Bere barnean (x² +a²)-ren erroa daukaten funtzioen integralak
r
=
x
2
+
a
2
{\displaystyle r={\sqrt {x^{2}+a^{2}}}}
aldatu
∫
r
d
x
=
1
2
(
x
r
+
a
2
ln
(
x
+
r
)
)
+
K
{\displaystyle \int r\;dx={\frac {1}{2}}\left(xr+a^{2}\,\ln \left(x+r\right)\right)+K}
∫
r
3
d
x
=
1
4
x
r
3
+
1
8
3
a
2
x
r
+
3
8
a
4
ln
(
x
+
r
)
+
K
{\displaystyle \int r^{3}\;dx={\frac {1}{4}}xr^{3}+{\frac {1}{8}}3a^{2}xr+{\frac {3}{8}}a^{4}\ln \left(x+r\right)+K}
∫
r
5
d
x
=
1
6
x
r
5
+
5
24
a
2
x
r
3
+
5
16
a
4
x
r
+
5
16
a
6
ln
(
x
+
r
)
+
K
{\displaystyle \int r^{5}\;dx={\frac {1}{6}}xr^{5}+{\frac {5}{24}}a^{2}xr^{3}+{\frac {5}{16}}a^{4}xr+{\frac {5}{16}}a^{6}\ln \left(x+r\right)+K}
∫
x
r
d
x
=
r
3
3
+
K
{\displaystyle \int xr\;dx={\frac {r^{3}}{3}}+K}
∫
x
r
3
d
x
=
r
5
5
+
K
{\displaystyle \int xr^{3}\;dx={\frac {r^{5}}{5}}+K}
∫
x
r
2
n
+
1
d
x
=
r
2
n
+
3
2
n
+
3
+
K
{\displaystyle \int xr^{2n+1}\;dx={\frac {r^{2n+3}}{2n+3}}+K}
∫
x
2
r
d
x
=
x
r
3
4
−
a
2
x
r
8
−
a
4
8
ln
(
x
+
r
)
+
K
{\displaystyle \int x^{2}r\;dx={\frac {xr^{3}}{4}}-{\frac {a^{2}xr}{8}}-{\frac {a^{4}}{8}}\ln \left(x+r\right)+K}
∫
x
2
r
3
d
x
=
x
r
5
6
−
a
2
x
r
3
24
−
a
4
x
r
16
−
a
6
16
ln
(
x
+
r
)
+
K
{\displaystyle \int x^{2}r^{3}\;dx={\frac {xr^{5}}{6}}-{\frac {a^{2}xr^{3}}{24}}-{\frac {a^{4}xr}{16}}-{\frac {a^{6}}{16}}\ln \left(x+r\right)+K}
∫
x
3
r
d
x
=
r
5
5
−
a
2
r
3
3
+
K
{\displaystyle \int x^{3}r\;dx={\frac {r^{5}}{5}}-{\frac {a^{2}r^{3}}{3}}+K}
∫
x
3
r
3
d
x
=
r
7
7
−
a
2
r
5
5
+
K
{\displaystyle \int x^{3}r^{3}\;dx={\frac {r^{7}}{7}}-{\frac {a^{2}r^{5}}{5}}+K}
∫
x
3
r
2
n
+
1
d
x
=
r
2
n
+
5
2
n
+
5
−
a
3
r
2
n
+
3
2
n
+
3
+
K
{\displaystyle \int x^{3}r^{2n+1}\;dx={\frac {r^{2n+5}}{2n+5}}-{\frac {a^{3}r^{2n+3}}{2n+3}}+K}
∫
x
4
r
d
x
=
x
3
r
3
6
−
a
2
x
r
3
8
+
a
4
x
r
16
+
a
6
16
ln
(
x
+
r
)
+
K
{\displaystyle \int x^{4}r\;dx={\frac {x^{3}r^{3}}{6}}-{\frac {a^{2}xr^{3}}{8}}+{\frac {a^{4}xr}{16}}+{\frac {a^{6}}{16}}\ln \left(x+r\right)+K}
∫
x
4
r
3
d
x
=
x
3
r
5
8
−
a
2
x
r
5
16
+
a
4
x
r
3
64
+
3
a
6
x
r
128
+
3
a
8
128
ln
(
x
+
r
)
+
K
{\displaystyle \int x^{4}r^{3}\;dx={\frac {x^{3}r^{5}}{8}}-{\frac {a^{2}xr^{5}}{16}}+{\frac {a^{4}xr^{3}}{64}}+{\frac {3a^{6}xr}{128}}+{\frac {3a^{8}}{128}}\ln \left(x+r\right)+K}
∫
x
5
r
d
x
=
r
7
7
−
2
a
2
r
5
5
+
a
4
r
3
3
+
K
{\displaystyle \int x^{5}r\;dx={\frac {r^{7}}{7}}-{\frac {2a^{2}r^{5}}{5}}+{\frac {a^{4}r^{3}}{3}}+K}
∫
x
5
r
3
