
−
cos
x
+
K
{\displaystyle \int \sin {x}\,dx=-\cos {x}+K}
∫
cos
x
d
x
=
sin
x
+
K
{\displaystyle \int \cos {x}\,dx=\sin {x}+K}
∫
tan
x
d
x
=
ln
|
sec
x
|
+
K
{\displaystyle \int \tan {x}\,dx=\ln {\left|\sec {x}\right|}+K}
∫
cot
x
d
x
=
−
ln
|
csc
x
|
+
K
{\displaystyle \int \cot {x}\,dx=-\ln {\left|\csc {x}\right|}+K}
∫
sec
x
d
x
=
ln
|
sec
x
+
tan
x
|
+
K
{\displaystyle \int \sec {x}\,dx=\ln {\left|\sec {x}+\tan {x}\right|}+K}
∫
csc
x
d
x
=
−
ln
|
csc
x
+
cot
x
|
+
K
{\displaystyle \int \csc {x}\,dx=-\ln {\left|\csc {x}+\cot {x}\right|}+K}
∫
sec
2
x
d
x
=
tan
x
+
K
{\displaystyle \int \sec ^{2}x\,dx=\tan x+K}
∫
csc
2
x
d
x
=
−
cot
x
+
K
{\displaystyle \int \csc ^{2}x\,dx=-\cot x+K}
∫
sec
x
tan
x
d
x
=
sec
x
+
K
{\displaystyle \int \sec {x}\,\tan {x}\,dx=\sec {x}+K}
∫
csc
x
cot
x
d
x
=
−
csc
x
+
K
{\displaystyle \int \csc {x}\,\cot {x}\,dx=-\csc {x}+K}
∫
sin
2
x
d
x
=
1
2
(
x
−
sin
x
cos
x
)
+
K
{\displaystyle \int \sin ^{2}x\,dx={\frac {1}{2}}(x-\sin x\cos x)+K}
∫
cos
2
x
d
x
=
1
2
(
x
+
sin
x
cos
x
)
+
K
{\displaystyle \int \cos ^{2}x\,dx={\frac {1}{2}}(x+\sin x\cos x)+K}
∫
sec
3
x
d
x
=
1
2
sec
x
tan
x
+
1
2
ln
|
sec
x
+
tan
x
|
+
K
{\displaystyle \int \sec ^{3}x\,dx={\frac {1}{2}}\sec x\tan x+{\frac {1}{2}}\ln |\sec x+\tan x|+K}
∫
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}
∫
bgtan
x
d
x
=
x
bgtan
x
−
1
2
ln
|
1
+
x
2
|
+
K
{\displaystyle \int {\mbox{bgtan}}{x}\,dx=x\,{\mbox{bgtan}}{x}-{\frac {1}{2}}\ln {\left|1+x^{2}\right|}+K}
Hiperboliese funksies
∫
sinh
x
d
x
=
cosh
x
+
K
{\displaystyle \int \sinh x\,dx=\cosh x+K}
∫
cosh
x
d
x
=
sinh
x
+
K
{\displaystyle \int \cosh x\,dx=\sinh x+K}
∫
tanh
x
d
x
=
ln
|
cosh
x
|
+
K
{\displaystyle \int \tanh x\,dx=\ln |\cosh x|+K}
∫
csch
x
d
x
=
ln
|
tanh
x
2
|
+
K
{\displaystyle \int {\mbox{csch}}\,x\,dx=\ln \left|\tanh {x \over 2}\right|+K}
∫
sech
x
d
x
=
bgtan
(
sinh
x
)
+
K
{\displaystyle \int {\mbox{sech}}\,x\,dx={\mbox{bgtan}}(\sinh x)+K}
∫
coth
x
d
x
=
ln
|
sinh
x
|
+
K
{\displaystyle \int \coth x\,dx=\ln |\sinh x|+K}
∫
sech
2
x
d
x
=
tanh
x
+
K
{\displaystyle \int {\mbox{sech}}^{2}x\,dx=\tanh x+K}
Inverse hiperboliese funksies
∫
sinh
−
1
x
d
x
=
x
sinh
−
1
x
−
x
2
+
1
+
K
{\displaystyle \int \sinh ^{-1}x\,dx=x\sinh ^{-1}x-{\sqrt {x^{2}+1}}+K}
∫
cosh
−
1
x
d
x
=
x
cosh
−
1
x
−
x
2
−
1
+
K
{\displaystyle \int \cosh ^{-1}x\,dx=x\cosh ^{-1}x-{\sqrt {x^{2}-1}}+K}
∫
tanh
−
1
x
d
x
=
x
tanh
−
1
x
+
1
2
log
(
1
−
x
2
)
+
K
{\displaystyle \int \tanh ^{-1}x\,dx=x\tanh ^{-1}x+{\frac {1}{2}}\log {(1-x^{2})}+K}
∫
csch
−
1
x
d
x
=
x
csch
−
1
x
+
log
[
x
(
1
+
1
x
2
+
1
)
]
+
K
