من ويكيبيديا، الموسوعة الحرة
هذه قائمة بتكاملات لمختلف الدوال في الرياضيات .[1] [2]
قواعد مكاملة الدوال العامة
∫
a
f
(
x
)
d
x
=
a
∫
f
(
x
)
d
x
{\displaystyle \int af(x)\,dx=a\int f(x)\,dx}
∫
[
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
−
∫
(
d
[
f
(
x
)
]
∫
g
(
x
)
d
x
)
d
x
{\displaystyle \int f(x)g(x)\,dx=f(x)\int g(x)\,dx-\int \left(d[f(x)]\int g(x)\,dx\right)\,dx}
∫
a
f
(
y
)
d
y
=
a
∫
f
(
y
)
d
y
{\displaystyle \int af(y)\,dy=a\int f(y)\,dy}
∫
[
f
(
y
)
+
g
(
y
)
]
d
y
=
∫
f
(
y
)
d
y
+
∫
g
(
y
)
d
y
{\displaystyle \int [f(y)+g(y)]\,dy=\int f(y)\,dy+\int g(y)\,dy}
∫
f
(
y
)
g
(
y
)
d
y
=
f
(
y
)
∫
g
(
y
)
d
y
−
∫
(
d
[
f
(
y
)
]
∫
g
(
y
)
d
y
)
d
y
{\displaystyle \int f(y)g(y)\,dy=f(y)\int g(y)\,dy-\int \left(d[f(y)]\int g(y)\,dy\right)\,dy}
تكاملات الدوال البسيطة
قائمة تكاملات الدوال غير النسبية
∫
d
u
a
2
−
u
2
=
arcsin
u
a
+
C
{\displaystyle \int {du \over {\sqrt {a^{2}-u^{2}}}}=\arcsin {u \over a}+C}
∫
−
d
u
a
2
−
u
2
=
arccos
u
a
+
C
{\displaystyle \int {-du \over {\sqrt {a^{2}-u^{2}}}}=\arccos {u \over a}+C}
∫
d
u
u
u
2
−
a
2
=
1
a
arcsec
|
u
|
a
+
C
{\displaystyle \int {du \over u{\sqrt {u^{2}-a^{2}}}}={1 \over a}{\mbox{arcsec}}\,{|u| \over a}+C}
اللوغاريتمات
∫
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}
الدوال الأسية
∫
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}
∫
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
|
sec
x
|
+
C
{\displaystyle \int \tan {x}\,dx=\ln {\left|\sec {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}
∫
arctan
x
d
x
=
x
arctan
x
−
1
2
ln
|
1
+
x
2
|
+
C
{\displaystyle \int \arctan {x}\,dx=x\,\arctan {x}-{\frac {1}{2}}\ln {\left|1+x^{2}\right|}+C}
دوال القطع الزائد
∫
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}
تكاملات محددة
∫
0
∞
x
e
−
x
d
x
=
1
2
π
{\displaystyle \int _{0}^{\infty }{{\sqrt {x}}\,e^{-x}\,dx}={\frac {1}{2}}{\sqrt {\pi }}}
(أنظر أيضا دالة غاما )
∫
0
∞
e
−
x
2
d
x
=
1
2
π
{\displaystyle \int _{0}^{\infty }{e^{-x^{2}}\,dx}={\frac {1}{2}}{\sqrt {\pi }}}
∫
0
∞
x
e
x
−
1
d
x
=
π
2
6
{\displaystyle \int _{0}^{\infty }{{\frac {x}{e^{x}-1}}\,dx}={\frac {\pi ^{2}}{6}}}
(أنظر أيضا عدد بيرنولي )
∫
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
∞
x
z
−
1
e
−
x
d
x
=
Γ
(
z
)
{\displaystyle \int _{0}^{\infty }x^{z-1}\,e^{-x}\,dx=\Gamma (z)}
(حيث
Γ
(
z
)
{\displaystyle \Gamma (z)}
هي دالة غاما .)
∫
−
∞
∞
e
−
(
a
x
2
+
b
x
+
c
)
d
x
=
π
a
e
b
2
−
4
a
c
4
a
{\displaystyle \int _{-\infty }^{\infty }e^{-(ax^{2}+bx+c)}\,dx={\sqrt {\frac {\pi }{a}}}e^{\frac {b^{2}-4ac}{4a}}}
انظر
مراجع