d
x
=
r
9
9
−
2
a
2
r
7
7
+
a
4
r
5
5
+
K
{\displaystyle \int x^{5}r^{3}\;dx={\frac {r^{9}}{9}}-{\frac {2a^{2}r^{7}}{7}}+{\frac {a^{4}r^{5}}{5}}+K}
∫
x
5
r
2
n
+
1
d
x
=
r
2
n
+
7
2
n
+
7
−
2
a
2
r
2
n
+
5
2
n
+
5
+
a
4
r
2
n
+
3
2
n
+
3
+
K
{\displaystyle \int x^{5}r^{2n+1}\;dx={\frac {r^{2n+7}}{2n+7}}-{\frac {2a^{2}r^{2n+5}}{2n+5}}+{\frac {a^{4}r^{2n+3}}{2n+3}}+K}
∫
r
d
x
x
=
r
−
a
ln
|
a
+
r
x
|
=
r
−
a
sinh
−
1
a
x
+
K
{\displaystyle \int {\frac {r\;dx}{x}}=r-a\ln \left|{\frac {a+r}{x}}\right|=r-a\sinh ^{-1}{\frac {a}{x}}+K}
∫
r
3
d
x
x
=
r
3
3
+
a
2
r
−
a
3
ln
|
a
+
r
x
|
+
K
{\displaystyle \int {\frac {r^{3}\;dx}{x}}={\frac {r^{3}}{3}}+a^{2}r-a^{3}\ln \left|{\frac {a+r}{x}}\right|+K}
∫
r
5
d
x
x
=
r
5
5
+
a
2
r
3
3
+
a
4
r
−
a
5
ln
|
a
+
r
x
|
+
K
{\displaystyle \int {\frac {r^{5}\;dx}{x}}={\frac {r^{5}}{5}}+{\frac {a^{2}r^{3}}{3}}+a^{4}r-a^{5}\ln \left|{\frac {a+r}{x}}\right|+K}
∫
r
7
d
x
x
=
r
7
7
+
a
2
r
5
5
+
a
4
r
3
3
+
a
6
r
−
a
7
ln
|
a
+
r
x
|
+
K
{\displaystyle \int {\frac {r^{7}\;dx}{x}}={\frac {r^{7}}{7}}+{\frac {a^{2}r^{5}}{5}}+{\frac {a^{4}r^{3}}{3}}+a^{6}r-a^{7}\ln \left|{\frac {a+r}{x}}\right|+K}
∫
d
x
r
=
sinh
−
1
x
a
=
ln
|
x
+
r
|
+
K
{\displaystyle \int {\frac {dx}{r}}=\sinh ^{-1}{\frac {x}{a}}=\ln \left|x+r\right|+K}
∫
d
x
r
3
=
x
a
2
r
+
K
{\displaystyle \int {\frac {dx}{r^{3}}}={\frac {x}{a^{2}r}}+K}
∫
x
d
x
r
=
r
+
K
{\displaystyle \int {\frac {x\,dx}{r}}=r+K}
∫
x
d
x
r
3
=
−
1
r
+
K
{\displaystyle \int {\frac {x\,dx}{r^{3}}}=-{\frac {1}{r}}+K}
∫
x
2
d
x
r
=
x
2
r
−
a
2
2
sinh
−
1
x
a
=
x
2
r
−
a
2
2
ln
|
x
+
r
|
+
K
{\displaystyle \int {\frac {x^{2}\;dx}{r}}={\frac {x}{2}}r-{\frac {a^{2}}{2}}\,\sinh ^{-1}{\frac {x}{a}}={\frac {x}{2}}r-{\frac {a^{2}}{2}}\ln \left|x+r\right|+K}
∫
d
x
x
r
=
−
1
a
sinh
−
1
a
x
=
−
1
a
ln
|
a
+
r
x
|
+
K
{\displaystyle \int {\frac {dx}{xr}}=-{\frac {1}{a}}\,\sinh ^{-1}{\frac {a}{x}}=-{\frac {1}{a}}\ln \left|{\frac {a+r}{x}}\right|+K}
Bere barnean (x² -a²)-ren erroa daukaten funtzioen integralak
s
=
x
2
−
a
2
{\displaystyle s={\sqrt {x^{2}-a^{2}}}}
aldatu
Bere barnean (a²-x²)-ren erroa daukaten funtzioen integralak
u
=
a
2
−
x
2
{\displaystyle u={\sqrt {a^{2}-x^{2}}}}
aldatu
∫
u
d
x
=
1
2
(
x
u
+
a
2
arcsin
x
a
)
+
K
(
|
x
|
≤
|
a
|
)
{\displaystyle \int u\;dx={\frac {1}{2}}\left(xu+a^{2}\arcsin {\frac {x}{a}}\right)+K\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫
x
u