{\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]}+K}
∫
sech
−
1
x
d
x
=
x
sech
−
1
x
−
bgtan
(
x
x
−
1
1
−
x
1
+
x
)
+
K
{\displaystyle \int {\mbox{sech}}^{-1}\,x\,dx=x{\mbox{sech}}^{-1}x-{\mbox{bgtan}}{\left({\frac {x}{x-1}}{\sqrt {\frac {1-x}{1+x}}}\right)}+K}
∫
coth
−
1
x
d
x
=
x
coth
−
1
x
+
1
2
log
(
x
2
−
1
)
+
K
{\displaystyle \int \coth ^{-1}x\,dx=x\coth ^{-1}x+{\frac {1}{2}}\log {(x^{2}-1)}+K}
Bepaalde integrale sonder geslote-vorm afgeleides
∫
0
∞
x
e
−
x
d
x
=
1
2
π
{\displaystyle \int _{0}^{\infty }{{\sqrt {x}}\,e^{-x}\,dx}={\frac {1}{2}}{\sqrt {\pi }}}
∫
0
∞
e
−
x
2
d
x
=
1
2
π
{\displaystyle \int _{0}^{\infty }{e^{-x^{2}}\,dx}={\frac {1}{2}}{\sqrt {\pi }}}
(die Gaussiese integraal )
∫
0
∞
x
e
x
−
1
d
x
=
π
2
6
{\displaystyle \int _{0}^{\infty }{{\frac {x}{e^{x}-1}}\,dx}={\frac {\pi ^{2}}{6}}}
∫
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}}}
(mits n `n ewe heelgetal en
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}}}
(mits n `n onewe heelgetal en
n
≥
3
{\displaystyle \scriptstyle {n\geq 3}}
)
∫
0
∞
sin
2
x
x
2
d
x
=
π
2
{\displaystyle \int _{0}^{\infty }{\frac {\sin ^{2}{x}}{x^{2}}}\,dx={\frac {\pi }{2}}}
∫
0
∞
x
z
−
1
e
−
x
d
x
=
Γ
(
z
)
{\displaystyle \int _{0}^{\infty }x^{z-1}\,e^{-x}\,dx=\Gamma (z)}
(waar
Γ
(
z
)
{\displaystyle \Gamma (z)}
die Gamma funksie is)
∫
−
∞
∞
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]}
(waar
exp
[
u
]
{\displaystyle \exp[u]}
die eksponensiaalfunksie
e
u
{\displaystyle e^{u}}
is.)
∫
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)}
(waar
I
0
(
x
)
{\displaystyle I_{0}(x)}
die gewysigde Bessel funksie van die eerste tipe is.)
∫
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}}}}
∫
−
∞
∞
(
1
+
x
2
/
ν
)
−
(
ν
+
1
)
/
2
d
x
=
ν
π
Γ
(
ν
/
2
)
Γ
(
(
ν
+
1
)
/
2
)
)
{\displaystyle \int _{-\infty }^{\infty }{(1+x^{2}/\nu )^{-(\nu +1)/2}dx}={\frac {{\sqrt {\nu \pi }}\ \Gamma (\nu /2)}{\Gamma ((\nu +1)/2))}}\,}
(
ν
>
0
{\displaystyle \nu >0\,}
.
∫
a
b
f
(
x
)
d
x
=
(
b
−
a
)
∑
n
=
1
∞
∑
m
=
1
2
n
−
1
(
−
1
)
m
+
1
2
−
n
f
(
a
+
m
(
b
−
a
)
2
−
n
)
{\displaystyle \int _{a}^{b}{f(x)\,dx}=(b-a)\sum \limits _{n=1}^{\infty }{\sum \limits _{m=1}^{2^{n}-1}{\left({-1}\right)^{m+1}}}2^{-n}f(a+m\left({b-a}\right)2^{-n})}
Verwysings
Stewart, J. (2003). Single Variable Calculus . (5th ed.). Belmont, USA: Thomson Learning.
Groenewald, G.J., Hitge, M. (2005). Analise II Studiegids vir WISK121A . Potchefstroom: Noordwes-Universiteit.
Jordan, D.W., Smith, P. (2002). Mathematical techniques: An introduction for the engineering, physical and mathematical sciences . USA: Oxford University Press.
Aantekeninge