मुक्त ज्ञानकोश विकिपीडिया से
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)
(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{)}}\,\!}
∫
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}
∫
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}
∫
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}
∫
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}
∫
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}
∫
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}
∫
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}
∫
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}
यह सम्पूर्ण पृष्ठ या इसके कुछ अनुभाग हिन्दी के अतिरिक्त अन्य भाषा(ओं) में भी लिखे गए हैं। आप इनका अनुवाद करके विकिपीडिया की सहायता कर सकते हैं।
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})}
∫
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 ).
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}