d
x
=
−
1
3
u
3
+
K
(
|
x
|
≤
|
a
|
)
{\displaystyle \int xu\;dx=-{\frac {1}{3}}u^{3}+K\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫
u
d
x
x
=
u
−
a
ln
|
a
+
u
x
|
+
K
(
|
x
|
≤
|
a
|
)
{\displaystyle \int {\frac {u\;dx}{x}}=u-a\ln \left|{\frac {a+u}{x}}\right|+K\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫
d
x
u
=
arcsin
x
a
+
K
(
|
x
|
≤
|
a
|
)
{\displaystyle \int {\frac {dx}{u}}=\arcsin {\frac {x}{a}}+K\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫
x
2
d
x
u
=
1
2
(
−
x
u
+
a
2
arcsin
x
a
)
+
K
(
|
x
|
≤
|
a
|
)
{\displaystyle \int {\frac {x^{2}\;dx}{u}}={\frac {1}{2}}\left(-xu+a^{2}\arcsin {\frac {x}{a}}\right)+K\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}
∫
u
d
x
=
1
2
(
x
u
−
sgn
x
cosh
−
1
|
x
a
|
)
+
K
(
|
x
|
≥
|
a
|
)
{\displaystyle \int u\;dx={\frac {1}{2}}\left(xu-\operatorname {sgn} x\,\cosh ^{-1}\left|{\frac {x}{a}}\right|\right)+K\qquad {\mbox{( }}|x|\geq |a|{\mbox{)}}}
Bere barnean (ax²+bx+c)-ren erroa daukaten funtzioen integralak
R
=
a
x
2
+
b
x
+
c
{\displaystyle R={\sqrt {ax^{2}+bx+c}}}
aldatu
∫
d
x
R
=
1
a
ln
|
2
a
R
+
2
a
x
+
b
|
+
K
(
a
>
0
)
{\displaystyle \int {\frac {dx}{R}}={\frac {1}{\sqrt {a}}}\ln \left|2{\sqrt {a}}R+2ax+b\right|+K\qquad {\mbox{( }}a>0{\mbox{)}}}
∫
d
x
R
=
1
a
sinh
−
1
2
a
x
+
b
4
a
c
−
b
2
+
K
(
a
>
0
,
4
a
c
−
b
2
>
0
)
{\displaystyle \int {\frac {dx}{R}}={\frac {1}{\sqrt {a}}}\,\sinh ^{-1}{\frac {2ax+b}{\sqrt {4ac-b^{2}}}}+K\qquad {\mbox{( }}a>0{\mbox{, }}4ac-b^{2}>0{\mbox{)}}}
∫
d
x
R
=
1
a
ln
|
2
a
x
+
b
|
+
K
(
a
>
0
,
4
a
c
−
b
2
=
0
)
{\displaystyle \int {\frac {dx}{R}}={\frac {1}{\sqrt {a}}}\ln |2ax+b|+K\quad {\mbox{( }}a>0{\mbox{, }}4ac-b^{2}=0{\mbox{)}}}
∫
d
x
R
=
−
1
−
a
arcsin
2
a
x
+
b
b
2
−
4
a
c
+
K
(
a
<
0
,
4
a
c
−
b
2
<
0
,
|
2
a
x
+
b
|
<
b
2
−
4
a
c
)
{\displaystyle \int {\frac {dx}{R}}=-{\frac {1}{\sqrt {-a}}}\arcsin {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+K\qquad {\mbox{( }}a<0{\mbox{, }}4ac-b^{2}<0{\mbox{, }}\left|2ax+b\right|<{\sqrt {b^{2}-4ac}}{\mbox{)}}}
∫
d
x
R
3
=
4
a
x
+
2
b
(
4
a
c
−
b
2
)
R
+
K
{\displaystyle \int {\frac {dx}{R^{3}}}={\frac {4ax+2b}{(4ac-b^{2})R}}+K}
∫
d
x
R
5
=
4
a
x
+
2
b
3
(
4
a
c
−
b
2
)
R
(
1
R
2
+
8
a
4
a
c
−
b
2
)
+
K
{\displaystyle \int {\frac {dx}{R^{5}}}={\frac {4ax+2b}{3(4ac-b^{2})R}}\left({\frac {1}{R^{2}}}+{\frac {8a}{4ac-b^{2}}}\right)+K}
∫
d
x
R
2
n
+
1
=
2
(
2
n
−
1
)
