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Integralų lentelė
Kalba
Stebėti
Keisti
Pagrindiniai ir dažniausiai pasitaikantys
integralai
:
∫
0
d
x
=
C
{\displaystyle \int 0\;{\mathsf {d}}x=C}
∫
a
d
x
=
a
x
+
C
{\displaystyle \int a\;{\mathsf {d}}x=ax+C}
∫
x
n
d
x
=
x
n
+
1
n
+
1
+
C
{\displaystyle \int x^{n}\;{\mathsf {d}}x={\frac {x^{n+1}}{n+1}}+C}
∫
n
d
x
x
=
n
ln
|
x
|
+
C
{\displaystyle \int {\frac {{\mathsf {nd}}x}{x}}=n\ln \left|x\right|+C}
∫
e
x
d
x
=
e
x
+
C
{\displaystyle \int {\mathsf {e}}^{x}\;{\mathsf {d}}x={\mathsf {e}}^{x}+C}
∫
a
x
d
x
=
a
x
ln
a
+
C
{\displaystyle \int a^{x}\;{\mathsf {d}}x={\frac {a^{x}}{\ln a}}+C}
∫
d
x
x
2
+
a
2
=
1
a
arctan
x
a
+
C
,
a
≠
0
{\displaystyle \int {\frac {{\mathsf {d}}x}{x^{2}+a^{2}}}={\frac {1}{a}}\arctan {\frac {x}{a}}+C,a\not =0}
∫
1
a
2
−
x
2
d
x
=
arcsin
x
a
+
C
,
a
>
0
{\displaystyle \int {\frac {1}{\sqrt {a^{2}-x^{2}}}}\;{\mathsf {d}}x=\arcsin {\frac {x}{a}}+C,\;a>0}
∫
d
x
x
2
−
a
2
=
1
2
a
ln
|
x
−
a
x
+
a
|
+
C
,
a
≠
0
{\displaystyle \int {\frac {{\mathsf {d}}x}{x^{2}-a^{2}}}={\frac {1}{2a}}\ln \left|{\frac {x-a}{x+a}}\right|+C,\;a\not =0}
∫
d
x
x
2
±
a
2
=
ln
|
x
+
x
2
±
a
2
|
+
C
,
a
≠
0
{\displaystyle \int {\frac {{\mathsf {d}}x}{\sqrt {x^{2}\pm a^{2}}}}=\ln \left|x+{\sqrt {x^{2}\pm a^{2}}}\right|+C,\;a\not =0}
∫
a
2
−
x
2
d
x
=
x
2
a
2
−
x
2
+
a
2
2
arcsin
x
a
+
C
{\displaystyle \int {\sqrt {a^{2}-x^{2}}}\;{\mathsf {d}}x={\frac {x}{2}}{\sqrt {a^{2}-x^{2}}}+{\frac {a^{2}}{2}}\arcsin {\frac {x}{a}}+C}
∫
x
2
±
a
2
d
x
=
x
2
x
2
±
a
2
±
a
2
2
ln
|
x
+
x
2
±
a
2
|
+
C
{\displaystyle \int {\sqrt {x^{2}\pm a^{2}}}\;{\mathsf {d}}x={\frac {x}{2}}{\sqrt {x^{2}\pm a^{2}}}\pm {\frac {a^{2}}{2}}\ln \left|x+{\sqrt {x^{2}\pm a^{2}}}\right|+C}
∫
a
x
+
b
d
x
=
(
2
b
3
a
+
2
x
3
)
a
x
+
b
+
C
{\displaystyle \int {\sqrt {ax+b}}\;{\mathsf {d}}x=\left({2b \over 3a}+{2x \over 3}\right){\sqrt {ax+b}}+C}
∫
a
x
+
b
d
x
=
2
3
a
(
a
x
+
b
)
3
/
2
+
C
{\displaystyle \int {\sqrt {ax+b}}dx={2 \over 3a}(ax+b)^{3/2}+C}
∫
(
a
+
b
)
2
−
x
d
x
=
−
2
(
a
2
+
2
a
b
+
b
2
−
x
)
a
2
+
2
a
b
+
b
2
−
x
3
+
C
{\displaystyle \int {\sqrt {(a+b)^{2}-x}}dx=-{2(a^{2}+2ab+b^{2}-x){\sqrt {a^{2}+2ab+b^{2}-x}} \over 3}+C}
∫
a
x
2
+
b
x
d
x
=
2
b
2
a
s
i
n
(
a
r
c
s
e
c
(
2
a
x
+
b
b
)
)
+
b
2
a
c
o
s
(
a
r
c
s
e
c
(
2
a
x
+
b
b
)
)
2
ln
(
|
s
i
n
(
a
r
c
s
e
c
(
2
a
x
+
b
b
)
)
−
1
s
i
n
(
a
r
c
s
e
c
(
2
a
x
+
b
b
)
)
+
1
|
)
16
a
2
×
c
o
s
(
a
r
c
s
e
c
(
2
a
x
+
b
b
)
)
2
+
C
{\displaystyle \int {\sqrt {ax^{2}+bx}}dx={{2b^{2}{\sqrt {a}}sin(arcsec({2ax+b \over b}))+b^{2}{\sqrt {a}}cos(arcsec({2ax+b \over b}))^{2}\ln(\left\vert {{sin(arcsec({{2ax+b} \over {b}}))-1} \over {sin(arcsec({{2ax+b} \over {b}}))+1}}\right\vert )} \over {16a^{2}\times cos(arcsec({{2ax+b} \over {b}}))^{2}}}+C}
Trigonometrinių reiškinių integralai
redaguoti
∫
sin
(
a
x
)
d
x
=
−
1
a
cos
(
a
x
)
+
C
{\displaystyle \int \sin(ax)\;{\mathsf {d}}x=-{\frac {1}{a}}\cos(ax)+C}
∫
cos
(
a
x
)
d
x
=
1
a
sin
(
a
x
)
+
C
{\displaystyle \int \cos(ax)\;{\mathsf {d}}x={\frac {1}{a}}\sin(ax)+C}
∫
tan
x
d
x
=
−
ln
|
cos
x
|
+
C
{\displaystyle \int \tan x\;{\mathsf {d}}x=-\ln |\cos x|+C}
∫
c
t
g
x
d
x
=
ln
|
sin
x
|
+
C
{\displaystyle \int ctgx\;{\mathsf {d}}x=\ln |\sin x|+C}
∫
d
x
sin
x
=
ln
|
tan
x
2
|
+
C
{\displaystyle \int {\frac {{\mathsf {d}}x}{\sin x}}=\ln \left|\tan {\frac {x}{2}}\right|+C}
∫
d
x
cos
x
=
ln
|
tan
(
x
2
+
π
4
)
|
+
C
{\displaystyle \int {\frac {{\mathsf {d}}x}{\cos x}}=\ln \left|\tan \left({\frac {x}{2}}+{\frac {\pi }{4}}\right)\right|+C}
∫
d
x
sin
2
x
=
−
c
t
g
x
+
C
{\displaystyle \int {\frac {{\mathsf {d}}x}{\sin ^{2}x}}=-ctgx+C}
∫
d
x
cos
2
x
=
tan
x
+
C
{\displaystyle \int {\frac {{\mathsf {d}}x}{\cos ^{2}x}}=\tan x+C}
Taip pat skaitykite
redaguoti
Integravimo metodai
Nuorodos
redaguoti
integralų lentelė
integralų lentelė su įrodymais
http://www.mathwords.com/i/integral_table.htm
