समाकल सूची
कलन
प्रवणता
डाइवर्जेन्स (अपसरण)
कर्ल
लाप्लास संकारक
प्रवणता प्रमेय
ग्रीन प्रमेय
स्टॉक प्रमेय
अपसरण प्रमेय
मेट्रिक्स कलन
आंशिक अवकलज
गुणित समाकल
रेखा समाकलन
पृष्ठ समाकल
आयतन समाकल
जेकोबियन
समाकलन , कलन की दो प्रमुख क्रियाओं में से एक है। अवकलन इस दृष्टि से समाकलन से भिन्न है कि अवकलज निकालने के लिये छोटे-छोटे और सरल नियम व उपाय हैं; जिनकी सहायता से कठिन से कठिन फलनों का भी अवकलज निकाला जा सकता है। समाकलन इस दृष्टि से कठिन है। इसलिये ज्ञात समाकलनों की सूची बहुत उपयोगी होती है।
नीचे कुछ अति सामान्य फलनों के समाकल दिये गये हैं:(x)
फलनों के समाकलन की सामान्य विधियाँ
(ये विधियाँ तभी लागू होंगी यदि दिया हुआ फलन समाकलनीय हो)
∫
a
f
(
x
)
d
x
=
a
∫
f
(
x
)
d
x
(
a
≠
0
, constant)
{\displaystyle \int af(x)\,dx=a\int f(x)\,dx\qquad {\mbox{(}a\neq 0{\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
)
−
∫
f
(
x
)
g
′
(
x
)
d
x
{\displaystyle \int f'(x)g(x)\,dx=f(x)g(x)-\int f(x)g'(x)\,dx}
-- ( खंडश: समाकलन (इण्टीग्रेशन बाई पार्ट्स) )
∫
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}
∫
[
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{)}\,\!}
सरल फलनों के समाकल
परिमेय फलन
∫
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
+
b
2
x
2
=
1
a
b
arctan
d
x
a
+
C
{\displaystyle \int {dx \over {a^{2}+b^{2}x^{2}={1 \over ab}\arctan {dx \over a}+C}
अपरिमेय फलन
∫
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}
लघुगणकीय फलन
∫
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}
चरघातांकी फलन
∫
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}
त्रिकोणमित्तीय फलन
∫
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}
∫
cosec
x
d
x
=
ln
|
cosec
x
−
cot
x
|
+
C
{\displaystyle \int {\mbox{cosec }{x}\,dx=\ln {\left|{\mbox{cosec }{x}-{\mbox{cot }{x}\right|}+C}
∫
sec
2
x
d
x
=
tan
x
+
C
{\displaystyle \int \sec ^{2}x\,dx=\tan x+C}
∫
cosec
x
cot
x
d
x
=
−
cosec
x
+
C
{\displaystyle \int {\mbox{cosec }{x}{\mbox{cot }{x}\,dx=-{\mbox{cosec }{x}+C}
∫
sec
x
tan
x
d
x
=
sec
x
+
C
{\displaystyle \int \sec {x}\,\tan {x}\,dx=\sec {x}+C}
∫
cosec
2
x
d
x
=
−
cot
x
+
C
{\displaystyle \int {\mbox{cosec}^{2}x\,dx=-{\mbox{cot }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}
∫
sec
3
x
d
x
=
1
2
sec
x
tan
x
+
1
2
ln
|
sec
x
+
tan
x
|
+
C
{\displaystyle \int \sec ^{3}x\,dx={\frac {1}{2}\sec x\tan x+{\frac {1}{2}\ln |\sec x+\tan x|+C}
(see integral of secant cubed)
∫
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}
हाइपरबोलिक फलन
∫
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}
∫
sech
2
x
d
x
=
tanh
x
+
C
{\displaystyle \int {\mbox{sech}^{2}x\,dx=\tanh x+C}
इन्वर्स हाइपरबोलिक फलन
∫
arcsinh
x
d
x
=
x
arcsinh
x
−
x
2
+
1
+
C
{\displaystyle \int \operatorname {arcsinh} x\,dx=x\operatorname {arcsinh} x-{\sqrt {x^{2}+1}+C}
∫
arccosh
x
d
x
=
x
arccosh
x
−
x
2
−
1
+
C
{\displaystyle \int \operatorname {arccosh} x\,dx=x\operatorname {arccosh} x-{\sqrt {x^{2}-1}+C}
∫
arctanh
x
d
x
=
x
arctanh
x
+
1
2
log
(
1
−
x
2
)
+
C
{\displaystyle \int \operatorname {arctanh} x\,dx=x\operatorname {arctanh} x+{\frac {1}{2}\log {(1-x^{2})}+C}
∫
arccsch
x
d
x
=
x
arccsch
x
+
log
[
x
(
1
+
1
x
2
+
1
)
]
+
C
{\displaystyle \int \operatorname {arccsch} \,x\,dx=x\operatorname {arccsch} x+\log {\left[x\left({\sqrt {1+{\frac {1}{x^{2}+1\right)\right]}+C}
∫
arcsech
x
d
x
=
x
arcsech
x
−
arctan
(
x
x
−
1
1
−
x
1
+
x
)
+
C
{\displaystyle \int \operatorname {arcsech} \,x\,dx=x\operatorname {arcsech} x-\arctan {\left({\frac {x}{x-1}{\sqrt {\frac {1-x}{1+x}\right)}+C}
∫
arccoth
x
d
x
=
x
arccoth
x
+
1
2
log
(
x
2
−
1
)
+
C
{\displaystyle \int \operatorname {arccoth} \,x\,dx=x\operatorname {arccoth} x+{\frac {1}{2}\log {(x^{2}-1)}+C}
प्रतिवर्तन सूत्र (रिकर्सन फॉर्मूले)
∫
1
(
x
2
+
1
)
n
d
x
=
1
2
n
−
2
⋅