(
4
a
c
−
b
2
)
(
2
a
x
+
b
R
2
n
−
1
+
4
a
(
n
−
1
)
∫
d
x
R
2
n
−
1
)
{\displaystyle \int {\frac {dx}{R^{2n+1}}}={\frac {2}{(2n-1)(4ac-b^{2})}}\left({\frac {2ax+b}{R^{2n-1}}}+4a(n-1)\int {\frac {dx}{R^{2n-1}}}\right)}
∫
x
R
d
x
=
R
a
−
b
2
a
∫
d
x
R
{\displaystyle \int {\frac {x}{R}}\;dx={\frac {R}{a}}-{\frac {b}{2a}}\int {\frac {dx}{R}}}
∫
x
R
3
d
x
=
−
2
b
x
+
4
c
(
4
a
c
−
b
2
)
R
+
K
{\displaystyle \int {\frac {x}{R^{3}}}\;dx=-{\frac {2bx+4c}{(4ac-b^{2})R}}+K}
∫
x
R
2
n
+
1
d
x
=
−
1
(
2
n
−
1
)
a
R
2
n
−
1
−
b
2
a
∫
d
x
R
2
n
+
1
{\displaystyle \int {\frac {x}{R^{2n+1}}}\;dx=-{\frac {1}{(2n-1)aR^{2n-1}}}-{\frac {b}{2a}}\int {\frac {dx}{R^{2n+1}}}}
∫
d
x
x
R
=
−
1
c
ln
(
2
c
R
+
b
x
+
2
c
x
)
+
K
{\displaystyle \int {\frac {dx}{xR}}=-{\frac {1}{\sqrt {c}}}\ln \left({\frac {2{\sqrt {c}}R+bx+2c}{x}}\right)+K}
∫
d
x
x
R
=
−
1
c
sinh
−
1
(
b
x
+
2
c
|
x
|
4
a
c
−
b
2
)
+
K
{\displaystyle \int {\frac {dx}{xR}}=-{\frac {1}{\sqrt {c}}}\sinh ^{-1}\left({\frac {bx+2c}{|x|{\sqrt {4ac-b^{2}}}}}\right)+K}
Bere barnean (ax+b)-ren erroa daukaten funtzioen integralak
S
=
a
x
+
b
{\displaystyle S={\sqrt {ax+b}}}
aldatu
∫
d
x
a
x
+
b
=
2
a
x
+
b
a
+
K
{\displaystyle \int {\frac {dx}{\sqrt {ax+b}}}\,=\,{\frac {2{\sqrt {ax+b}}}{a}}+K}
∫
d
x
x
a
x
+
b
=
−
2
b
tanh
−
1
a
x
+
b
b
+
K
{\displaystyle \int {\frac {dx}{x{\sqrt {ax+b}}}}\,=\,{\frac {-2}{\sqrt {b}}}\tanh ^{-1}{\sqrt {\frac {ax+b}{b}}}+K}
∫
a
x
+
b
x
d
x
=
2
(
a
x
+
b
−
b
tanh
−
1
a
x
+
b
b
)
+
K
{\displaystyle \int {\frac {\sqrt {ax+b}}{x}}\,dx\;=\;2\left({\sqrt {ax+b}}-{\sqrt {b}}\tanh ^{-1}{\sqrt {\frac {ax+b}{b}}}\right)+K}
∫
x
n
a
x
+
b
d
x
=
2
a
(
2
n
+
1
)
(
x
n
a
x
+
b
−
b
n
∫
x
n
−
1
a
x
+
b
d
x
)
{\displaystyle \int {\frac {x^{n}}{\sqrt {ax+b}}}\,dx\;=\;{\frac {2}{a(2n+1)}}\left(x^{n}{\sqrt {ax+b}}-bn\int {\frac {x^{n-1}}{\sqrt {ax+b}}}\,dx\right)}
∫
x
n
a
x
+
b
d
x
=
2
2
n
+
1
(
x
n
+
1
a
x
+
b
+
b
x
n
a
x
+
b
−
n
b
∫
x
n
−
1
a
x
+
b
d
x
)
{\displaystyle \int x^{n}{\sqrt {ax+b}}\,dx\;=\;{\frac {2}{2n+1}}\left(x^{n+1}{\sqrt {ax+b}}+bx^{n}{\sqrt {ax+b}}-nb\int x^{n-1}{\sqrt {ax+b}}\,dx\right)}
Bibliografia
aldatu
Peirce, Benjamin Osgood. (1929). «3. atalaburua» A Short Table of Integrals. (3. argitaraldia) Boston: Ginn and Co, 16.–30 or. .
Milton Abramowitz & Irene A. Stegun, eds., Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables 1972, Dover: New York. (Ikusi: capítol 3 .)