x
(
x
2
+
1
)
n
−
1
+
2
n
−
3
2
n
−
2
⋅
∫
1
(
x
2
+
1
)
n
−
1
d
x
,
n
≥
2
{\displaystyle \int {\frac {1}{(x^{2}+1)^{n}\,\mathrm {d} x={\frac {1}{2n-2}\cdot {\frac {x}{(x^{2}+1)^{n-1}+{\frac {2n-3}{2n-2}\cdot \int {\frac {1}{(x^{2}+1)^{n-1}\,\mathrm {d} x,\quad n\geq 2}
∫
sin
n
(
x
)
d
x
=
n
−
1
n
∫
sin
n
−
2
(
x
)
d
x
−
1
n
cos
(
x
)
sin
n
−
1
(
x
)
,
n
≥
2
{\displaystyle \int \sin ^{n}(x)dx={\frac {n-1}{n}\int \sin ^{n-2}(x)dx-{\frac {1}{n}\cos(x)\sin ^{n-1}(x),\quad n\geq 2}
∫
cos
n
(
x
)
d
x
=
n
−
1
n
∫
cos
n
−
2
(
x
)
d
x
+
1
n
sin
(
x
)
cos
n
−
1
(
x
)
,
n
≥
2
{\displaystyle \int \cos ^{n}(x)dx={\frac {n-1}{n}\int \cos ^{n-2}(x)dx+{\frac {1}{n}\sin(x)\cos ^{n-1}(x),\quad n\geq 2}
There are some functions whose antiderivatives cannot be expressed in closed form. However, the values of the definite integrals of some of these functions over some common intervals can be calculated. A few useful integrals are given below.
∫
0
∞
x
e
−
x
d
x
=
1
2
π
{\displaystyle \int _{0}^{\infty }{\sqrt {x}\,e^{-x}\,dx}={\frac {1}{2}{\sqrt {\pi }
(see also Gamma function)
∫
0
∞
e
−
x
2
d
x
=
1
2
π
{\displaystyle \int _{0}^{\infty }{e^{-x^{2}\,dx}={\frac {1}{2}{\sqrt {\pi }
(the Gaussian 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}
(see also Bernoulli number)
∫
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
∞
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)}
(where
Γ
(
z
)
{\displaystyle \Gamma (z)}
is the Gamma function)
∫
−
∞
∞
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]}
(where
exp
[
u
]
{\displaystyle \exp[u]}
is the exponential function
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)}
(where
I
0
(
x
)
{\displaystyle I_{0}(x)}
is the modified Bessel function of the first kind)
∫
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}\left({\sqrt {x^{2}+y^{2}\right)}
∫
−
∞
∞
(
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\,}
, this is related to the probability density function of the Student's t-distribution)
The method of exhaustion provides a formula for the general case when no antiderivative exists:
∫
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})}
The "sophomore's dream"
∫
0
1
x
−
x
d
x
=
∑
n
=
1
∞
n
−
n
(
=
1.291285997
…
)
∫
0
1
x
x
d
x
=
∑
n
=
1
∞
−
(
−
1
)
n
n
−
n
(
=
0.783430510712
…
)
{\displaystyle {\begin{aligned}\int _{0}^{1}x^{-x}\,dx&=\sum _{n=1}^{\infty }n^{-n}&&(=1.291285997\dots )\\\int _{0}^{1}x^{x}\,dx&=\sum _{n=1}^{\infty }-(-1)^{n}n^{-n}&&(=0.783430510712\dots )\end{aligned}
(जॉन बर्नौली के नाम से प्रसिद्ध; see sophomore's dream).
विशेष फलन
गामा फलन :
Γ
(
z
)
=
∫
0
∞
x
z
−
1
e
−
x
d
x
{\displaystyle \Gamma (z)=\int _{0}^{\infty }x^{z-1}\,e^{-x}\,dx}
एरर फलन:
erf
(
x
)
=
2
π
∫
0
x
e
−
t
2
d
t
{\displaystyle {\text{erf}(x)={\frac {2}{\sqrt {\pi }\int _{0}^{x}e^{-t^{2}dt}
गघुगणकीय समाकल:
Li
(
x
)
=
∫
0
x
d
x
ln
x
{\displaystyle {\text{Li}(x)=\int _{0}^{x}{\frac {dx}{\ln x}
एलिप्टिक समाकल :
F
(
a
,
θ
)
=
∫
0
sin
θ
d
x
(
1
−
x
2
)
(
1
−
a
2
x
2
)
{\displaystyle F(a,\theta )=\int _{0}^{\text{sin }\theta }{\frac {dx}{\sqrt {(1-x^{2})(1-a^{2}x^{2})}
ज्या समाकल:
Si
(
x
)
=
∫
0
x
sin
t
t
d
t
{\displaystyle {\text{Si}(x)=\int _{0}^{x}{\frac {\text{sin }t}{t}dt}
कोज्या समाकल:
Ci
(
x
)
=
−
∫
x
∞
cos
t
t
d
t
{\displaystyle {\text{Ci}(x)=-\int _{x}^{\infty }{\frac {\text{cos}t}{t}dt}
इन्हें भी देखें
बाहरी कड़ियाँ
समाकलजों की सूची
उपपत्तियाँ
आनलाइन सेवाएँ
मुक्त स्रोत प्रोग्राम
Rational functions
Irrational functions
Trigonometric functions
Inverse trigonometric functions
Hyperbolic functions
Inverse hyperbolic functions
Exponential functions
Logarithmic functions
Gaussian functions
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